Mixing
Pythagorean triples where the sum of the two cubes is also a square
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Are there any Primitive Pythagorean triple solutions $(a,b,c)$ where the sum of the two cubes is also a square? In other words are there coprime $a,b>0 in mathbb{N} ;, (a,b)=1$ where $a^2+b^2=c^2$ and $a^3+b^3=d^2$ for some $c,d in mathbb{N}$
number-theory
asked Jan 17 at 15:09
user1035795user1035795
562
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|
show 1 more comment
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Are there any Primitive Pythagorean triple solutions $(a,b,c)$ where the sum of the two cubes is also a square? In other words are there coprime $a,b>0 in mathbb{N} ;, (a,b)=1$ where $a^2+b^2=c^2$ and $a^3+b^3=d^2$ for some $c,d in mathbb{N}$
number-theory
asked Jan 17 at 15:09
user1035795user1035795
562
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Is there a particular reason that you're requiring primitivity? The answer might well be different without it, since your last equation is no longer homogeneous.
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– user3482749
Jan 17 at 15:20
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You can just multiply $a$, $b$ and $c$ by any $n^2$ and $d$ by $n^3$ (for any $ninmathbb N$) to form a new solution if you have one, so it makes some kind of sense to rule that out, but that's not the same as requiring primitivity, which I also don't see a reason for.
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– Henrik
Jan 17 at 15:30
2
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Because we can always find a trivial solution i.e. $3^3+4^3=91$, if use that as a common factor $a=3.91, b=4.91$ then $a^3+b^3 = 91^3(3^3+4^3)=91^4$ which is square
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– user1035795
Jan 17 at 15:34
1
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But doesn't that just mean that we can find solutions where the triple isn't primitive, not that all solutions where the triple isn't primitive is trivial (in that sense)? I.e. might solutions where the triple isn't primitive not still be interesting?
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– Henrik
Jan 17 at 16:27
1
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Yes, they might, I guess "Interest" is subjective. My interest is the existence of a coprime solution
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– user1035795
Jan 17 at 16:36
|
show 1 more comment
$begingroup$
Are there any Primitive Pythagorean triple solutions $(a,b,c)$ where the sum of the two cubes is also a square? In other words are there coprime $a,b>0 in mathbb{N} ;, (a,b)=1$ where $a^2+b^2=c^2$ and $a^3+b^3=d^2$ for some $c,d in mathbb{N}$
number-theory
asked Jan 17 at 15:09
user1035795user1035795
562
$endgroup$
Are there any Primitive Pythagorean triple solutions $(a,b,c)$ where the sum of the two cubes is also a square? In other words are there coprime $a,b>0 in mathbb{N} ;, (a,b)=1$ where $a^2+b^2=c^2$ and $a^3+b^3=d^2$ for some $c,d in mathbb{N}$
number-theory
number-theory
asked Jan 17 at 15:09
user1035795user1035795
562
asked Jan 17 at 15:09
user1035795user1035795
562
asked Jan 17 at 15:09
user1035795user1035795
562
asked Jan 17 at 15:09
user1035795user1035795
562
asked Jan 17 at 15:09
user1035795user1035795
562
562
$begingroup$
Is there a particular reason that you're requiring primitivity? The answer might well be different without it, since your last equation is no longer homogeneous.
$endgroup$
– user3482749
Jan 17 at 15:20
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You can just multiply $a$, $b$ and $c$ by any $n^2$ and $d$ by $n^3$ (for any $ninmathbb N$) to form a new solution if you have one, so it makes some kind of sense to rule that out, but that's not the same as requiring primitivity, which I also don't see a reason for.
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– Henrik
Jan 17 at 15:30
2
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Because we can always find a trivial solution i.e. $3^3+4^3=91$, if use that as a common factor $a=3.91, b=4.91$ then $a^3+b^3 = 91^3(3^3+4^3)=91^4$ which is square
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– user1035795
Jan 17 at 15:34
1
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But doesn't that just mean that we can find solutions where the triple isn't primitive, not that all solutions where the triple isn't primitive is trivial (in that sense)? I.e. might solutions where the triple isn't primitive not still be interesting?
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– Henrik
Jan 17 at 16:27
1
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Yes, they might, I guess "Interest" is subjective. My interest is the existence of a coprime solution
$endgroup$
– user1035795
Jan 17 at 16:36
|
show 1 more comment
$begingroup$
Is there a particular reason that you're requiring primitivity? The answer might well be different without it, since your last equation is no longer homogeneous.
$endgroup$
– user3482749
Jan 17 at 15:20
$begingroup$
You can just multiply $a$, $b$ and $c$ by any $n^2$ and $d$ by $n^3$ (for any $ninmathbb N$) to form a new solution if you have one, so it makes some kind of sense to rule that out, but that's not the same as requiring primitivity, which I also don't see a reason for.
$endgroup$
– Henrik
Jan 17 at 15:30
2
$begingroup$
Because we can always find a trivial solution i.e. $3^3+4^3=91$, if use that as a common factor $a=3.91, b=4.91$ then $a^3+b^3 = 91^3(3^3+4^3)=91^4$ which is square
$endgroup$
– user1035795
Jan 17 at 15:34
1
$begingroup$
But doesn't that just mean that we can find solutions where the triple isn't primitive, not that all solutions where the triple isn't primitive is trivial (in that sense)? I.e. might solutions where the triple isn't primitive not still be interesting?
$endgroup$
– Henrik
Jan 17 at 16:27
1
$begingroup$
Yes, they might, I guess "Interest" is subjective. My interest is the existence of a coprime solution
$endgroup$
– user1035795
Jan 17 at 16:36
$begingroup$
Is there a particular reason that you're requiring primitivity? The answer might well be different without it, since your last equation is no longer homogeneous.
$endgroup$
– user3482749
Jan 17 at 15:20
$begingroup$
Is there a particular reason that you're requiring primitivity? The answer might well be different without it, since your last equation is no longer homogeneous.
$endgroup$
– user3482749
Jan 17 at 15:20
$begingroup$
You can just multiply $a$, $b$ and $c$ by any $n^2$ and $d$ by $n^3$ (for any $ninmathbb N$) to form a new solution if you have one, so it makes some kind of sense to rule that out, but that's not the same as requiring primitivity, which I also don't see a reason for.
$endgroup$
– Henrik
Jan 17 at 15:30
$begingroup$
You can just multiply $a$, $b$ and $c$ by any $n^2$ and $d$ by $n^3$ (for any $ninmathbb N$) to form a new solution if you have one, so it makes some kind of sense to rule that out, but that's not the same as requiring primitivity, which I also don't see a reason for.
$endgroup$
– Henrik
Jan 17 at 15:30
2
2
$begingroup$
Because we can always find a trivial solution i.e. $3^3+4^3=91$, if use that as a common factor $a=3.91, b=4.91$ then $a^3+b^3 = 91^3(3^3+4^3)=91^4$ which is square
$endgroup$
– user1035795
Jan 17 at 15:34
$begingroup$
Because we can always find a trivial solution i.e. $3^3+4^3=91$, if use that as a common factor $a=3.91, b=4.91$ then $a^3+b^3 = 91^3(3^3+4^3)=91^4$ which is square
$endgroup$
– user1035795
Jan 17 at 15:34
1
1
$begingroup$
But doesn't that just mean that we can find solutions where the triple isn't primitive, not that all solutions where the triple isn't primitive is trivial (in that sense)? I.e. might solutions where the triple isn't primitive not still be interesting?
$endgroup$
– Henrik
Jan 17 at 16:27
$begingroup$
But doesn't that just mean that we can find solutions where the triple isn't primitive, not that all solutions where the triple isn't primitive is trivial (in that sense)? I.e. might solutions where the triple isn't primitive not still be interesting?
$endgroup$
– Henrik
Jan 17 at 16:27
1
1
$begingroup$
Yes, they might, I guess "Interest" is subjective. My interest is the existence of a coprime solution
$endgroup$
– user1035795
Jan 17 at 16:36
$begingroup$
Yes, they might, I guess "Interest" is subjective. My interest is the existence of a coprime solution
$endgroup$
– user1035795
Jan 17 at 16:36
|
show 1 more comment
4 Answers
4
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Disclaimer: These are some unfinished thoughts I will leave here to work on later, or for others to continue.
Given that $a$ and $b$ are coprime, it follows that $gcd(a+b,a^2-ab+b^2)$ divides $3$ because
$$gcd(a+b,a^2-ab+b^2)=gcd(a+b,3b^2)=gcd(a+b,3).$$
Suppose towards a contradiction that the gcd equals $3$: Then the factorization
$$d^2=a^3+b^3=(a+b)(a^2-ab+b^2),$$
shows that there exist $e,finBbb{Z}$ such that
$$a+b=3e^2qquadtext{ and }qquad a^2-ab+b^2=3f^2,$$
from which it quickly follows that
$$9e^4=(a+b)^2=a^2+2ab+b^2=3c^2-6f^2,$$
and reducing mod $8$ yields a contradiction, so the gcd is $1$. Hence there exist $e,finBbb{Z}$ such that
$$a+b=e^2qquadtext{ and }qquad a^2-ab+b^2=f^2,$$
and in the same way as before we find that
$$e^4=(a+b)^2=a^2+2ab+b^2=3c^2-2f^2.$$
Luckily $Bbb{Z}[sqrt{6}]$ is a UFD, and we have
$$N((3c-2f)+(c-f)sqrt{6}):=
left((3c-2f)+(c-f)sqrt{6}right)left((3c-2f)-(c-f)sqrt{6}right)
=3c^2-2f^2=e^4.$$
The gcd of two conjugate factors divides $2(3c-2f)$ and $2(c-f)$, and because $c$ and $f$ are coprime it follows that the gcd divides $2$. Because their product $e^4=(a+b)^2$ is odd, the two conjugate factors are in fact coprime. This means there exists some $x+ysqrt{6}inBbb{Z}[sqrt{6}]$ such that
$$(3c-2f)+(c-f)sqrt{6}=(x+ysqrt{6})^4.tag{1}$$
This immediately tells us that
$$a+b=e^2=sqrt{N((3c-2f)+(c-f)sqrt{6})}=(x+ysqrt{6})^2(x-ysqrt{6})^2=(x^2-6y^2)^2.tag{2}$$
Furthermore, expanding equation $(1)$ yields the two equations
$$3c-2f=x^4+36x^2y^2+36y^4qquadtext{ and }qquad c-f=4x^3y+24xy^3.$$
Because $c-f>0$, without loss of generality $x,y>0$. The above tells us that
begin{eqnarray*}
c&=&x^4- 8x^3y+36x^2y^2-48xy^3+36y^4,\
f&=&x^4- 12x^3y+36x^2y^2-72xy^3+36y^4,
end{eqnarray*}
and hence that
$$ab=c^2-f^2=(c-f)(c+f)=8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).tag{3}$$
This means $a$ and $b$ are the roots of the quadratic polynomial
$$Z^2-(x^2-6y^2)^2Z+8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).$$
This polynomial has integer roots if and only if its discriminant $Delta$ is a square, where
$$Delta=(x^2-6y^2)^4-32xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4),$$
which leaves me with the question of when this homogeneous octic polynomial takes on square values.
answered Jan 18 at 15:01
ServaesServaes
25.8k33996
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A partial, in progress answer: $a,b$ satisfies
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
for integers $m,n,r,s$ such that
$$
gcd(m,n)=gcd(r,s)=1
$$
and $r,s$ are odd, $snotequiv 0pmod 3$ and exactly one of $m,n$ is odd. It can be shown that $gcd(a,b)=1$ using either $m,n$ or $r,s$. Perhaps this system is already not solvable.
The only other restriction remaining is
$$
a+b=u^2
$$
for some integer $u$.
A sanity check:
$$
a^2+b^2 = (m^2+n^2)^2
$$
and
$$
a^2-ab+b^2 = left(frac{3r^2+s^2}{4}right)^2 = v^2 in mathbb Z
$$
Hence if $a+b=u^2$ then
$$
a^3+b^3 = (a+b)(a^2-ab+b^2) = (uv)^2
$$
Let $a,b,c$ be a primitive Pythagorean triplet. Then we know that for some integers $m>n$ and $gcd(m,n)=1$,
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
In particular, $m,n$ also have different parity.
Proposition 1. Let $a,b,c$ be a primitive Pythagorean triplet such that
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
and $m>n,;;gcd(m,n)=1$. If
$$
a^3+b^3=d^2,
$$
for some integer $d$ then
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
for some integers $u,v$.
Proof. We start with
$$
(a+b)(a^2-ab+b^2) = a^3+b^3 = d^2
$$
Since
$$
3a^2 = (2a-b)(a+b) + (a^2-ab+b^2)
$$
This shows that
$
gcd(a+b,a^2-ab+b^2)
$
divides $3a^2$. Checking $pmod 3$, the equation
$$
a+b = m^2+2mn-n^2 equiv 0 pmod 3
$$
is possible only if $m,nequiv 0pmod 3$. This contradicts $gcd(m,n)=1$, therefore $3nmid a+b$. Hence $gcd(a+b,a^2-ab+b^2)$ divides $a^2$. Now if a prime $p$ divides $gcd(a+b,a^2-ab+b^2)$, then $p$ divides $a^2$ and hence $pmid a$. But that means $p$ divides $b$, contradicting $gcd(a,b)=1$. Hence we conclude that
$$
gcd(a+b,a^2-ab+b^2)=1
$$
As a result, we can write
$$
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
$$
for some integers $u,v$. (It cannot have been $-u^2,-v^2$ instead since $a,b>0$.)
$$tag*{$square$}$$
For the equation
$$
a^2-ab+b^2=v^2,
$$
since $gcd(a,b)=1$ we must have $gcd(a,v)=gcd(b,v)=1$.
Proposition 2. The primitive integer solutions to
$$
a^2-ab+b^2 = v^2
$$
are
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
where $r,s$ are odd integers satisfying $gcd(r,s)=1$.
Proof. We first convert the equation to
$$
a^2-ab+b^2=v^2 Longleftrightarrow (2b-a)^2 + 3a^2 = (2v)^2
$$
Following The Solution of the Diophantine Equation$X^2+3Y^2=Z^2$,
Theorem 2.2 Let $E:x^2+3y^2=z^2$ be the diophantine equation and $(x,y,z)inmathbb Z^3$ with $gcd(x,y)=1$, $y$ is odd and $gcd(xz,3)=1$. Then
$$
begin{align}
|x| &= frac{3r^2-s^2}{2}\
|y| &= rs\
|z| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd integers $r,s$ and $gcd(r,s)=1$. (Not written but implied that $snotequiv 0pmod 3$.
Since $a$ is already odd, to use this result we need to check that
$$
gcd(2b-a,a)=1,quad gcd((2b-a)(2v),3)=1
$$
The first part is immediate since $gcd(a,b)=1$ and $a$ is odd. For the second part, if $3mid v$ or $3mid 2b-a$ then from
$$
(2b-a)^2+3a^2=v^2
$$
we get $3mid a$. For both cases $3$ divides $2b-a,a$ so $3$ divides $b$. This contradicts $gcd(a,b)=1$. Therefore indeed $gcd((2b-a)(2b),3)=1$, so we obtain the solutions
$$
begin{align}
|2b-a| &= frac{3r^2-s^2}{2}\
|a| &= rs\
|2v| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd $r,s$ satisfying $gcd(r,s)=1$. Rearranging:
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
which is what we want.
$$tag*{$square$}$$
This gives us a new restriction:
$$
u^2 = a+b = frac{3r^2+6rs-s^2}{4} text{ or } frac{s^2+6rs-3r^2}{4}
$$
We do a substitution for the odd $r=2f+1,s=2g+1$, giving
$$
u^2 = 2 + 6 f + 3 f^2 + 2 g + 6 f g - g^2
$$
or
$$
u^2 = 1 - 3 f^2 + 4 g + 6 f g + g^2
$$
For the first equation, LHS $equiv 0,1pmod 4$ but RHS $equiv 2,3pmod 4$, which is impossible. Therefore it must have been
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
answered Jan 18 at 15:17
Yong Hao NgYong Hao Ng
3,5691222
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add a comment |
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If we suppose that such integers exist and write, say, $a=r^2-s^2$ and $b=2rs$ for coprime (positive) integers $r$ and $s$, then
$$
d^2=(r^2+2rs-s^2)(r^4-2r^3s++2r^2s^2+2rs^3+s^4)
$$
and hence a solution would correspond to a (nontrivial) rational point on the (genus $2$) curve
$$
y^2 = x^6 -3x^4+8x^3+3x^2-1.
$$
The Jacobian of this curve has rank $1$ and a Chabauty argument in Magma using the prime $17$ shows that there are no such points. There may be an easier way to see this, but I'm afraid it's not obvious to me.
answered Jan 20 at 5:39
Mike BennettMike Bennett
2,39478
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What area of maths do I need to investigate in order to understand how you derived the equation in $x,y$ and the subsequent claims.
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– user1035795
Jan 20 at 12:51
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Multiply out the first equation (in $r, s$ and $d$) and then divide by $s^6$. The machinery needed to understand Chabauty's method is quite advanced. A book by Cassels and Flynn ("Prolegomena....") is a good place to start, but assumes a solid undergraduate background.
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– Mike Bennett
Jan 20 at 17:11
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@user1035795 From $a^2+b^2=c^2$ you get the primitive pythagorean parameterization $a=r^2-s^2,b=2rs$, then a substitution into $d^2=a^3+b^3$ gives you equation 1. Next since $sneq 0$, let $y=d/s^3$ and $x=r/s$ to get the curve over $mathbb Q$ as in equation 2.
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– Yong Hao Ng
Jan 21 at 4:52
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Thank you Mike, yes plain as day now you (and Yong Hao Ng) pointed out. I'll check out the Chabauty method, thanks.
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– user1035795
Jan 21 at 10:27
add a comment |
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equations $$X^3+Y^3=Z^2$$
Can be expressed by integers $p,s,a,b,c$ . where the number of $c$ characterizes the degree of primitiveness.
$$X=(3a^2+4ab+b^2)(3a^2+b^2)c^2$$
$$Y=2b(a+b)(3a^2+b^2)c^2$$
$$Z=3(a+b)^2(3a^2+b^2)^2c^3$$
And more.
$$X=2b(b-a)(3a^2+b^2)c^2$$
$$Y=2b(b+a)(3a^2+b^2)c^2$$
$$Z=4b^2(3a^2+b^2)^2c^3$$
If we decide to factor $$X^3+Y^3=qZ^2$$
For a compact notation we replace :
$$a=s(2p-s)$$
$$b=p^2-s^2$$
$$t=p^2-ps+s^2$$
then:
$$X=qb(a+b)c^2$$
$$Y=qa(a+b)c^2$$
$$Z=qt(a+b)^2c^3$$
And the most beautiful solution. If we use the solutions of Pell's equation: $p^2-3a^2s^2=1$
by the way $a$ May appear as a factor in the decision and.
Then the solutions are of the form::
$$X=qa(2p-3as)sc^2$$
$$Y=q(p-2as)pc^2$$
$$Z=q(p^2-3aps+3a^2s^2)c^3$$
If we change the sign : $$Y^3-X^3=qZ^2$$
Then the solutions are of the form:
$$X=qa(2p+3as)sc^2$$
$$Y=q(p+2as)pc^2$$
$$Z=q(p^2+3aps+3a^2s^2)c^3$$
Another solution of the equation: $$X^3+Y^3=qZ^2$$
$p,s$ - integers asked us.
To facilitate the calculations we make the change. $a,b,c$
If the ratio is as follows : $q=3t^2+1$
$$b=2q(q+2mp{6t})p^2+6p(tmp1)ps+(q-1mp{3t})s^2$$
$$c=6q(q-2(1pm{t}))p^2+6q(tmp1)ps+3(1mp{t})s^2$$
$$a=12q(1mp{t})p^2+6(4tmp{q})ps+3(1mp{t})s^2$$
If the ratio is as follows: $q=t^2+3$
$$b=3(q-1)(1pm{t})s^2+2(3pm{(q-1)t})ps+(1pm{t})p^2$$
$$c=3(6-(q-1)(q-3mp{t}))s^2+6(1pm{t})ps+(q-3pm{t})p^2$$
$$a=3(6-(q-1)(1mp{t}))s^2+6(1pm{t})ps+(1pm{t})p^2$$
Then the solutions are of the form:
$$X=2c(c-b)$$
$$Y=(c-3b)(c-b)$$
$$Z=3a(c-b)^2$$
Then the solutions are of the form:
$$X=2(c-b)c$$
$$Y=2(c+b)c$$
$$Z=4ac^2$$
If the ratio is as follows : $q=t^2+3$
$$c=6(q-4)(2pm{t})p^2+4(6pm{(q-4)t})ps+2(2pm{t})s^2$$
$$b=3(24-(q-4)(q-3mp{2t}))p^2+12(2pm{t})ps+(q-3pm{2t})s^2$$
$$a=3(24-(q-4)(4mp{2t}))p^2+12(2pm{t})ps+2(2pm{t})s^2$$
If the ratio is as follows $q=3t^2+4$
$$c=q(-q+7(4mp{3t}))p^2+6q(tmp{1})ps+(q-4mp{3t})s^2$$
$$b=3q(2q-7(1pm{t}))p^2+6q(tmp{1})ps+3(1mp{t})s^2$$
$$a=21q(1mp{t})p^2+6(7tmp{q})ps+3(1mp{t})s^2$$
Then the solutions are of the form :
$$X=2(3c-2b)c$$
$$Y=2(3c+2b)c$$
$$Z=12ac^2$$
Then the solutions are of the form :
$$X=(2b-c)b$$
$$Y=(2b+c)b$$
$$Z=2ab^2$$
answered Jan 18 at 16:01
individindivid
3,2621916
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So with all of that can you give an example solution?
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– user1035795
Jan 20 at 12:53
add a comment |
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$begingroup$
Disclaimer: These are some unfinished thoughts I will leave here to work on later, or for others to continue.
Given that $a$ and $b$ are coprime, it follows that $gcd(a+b,a^2-ab+b^2)$ divides $3$ because
$$gcd(a+b,a^2-ab+b^2)=gcd(a+b,3b^2)=gcd(a+b,3).$$
Suppose towards a contradiction that the gcd equals $3$: Then the factorization
$$d^2=a^3+b^3=(a+b)(a^2-ab+b^2),$$
shows that there exist $e,finBbb{Z}$ such that
$$a+b=3e^2qquadtext{ and }qquad a^2-ab+b^2=3f^2,$$
from which it quickly follows that
$$9e^4=(a+b)^2=a^2+2ab+b^2=3c^2-6f^2,$$
and reducing mod $8$ yields a contradiction, so the gcd is $1$. Hence there exist $e,finBbb{Z}$ such that
$$a+b=e^2qquadtext{ and }qquad a^2-ab+b^2=f^2,$$
and in the same way as before we find that
$$e^4=(a+b)^2=a^2+2ab+b^2=3c^2-2f^2.$$
Luckily $Bbb{Z}[sqrt{6}]$ is a UFD, and we have
$$N((3c-2f)+(c-f)sqrt{6}):=
left((3c-2f)+(c-f)sqrt{6}right)left((3c-2f)-(c-f)sqrt{6}right)
=3c^2-2f^2=e^4.$$
The gcd of two conjugate factors divides $2(3c-2f)$ and $2(c-f)$, and because $c$ and $f$ are coprime it follows that the gcd divides $2$. Because their product $e^4=(a+b)^2$ is odd, the two conjugate factors are in fact coprime. This means there exists some $x+ysqrt{6}inBbb{Z}[sqrt{6}]$ such that
$$(3c-2f)+(c-f)sqrt{6}=(x+ysqrt{6})^4.tag{1}$$
This immediately tells us that
$$a+b=e^2=sqrt{N((3c-2f)+(c-f)sqrt{6})}=(x+ysqrt{6})^2(x-ysqrt{6})^2=(x^2-6y^2)^2.tag{2}$$
Furthermore, expanding equation $(1)$ yields the two equations
$$3c-2f=x^4+36x^2y^2+36y^4qquadtext{ and }qquad c-f=4x^3y+24xy^3.$$
Because $c-f>0$, without loss of generality $x,y>0$. The above tells us that
begin{eqnarray*}
c&=&x^4- 8x^3y+36x^2y^2-48xy^3+36y^4,\
f&=&x^4- 12x^3y+36x^2y^2-72xy^3+36y^4,
end{eqnarray*}
and hence that
$$ab=c^2-f^2=(c-f)(c+f)=8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).tag{3}$$
This means $a$ and $b$ are the roots of the quadratic polynomial
$$Z^2-(x^2-6y^2)^2Z+8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).$$
This polynomial has integer roots if and only if its discriminant $Delta$ is a square, where
$$Delta=(x^2-6y^2)^4-32xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4),$$
which leaves me with the question of when this homogeneous octic polynomial takes on square values.
answered Jan 18 at 15:01
ServaesServaes
25.8k33996
$endgroup$
add a comment |
$begingroup$
Disclaimer: These are some unfinished thoughts I will leave here to work on later, or for others to continue.
Given that $a$ and $b$ are coprime, it follows that $gcd(a+b,a^2-ab+b^2)$ divides $3$ because
$$gcd(a+b,a^2-ab+b^2)=gcd(a+b,3b^2)=gcd(a+b,3).$$
Suppose towards a contradiction that the gcd equals $3$: Then the factorization
$$d^2=a^3+b^3=(a+b)(a^2-ab+b^2),$$
shows that there exist $e,finBbb{Z}$ such that
$$a+b=3e^2qquadtext{ and }qquad a^2-ab+b^2=3f^2,$$
from which it quickly follows that
$$9e^4=(a+b)^2=a^2+2ab+b^2=3c^2-6f^2,$$
and reducing mod $8$ yields a contradiction, so the gcd is $1$. Hence there exist $e,finBbb{Z}$ such that
$$a+b=e^2qquadtext{ and }qquad a^2-ab+b^2=f^2,$$
and in the same way as before we find that
$$e^4=(a+b)^2=a^2+2ab+b^2=3c^2-2f^2.$$
Luckily $Bbb{Z}[sqrt{6}]$ is a UFD, and we have
$$N((3c-2f)+(c-f)sqrt{6}):=
left((3c-2f)+(c-f)sqrt{6}right)left((3c-2f)-(c-f)sqrt{6}right)
=3c^2-2f^2=e^4.$$
The gcd of two conjugate factors divides $2(3c-2f)$ and $2(c-f)$, and because $c$ and $f$ are coprime it follows that the gcd divides $2$. Because their product $e^4=(a+b)^2$ is odd, the two conjugate factors are in fact coprime. This means there exists some $x+ysqrt{6}inBbb{Z}[sqrt{6}]$ such that
$$(3c-2f)+(c-f)sqrt{6}=(x+ysqrt{6})^4.tag{1}$$
This immediately tells us that
$$a+b=e^2=sqrt{N((3c-2f)+(c-f)sqrt{6})}=(x+ysqrt{6})^2(x-ysqrt{6})^2=(x^2-6y^2)^2.tag{2}$$
Furthermore, expanding equation $(1)$ yields the two equations
$$3c-2f=x^4+36x^2y^2+36y^4qquadtext{ and }qquad c-f=4x^3y+24xy^3.$$
Because $c-f>0$, without loss of generality $x,y>0$. The above tells us that
begin{eqnarray*}
c&=&x^4- 8x^3y+36x^2y^2-48xy^3+36y^4,\
f&=&x^4- 12x^3y+36x^2y^2-72xy^3+36y^4,
end{eqnarray*}
and hence that
$$ab=c^2-f^2=(c-f)(c+f)=8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).tag{3}$$
This means $a$ and $b$ are the roots of the quadratic polynomial
$$Z^2-(x^2-6y^2)^2Z+8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).$$
This polynomial has integer roots if and only if its discriminant $Delta$ is a square, where
$$Delta=(x^2-6y^2)^4-32xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4),$$
which leaves me with the question of when this homogeneous octic polynomial takes on square values.
answered Jan 18 at 15:01
ServaesServaes
25.8k33996
$endgroup$
add a comment |
$begingroup$
Disclaimer: These are some unfinished thoughts I will leave here to work on later, or for others to continue.
Given that $a$ and $b$ are coprime, it follows that $gcd(a+b,a^2-ab+b^2)$ divides $3$ because
$$gcd(a+b,a^2-ab+b^2)=gcd(a+b,3b^2)=gcd(a+b,3).$$
Suppose towards a contradiction that the gcd equals $3$: Then the factorization
$$d^2=a^3+b^3=(a+b)(a^2-ab+b^2),$$
shows that there exist $e,finBbb{Z}$ such that
$$a+b=3e^2qquadtext{ and }qquad a^2-ab+b^2=3f^2,$$
from which it quickly follows that
$$9e^4=(a+b)^2=a^2+2ab+b^2=3c^2-6f^2,$$
and reducing mod $8$ yields a contradiction, so the gcd is $1$. Hence there exist $e,finBbb{Z}$ such that
$$a+b=e^2qquadtext{ and }qquad a^2-ab+b^2=f^2,$$
and in the same way as before we find that
$$e^4=(a+b)^2=a^2+2ab+b^2=3c^2-2f^2.$$
Luckily $Bbb{Z}[sqrt{6}]$ is a UFD, and we have
$$N((3c-2f)+(c-f)sqrt{6}):=
left((3c-2f)+(c-f)sqrt{6}right)left((3c-2f)-(c-f)sqrt{6}right)
=3c^2-2f^2=e^4.$$
The gcd of two conjugate factors divides $2(3c-2f)$ and $2(c-f)$, and because $c$ and $f$ are coprime it follows that the gcd divides $2$. Because their product $e^4=(a+b)^2$ is odd, the two conjugate factors are in fact coprime. This means there exists some $x+ysqrt{6}inBbb{Z}[sqrt{6}]$ such that
$$(3c-2f)+(c-f)sqrt{6}=(x+ysqrt{6})^4.tag{1}$$
This immediately tells us that
$$a+b=e^2=sqrt{N((3c-2f)+(c-f)sqrt{6})}=(x+ysqrt{6})^2(x-ysqrt{6})^2=(x^2-6y^2)^2.tag{2}$$
Furthermore, expanding equation $(1)$ yields the two equations
$$3c-2f=x^4+36x^2y^2+36y^4qquadtext{ and }qquad c-f=4x^3y+24xy^3.$$
Because $c-f>0$, without loss of generality $x,y>0$. The above tells us that
begin{eqnarray*}
c&=&x^4- 8x^3y+36x^2y^2-48xy^3+36y^4,\
f&=&x^4- 12x^3y+36x^2y^2-72xy^3+36y^4,
end{eqnarray*}
and hence that
$$ab=c^2-f^2=(c-f)(c+f)=8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).tag{3}$$
This means $a$ and $b$ are the roots of the quadratic polynomial
$$Z^2-(x^2-6y^2)^2Z+8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).$$
This polynomial has integer roots if and only if its discriminant $Delta$ is a square, where
$$Delta=(x^2-6y^2)^4-32xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4),$$
which leaves me with the question of when this homogeneous octic polynomial takes on square values.
answered Jan 18 at 15:01
ServaesServaes
25.8k33996
$endgroup$
Disclaimer: These are some unfinished thoughts I will leave here to work on later, or for others to continue.
Given that $a$ and $b$ are coprime, it follows that $gcd(a+b,a^2-ab+b^2)$ divides $3$ because
$$gcd(a+b,a^2-ab+b^2)=gcd(a+b,3b^2)=gcd(a+b,3).$$
Suppose towards a contradiction that the gcd equals $3$: Then the factorization
$$d^2=a^3+b^3=(a+b)(a^2-ab+b^2),$$
shows that there exist $e,finBbb{Z}$ such that
$$a+b=3e^2qquadtext{ and }qquad a^2-ab+b^2=3f^2,$$
from which it quickly follows that
$$9e^4=(a+b)^2=a^2+2ab+b^2=3c^2-6f^2,$$
and reducing mod $8$ yields a contradiction, so the gcd is $1$. Hence there exist $e,finBbb{Z}$ such that
$$a+b=e^2qquadtext{ and }qquad a^2-ab+b^2=f^2,$$
and in the same way as before we find that
$$e^4=(a+b)^2=a^2+2ab+b^2=3c^2-2f^2.$$
Luckily $Bbb{Z}[sqrt{6}]$ is a UFD, and we have
$$N((3c-2f)+(c-f)sqrt{6}):=
left((3c-2f)+(c-f)sqrt{6}right)left((3c-2f)-(c-f)sqrt{6}right)
=3c^2-2f^2=e^4.$$
The gcd of two conjugate factors divides $2(3c-2f)$ and $2(c-f)$, and because $c$ and $f$ are coprime it follows that the gcd divides $2$. Because their product $e^4=(a+b)^2$ is odd, the two conjugate factors are in fact coprime. This means there exists some $x+ysqrt{6}inBbb{Z}[sqrt{6}]$ such that
$$(3c-2f)+(c-f)sqrt{6}=(x+ysqrt{6})^4.tag{1}$$
This immediately tells us that
$$a+b=e^2=sqrt{N((3c-2f)+(c-f)sqrt{6})}=(x+ysqrt{6})^2(x-ysqrt{6})^2=(x^2-6y^2)^2.tag{2}$$
Furthermore, expanding equation $(1)$ yields the two equations
$$3c-2f=x^4+36x^2y^2+36y^4qquadtext{ and }qquad c-f=4x^3y+24xy^3.$$
Because $c-f>0$, without loss of generality $x,y>0$. The above tells us that
begin{eqnarray*}
c&=&x^4- 8x^3y+36x^2y^2-48xy^3+36y^4,\
f&=&x^4- 12x^3y+36x^2y^2-72xy^3+36y^4,
end{eqnarray*}
and hence that
$$ab=c^2-f^2=(c-f)(c+f)=8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).tag{3}$$
This means $a$ and $b$ are the roots of the quadratic polynomial
$$Z^2-(x^2-6y^2)^2Z+8xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4).$$
This polynomial has integer roots if and only if its discriminant $Delta$ is a square, where
$$Delta=(x^2-6y^2)^4-32xy(x^2+6y^2)(x^4-10x^3y+36x^2y^2-60xy^3+36y^4),$$
which leaves me with the question of when this homogeneous octic polynomial takes on square values.
answered Jan 18 at 15:01
ServaesServaes
25.8k33996
answered Jan 18 at 15:01
ServaesServaes
25.8k33996
answered Jan 18 at 15:01
ServaesServaes
25.8k33996
answered Jan 18 at 15:01
ServaesServaes
25.8k33996
25.8k33996
add a comment |
add a comment |
$begingroup$
A partial, in progress answer: $a,b$ satisfies
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
for integers $m,n,r,s$ such that
$$
gcd(m,n)=gcd(r,s)=1
$$
and $r,s$ are odd, $snotequiv 0pmod 3$ and exactly one of $m,n$ is odd. It can be shown that $gcd(a,b)=1$ using either $m,n$ or $r,s$. Perhaps this system is already not solvable.
The only other restriction remaining is
$$
a+b=u^2
$$
for some integer $u$.
A sanity check:
$$
a^2+b^2 = (m^2+n^2)^2
$$
and
$$
a^2-ab+b^2 = left(frac{3r^2+s^2}{4}right)^2 = v^2 in mathbb Z
$$
Hence if $a+b=u^2$ then
$$
a^3+b^3 = (a+b)(a^2-ab+b^2) = (uv)^2
$$
Let $a,b,c$ be a primitive Pythagorean triplet. Then we know that for some integers $m>n$ and $gcd(m,n)=1$,
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
In particular, $m,n$ also have different parity.
Proposition 1. Let $a,b,c$ be a primitive Pythagorean triplet such that
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
and $m>n,;;gcd(m,n)=1$. If
$$
a^3+b^3=d^2,
$$
for some integer $d$ then
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
for some integers $u,v$.
Proof. We start with
$$
(a+b)(a^2-ab+b^2) = a^3+b^3 = d^2
$$
Since
$$
3a^2 = (2a-b)(a+b) + (a^2-ab+b^2)
$$
This shows that
$
gcd(a+b,a^2-ab+b^2)
$
divides $3a^2$. Checking $pmod 3$, the equation
$$
a+b = m^2+2mn-n^2 equiv 0 pmod 3
$$
is possible only if $m,nequiv 0pmod 3$. This contradicts $gcd(m,n)=1$, therefore $3nmid a+b$. Hence $gcd(a+b,a^2-ab+b^2)$ divides $a^2$. Now if a prime $p$ divides $gcd(a+b,a^2-ab+b^2)$, then $p$ divides $a^2$ and hence $pmid a$. But that means $p$ divides $b$, contradicting $gcd(a,b)=1$. Hence we conclude that
$$
gcd(a+b,a^2-ab+b^2)=1
$$
As a result, we can write
$$
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
$$
for some integers $u,v$. (It cannot have been $-u^2,-v^2$ instead since $a,b>0$.)
$$tag*{$square$}$$
For the equation
$$
a^2-ab+b^2=v^2,
$$
since $gcd(a,b)=1$ we must have $gcd(a,v)=gcd(b,v)=1$.
Proposition 2. The primitive integer solutions to
$$
a^2-ab+b^2 = v^2
$$
are
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
where $r,s$ are odd integers satisfying $gcd(r,s)=1$.
Proof. We first convert the equation to
$$
a^2-ab+b^2=v^2 Longleftrightarrow (2b-a)^2 + 3a^2 = (2v)^2
$$
Following The Solution of the Diophantine Equation$X^2+3Y^2=Z^2$,
Theorem 2.2 Let $E:x^2+3y^2=z^2$ be the diophantine equation and $(x,y,z)inmathbb Z^3$ with $gcd(x,y)=1$, $y$ is odd and $gcd(xz,3)=1$. Then
$$
begin{align}
|x| &= frac{3r^2-s^2}{2}\
|y| &= rs\
|z| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd integers $r,s$ and $gcd(r,s)=1$. (Not written but implied that $snotequiv 0pmod 3$.
Since $a$ is already odd, to use this result we need to check that
$$
gcd(2b-a,a)=1,quad gcd((2b-a)(2v),3)=1
$$
The first part is immediate since $gcd(a,b)=1$ and $a$ is odd. For the second part, if $3mid v$ or $3mid 2b-a$ then from
$$
(2b-a)^2+3a^2=v^2
$$
we get $3mid a$. For both cases $3$ divides $2b-a,a$ so $3$ divides $b$. This contradicts $gcd(a,b)=1$. Therefore indeed $gcd((2b-a)(2b),3)=1$, so we obtain the solutions
$$
begin{align}
|2b-a| &= frac{3r^2-s^2}{2}\
|a| &= rs\
|2v| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd $r,s$ satisfying $gcd(r,s)=1$. Rearranging:
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
which is what we want.
$$tag*{$square$}$$
This gives us a new restriction:
$$
u^2 = a+b = frac{3r^2+6rs-s^2}{4} text{ or } frac{s^2+6rs-3r^2}{4}
$$
We do a substitution for the odd $r=2f+1,s=2g+1$, giving
$$
u^2 = 2 + 6 f + 3 f^2 + 2 g + 6 f g - g^2
$$
or
$$
u^2 = 1 - 3 f^2 + 4 g + 6 f g + g^2
$$
For the first equation, LHS $equiv 0,1pmod 4$ but RHS $equiv 2,3pmod 4$, which is impossible. Therefore it must have been
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
answered Jan 18 at 15:17
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
add a comment |
$begingroup$
A partial, in progress answer: $a,b$ satisfies
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
for integers $m,n,r,s$ such that
$$
gcd(m,n)=gcd(r,s)=1
$$
and $r,s$ are odd, $snotequiv 0pmod 3$ and exactly one of $m,n$ is odd. It can be shown that $gcd(a,b)=1$ using either $m,n$ or $r,s$. Perhaps this system is already not solvable.
The only other restriction remaining is
$$
a+b=u^2
$$
for some integer $u$.
A sanity check:
$$
a^2+b^2 = (m^2+n^2)^2
$$
and
$$
a^2-ab+b^2 = left(frac{3r^2+s^2}{4}right)^2 = v^2 in mathbb Z
$$
Hence if $a+b=u^2$ then
$$
a^3+b^3 = (a+b)(a^2-ab+b^2) = (uv)^2
$$
Let $a,b,c$ be a primitive Pythagorean triplet. Then we know that for some integers $m>n$ and $gcd(m,n)=1$,
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
In particular, $m,n$ also have different parity.
Proposition 1. Let $a,b,c$ be a primitive Pythagorean triplet such that
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
and $m>n,;;gcd(m,n)=1$. If
$$
a^3+b^3=d^2,
$$
for some integer $d$ then
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
for some integers $u,v$.
Proof. We start with
$$
(a+b)(a^2-ab+b^2) = a^3+b^3 = d^2
$$
Since
$$
3a^2 = (2a-b)(a+b) + (a^2-ab+b^2)
$$
This shows that
$
gcd(a+b,a^2-ab+b^2)
$
divides $3a^2$. Checking $pmod 3$, the equation
$$
a+b = m^2+2mn-n^2 equiv 0 pmod 3
$$
is possible only if $m,nequiv 0pmod 3$. This contradicts $gcd(m,n)=1$, therefore $3nmid a+b$. Hence $gcd(a+b,a^2-ab+b^2)$ divides $a^2$. Now if a prime $p$ divides $gcd(a+b,a^2-ab+b^2)$, then $p$ divides $a^2$ and hence $pmid a$. But that means $p$ divides $b$, contradicting $gcd(a,b)=1$. Hence we conclude that
$$
gcd(a+b,a^2-ab+b^2)=1
$$
As a result, we can write
$$
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
$$
for some integers $u,v$. (It cannot have been $-u^2,-v^2$ instead since $a,b>0$.)
$$tag*{$square$}$$
For the equation
$$
a^2-ab+b^2=v^2,
$$
since $gcd(a,b)=1$ we must have $gcd(a,v)=gcd(b,v)=1$.
Proposition 2. The primitive integer solutions to
$$
a^2-ab+b^2 = v^2
$$
are
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
where $r,s$ are odd integers satisfying $gcd(r,s)=1$.
Proof. We first convert the equation to
$$
a^2-ab+b^2=v^2 Longleftrightarrow (2b-a)^2 + 3a^2 = (2v)^2
$$
Following The Solution of the Diophantine Equation$X^2+3Y^2=Z^2$,
Theorem 2.2 Let $E:x^2+3y^2=z^2$ be the diophantine equation and $(x,y,z)inmathbb Z^3$ with $gcd(x,y)=1$, $y$ is odd and $gcd(xz,3)=1$. Then
$$
begin{align}
|x| &= frac{3r^2-s^2}{2}\
|y| &= rs\
|z| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd integers $r,s$ and $gcd(r,s)=1$. (Not written but implied that $snotequiv 0pmod 3$.
Since $a$ is already odd, to use this result we need to check that
$$
gcd(2b-a,a)=1,quad gcd((2b-a)(2v),3)=1
$$
The first part is immediate since $gcd(a,b)=1$ and $a$ is odd. For the second part, if $3mid v$ or $3mid 2b-a$ then from
$$
(2b-a)^2+3a^2=v^2
$$
we get $3mid a$. For both cases $3$ divides $2b-a,a$ so $3$ divides $b$. This contradicts $gcd(a,b)=1$. Therefore indeed $gcd((2b-a)(2b),3)=1$, so we obtain the solutions
$$
begin{align}
|2b-a| &= frac{3r^2-s^2}{2}\
|a| &= rs\
|2v| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd $r,s$ satisfying $gcd(r,s)=1$. Rearranging:
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
which is what we want.
$$tag*{$square$}$$
This gives us a new restriction:
$$
u^2 = a+b = frac{3r^2+6rs-s^2}{4} text{ or } frac{s^2+6rs-3r^2}{4}
$$
We do a substitution for the odd $r=2f+1,s=2g+1$, giving
$$
u^2 = 2 + 6 f + 3 f^2 + 2 g + 6 f g - g^2
$$
or
$$
u^2 = 1 - 3 f^2 + 4 g + 6 f g + g^2
$$
For the first equation, LHS $equiv 0,1pmod 4$ but RHS $equiv 2,3pmod 4$, which is impossible. Therefore it must have been
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
answered Jan 18 at 15:17
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
add a comment |
$begingroup$
A partial, in progress answer: $a,b$ satisfies
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
for integers $m,n,r,s$ such that
$$
gcd(m,n)=gcd(r,s)=1
$$
and $r,s$ are odd, $snotequiv 0pmod 3$ and exactly one of $m,n$ is odd. It can be shown that $gcd(a,b)=1$ using either $m,n$ or $r,s$. Perhaps this system is already not solvable.
The only other restriction remaining is
$$
a+b=u^2
$$
for some integer $u$.
A sanity check:
$$
a^2+b^2 = (m^2+n^2)^2
$$
and
$$
a^2-ab+b^2 = left(frac{3r^2+s^2}{4}right)^2 = v^2 in mathbb Z
$$
Hence if $a+b=u^2$ then
$$
a^3+b^3 = (a+b)(a^2-ab+b^2) = (uv)^2
$$
Let $a,b,c$ be a primitive Pythagorean triplet. Then we know that for some integers $m>n$ and $gcd(m,n)=1$,
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
In particular, $m,n$ also have different parity.
Proposition 1. Let $a,b,c$ be a primitive Pythagorean triplet such that
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
and $m>n,;;gcd(m,n)=1$. If
$$
a^3+b^3=d^2,
$$
for some integer $d$ then
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
for some integers $u,v$.
Proof. We start with
$$
(a+b)(a^2-ab+b^2) = a^3+b^3 = d^2
$$
Since
$$
3a^2 = (2a-b)(a+b) + (a^2-ab+b^2)
$$
This shows that
$
gcd(a+b,a^2-ab+b^2)
$
divides $3a^2$. Checking $pmod 3$, the equation
$$
a+b = m^2+2mn-n^2 equiv 0 pmod 3
$$
is possible only if $m,nequiv 0pmod 3$. This contradicts $gcd(m,n)=1$, therefore $3nmid a+b$. Hence $gcd(a+b,a^2-ab+b^2)$ divides $a^2$. Now if a prime $p$ divides $gcd(a+b,a^2-ab+b^2)$, then $p$ divides $a^2$ and hence $pmid a$. But that means $p$ divides $b$, contradicting $gcd(a,b)=1$. Hence we conclude that
$$
gcd(a+b,a^2-ab+b^2)=1
$$
As a result, we can write
$$
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
$$
for some integers $u,v$. (It cannot have been $-u^2,-v^2$ instead since $a,b>0$.)
$$tag*{$square$}$$
For the equation
$$
a^2-ab+b^2=v^2,
$$
since $gcd(a,b)=1$ we must have $gcd(a,v)=gcd(b,v)=1$.
Proposition 2. The primitive integer solutions to
$$
a^2-ab+b^2 = v^2
$$
are
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
where $r,s$ are odd integers satisfying $gcd(r,s)=1$.
Proof. We first convert the equation to
$$
a^2-ab+b^2=v^2 Longleftrightarrow (2b-a)^2 + 3a^2 = (2v)^2
$$
Following The Solution of the Diophantine Equation$X^2+3Y^2=Z^2$,
Theorem 2.2 Let $E:x^2+3y^2=z^2$ be the diophantine equation and $(x,y,z)inmathbb Z^3$ with $gcd(x,y)=1$, $y$ is odd and $gcd(xz,3)=1$. Then
$$
begin{align}
|x| &= frac{3r^2-s^2}{2}\
|y| &= rs\
|z| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd integers $r,s$ and $gcd(r,s)=1$. (Not written but implied that $snotequiv 0pmod 3$.
Since $a$ is already odd, to use this result we need to check that
$$
gcd(2b-a,a)=1,quad gcd((2b-a)(2v),3)=1
$$
The first part is immediate since $gcd(a,b)=1$ and $a$ is odd. For the second part, if $3mid v$ or $3mid 2b-a$ then from
$$
(2b-a)^2+3a^2=v^2
$$
we get $3mid a$. For both cases $3$ divides $2b-a,a$ so $3$ divides $b$. This contradicts $gcd(a,b)=1$. Therefore indeed $gcd((2b-a)(2b),3)=1$, so we obtain the solutions
$$
begin{align}
|2b-a| &= frac{3r^2-s^2}{2}\
|a| &= rs\
|2v| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd $r,s$ satisfying $gcd(r,s)=1$. Rearranging:
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
which is what we want.
$$tag*{$square$}$$
This gives us a new restriction:
$$
u^2 = a+b = frac{3r^2+6rs-s^2}{4} text{ or } frac{s^2+6rs-3r^2}{4}
$$
We do a substitution for the odd $r=2f+1,s=2g+1$, giving
$$
u^2 = 2 + 6 f + 3 f^2 + 2 g + 6 f g - g^2
$$
or
$$
u^2 = 1 - 3 f^2 + 4 g + 6 f g + g^2
$$
For the first equation, LHS $equiv 0,1pmod 4$ but RHS $equiv 2,3pmod 4$, which is impossible. Therefore it must have been
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
answered Jan 18 at 15:17
Yong Hao NgYong Hao Ng
3,5691222
$endgroup$
A partial, in progress answer: $a,b$ satisfies
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
for integers $m,n,r,s$ such that
$$
gcd(m,n)=gcd(r,s)=1
$$
and $r,s$ are odd, $snotequiv 0pmod 3$ and exactly one of $m,n$ is odd. It can be shown that $gcd(a,b)=1$ using either $m,n$ or $r,s$. Perhaps this system is already not solvable.
The only other restriction remaining is
$$
a+b=u^2
$$
for some integer $u$.
A sanity check:
$$
a^2+b^2 = (m^2+n^2)^2
$$
and
$$
a^2-ab+b^2 = left(frac{3r^2+s^2}{4}right)^2 = v^2 in mathbb Z
$$
Hence if $a+b=u^2$ then
$$
a^3+b^3 = (a+b)(a^2-ab+b^2) = (uv)^2
$$
Let $a,b,c$ be a primitive Pythagorean triplet. Then we know that for some integers $m>n$ and $gcd(m,n)=1$,
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
In particular, $m,n$ also have different parity.
Proposition 1. Let $a,b,c$ be a primitive Pythagorean triplet such that
$$
a=m^2-n^2,quad b=2mn,quad c = m^2+n^2
$$
and $m>n,;;gcd(m,n)=1$. If
$$
a^3+b^3=d^2,
$$
for some integer $d$ then
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
for some integers $u,v$.
Proof. We start with
$$
(a+b)(a^2-ab+b^2) = a^3+b^3 = d^2
$$
Since
$$
3a^2 = (2a-b)(a+b) + (a^2-ab+b^2)
$$
This shows that
$
gcd(a+b,a^2-ab+b^2)
$
divides $3a^2$. Checking $pmod 3$, the equation
$$
a+b = m^2+2mn-n^2 equiv 0 pmod 3
$$
is possible only if $m,nequiv 0pmod 3$. This contradicts $gcd(m,n)=1$, therefore $3nmid a+b$. Hence $gcd(a+b,a^2-ab+b^2)$ divides $a^2$. Now if a prime $p$ divides $gcd(a+b,a^2-ab+b^2)$, then $p$ divides $a^2$ and hence $pmid a$. But that means $p$ divides $b$, contradicting $gcd(a,b)=1$. Hence we conclude that
$$
gcd(a+b,a^2-ab+b^2)=1
$$
As a result, we can write
$$
begin{align}
a+b &= u^2\
a^2-ab+b^2 &= v^2
end{align}
$$
for some integers $u,v$. (It cannot have been $-u^2,-v^2$ instead since $a,b>0$.)
$$tag*{$square$}$$
For the equation
$$
a^2-ab+b^2=v^2,
$$
since $gcd(a,b)=1$ we must have $gcd(a,v)=gcd(b,v)=1$.
Proposition 2. The primitive integer solutions to
$$
a^2-ab+b^2 = v^2
$$
are
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
where $r,s$ are odd integers satisfying $gcd(r,s)=1$.
Proof. We first convert the equation to
$$
a^2-ab+b^2=v^2 Longleftrightarrow (2b-a)^2 + 3a^2 = (2v)^2
$$
Following The Solution of the Diophantine Equation$X^2+3Y^2=Z^2$,
Theorem 2.2 Let $E:x^2+3y^2=z^2$ be the diophantine equation and $(x,y,z)inmathbb Z^3$ with $gcd(x,y)=1$, $y$ is odd and $gcd(xz,3)=1$. Then
$$
begin{align}
|x| &= frac{3r^2-s^2}{2}\
|y| &= rs\
|z| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd integers $r,s$ and $gcd(r,s)=1$. (Not written but implied that $snotequiv 0pmod 3$.
Since $a$ is already odd, to use this result we need to check that
$$
gcd(2b-a,a)=1,quad gcd((2b-a)(2v),3)=1
$$
The first part is immediate since $gcd(a,b)=1$ and $a$ is odd. For the second part, if $3mid v$ or $3mid 2b-a$ then from
$$
(2b-a)^2+3a^2=v^2
$$
we get $3mid a$. For both cases $3$ divides $2b-a,a$ so $3$ divides $b$. This contradicts $gcd(a,b)=1$. Therefore indeed $gcd((2b-a)(2b),3)=1$, so we obtain the solutions
$$
begin{align}
|2b-a| &= frac{3r^2-s^2}{2}\
|a| &= rs\
|2v| &= frac{3r^2+s^2}{2}
end{align}
$$
for some odd $r,s$ satisfying $gcd(r,s)=1$. Rearranging:
$$
begin{align}
a &= rs\
b &= frac{(3r-s)(r+s)}{4}text{ or }frac{(3r+s)(s-r)}{4}\
v &= frac{3r^2+s^2}{4}
end{align}
$$
which is what we want.
$$tag*{$square$}$$
This gives us a new restriction:
$$
u^2 = a+b = frac{3r^2+6rs-s^2}{4} text{ or } frac{s^2+6rs-3r^2}{4}
$$
We do a substitution for the odd $r=2f+1,s=2g+1$, giving
$$
u^2 = 2 + 6 f + 3 f^2 + 2 g + 6 f g - g^2
$$
or
$$
u^2 = 1 - 3 f^2 + 4 g + 6 f g + g^2
$$
For the first equation, LHS $equiv 0,1pmod 4$ but RHS $equiv 2,3pmod 4$, which is impossible. Therefore it must have been
$$
a=m^2-n^2=rs,quad b = 2mn=frac{(3r+s)(s-r)}{4}
$$
answered Jan 18 at 15:17
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 18 at 15:17
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 18 at 15:17
Yong Hao NgYong Hao Ng
3,5691222
answered Jan 18 at 15:17
Yong Hao NgYong Hao Ng
3,5691222
3,5691222
add a comment |
add a comment |
$begingroup$
If we suppose that such integers exist and write, say, $a=r^2-s^2$ and $b=2rs$ for coprime (positive) integers $r$ and $s$, then
$$
d^2=(r^2+2rs-s^2)(r^4-2r^3s++2r^2s^2+2rs^3+s^4)
$$
and hence a solution would correspond to a (nontrivial) rational point on the (genus $2$) curve
$$
y^2 = x^6 -3x^4+8x^3+3x^2-1.
$$
The Jacobian of this curve has rank $1$ and a Chabauty argument in Magma using the prime $17$ shows that there are no such points. There may be an easier way to see this, but I'm afraid it's not obvious to me.
answered Jan 20 at 5:39
Mike BennettMike Bennett
2,39478
$endgroup$
$begingroup$
What area of maths do I need to investigate in order to understand how you derived the equation in $x,y$ and the subsequent claims.
$endgroup$
– user1035795
Jan 20 at 12:51
$begingroup$
Multiply out the first equation (in $r, s$ and $d$) and then divide by $s^6$. The machinery needed to understand Chabauty's method is quite advanced. A book by Cassels and Flynn ("Prolegomena....") is a good place to start, but assumes a solid undergraduate background.
$endgroup$
– Mike Bennett
Jan 20 at 17:11
$begingroup$
@user1035795 From $a^2+b^2=c^2$ you get the primitive pythagorean parameterization $a=r^2-s^2,b=2rs$, then a substitution into $d^2=a^3+b^3$ gives you equation 1. Next since $sneq 0$, let $y=d/s^3$ and $x=r/s$ to get the curve over $mathbb Q$ as in equation 2.
$endgroup$
– Yong Hao Ng
Jan 21 at 4:52
$begingroup$
Thank you Mike, yes plain as day now you (and Yong Hao Ng) pointed out. I'll check out the Chabauty method, thanks.
$endgroup$
– user1035795
Jan 21 at 10:27
add a comment |
$begingroup$
If we suppose that such integers exist and write, say, $a=r^2-s^2$ and $b=2rs$ for coprime (positive) integers $r$ and $s$, then
$$
d^2=(r^2+2rs-s^2)(r^4-2r^3s++2r^2s^2+2rs^3+s^4)
$$
and hence a solution would correspond to a (nontrivial) rational point on the (genus $2$) curve
$$
y^2 = x^6 -3x^4+8x^3+3x^2-1.
$$
The Jacobian of this curve has rank $1$ and a Chabauty argument in Magma using the prime $17$ shows that there are no such points. There may be an easier way to see this, but I'm afraid it's not obvious to me.
answered Jan 20 at 5:39
Mike BennettMike Bennett
2,39478
$endgroup$
$begingroup$
What area of maths do I need to investigate in order to understand how you derived the equation in $x,y$ and the subsequent claims.
$endgroup$
– user1035795
Jan 20 at 12:51
$begingroup$
Multiply out the first equation (in $r, s$ and $d$) and then divide by $s^6$. The machinery needed to understand Chabauty's method is quite advanced. A book by Cassels and Flynn ("Prolegomena....") is a good place to start, but assumes a solid undergraduate background.
$endgroup$
– Mike Bennett
Jan 20 at 17:11
$begingroup$
@user1035795 From $a^2+b^2=c^2$ you get the primitive pythagorean parameterization $a=r^2-s^2,b=2rs$, then a substitution into $d^2=a^3+b^3$ gives you equation 1. Next since $sneq 0$, let $y=d/s^3$ and $x=r/s$ to get the curve over $mathbb Q$ as in equation 2.
$endgroup$
– Yong Hao Ng
Jan 21 at 4:52
$begingroup$
Thank you Mike, yes plain as day now you (and Yong Hao Ng) pointed out. I'll check out the Chabauty method, thanks.
$endgroup$
– user1035795
Jan 21 at 10:27
add a comment |
$begingroup$
If we suppose that such integers exist and write, say, $a=r^2-s^2$ and $b=2rs$ for coprime (positive) integers $r$ and $s$, then
$$
d^2=(r^2+2rs-s^2)(r^4-2r^3s++2r^2s^2+2rs^3+s^4)
$$
and hence a solution would correspond to a (nontrivial) rational point on the (genus $2$) curve
$$
y^2 = x^6 -3x^4+8x^3+3x^2-1.
$$
The Jacobian of this curve has rank $1$ and a Chabauty argument in Magma using the prime $17$ shows that there are no such points. There may be an easier way to see this, but I'm afraid it's not obvious to me.
answered Jan 20 at 5:39
Mike BennettMike Bennett
2,39478
$endgroup$
If we suppose that such integers exist and write, say, $a=r^2-s^2$ and $b=2rs$ for coprime (positive) integers $r$ and $s$, then
$$
d^2=(r^2+2rs-s^2)(r^4-2r^3s++2r^2s^2+2rs^3+s^4)
$$
and hence a solution would correspond to a (nontrivial) rational point on the (genus $2$) curve
$$
y^2 = x^6 -3x^4+8x^3+3x^2-1.
$$
The Jacobian of this curve has rank $1$ and a Chabauty argument in Magma using the prime $17$ shows that there are no such points. There may be an easier way to see this, but I'm afraid it's not obvious to me.
answered Jan 20 at 5:39
Mike BennettMike Bennett
2,39478
answered Jan 20 at 5:39
Mike BennettMike Bennett
2,39478
answered Jan 20 at 5:39
Mike BennettMike Bennett
2,39478
answered Jan 20 at 5:39
Mike BennettMike Bennett
2,39478
2,39478
$begingroup$
What area of maths do I need to investigate in order to understand how you derived the equation in $x,y$ and the subsequent claims.
$endgroup$
– user1035795
Jan 20 at 12:51
$begingroup$
Multiply out the first equation (in $r, s$ and $d$) and then divide by $s^6$. The machinery needed to understand Chabauty's method is quite advanced. A book by Cassels and Flynn ("Prolegomena....") is a good place to start, but assumes a solid undergraduate background.
$endgroup$
– Mike Bennett
Jan 20 at 17:11
$begingroup$
@user1035795 From $a^2+b^2=c^2$ you get the primitive pythagorean parameterization $a=r^2-s^2,b=2rs$, then a substitution into $d^2=a^3+b^3$ gives you equation 1. Next since $sneq 0$, let $y=d/s^3$ and $x=r/s$ to get the curve over $mathbb Q$ as in equation 2.
$endgroup$
– Yong Hao Ng
Jan 21 at 4:52
$begingroup$
Thank you Mike, yes plain as day now you (and Yong Hao Ng) pointed out. I'll check out the Chabauty method, thanks.
$endgroup$
– user1035795
Jan 21 at 10:27
add a comment |
$begingroup$
What area of maths do I need to investigate in order to understand how you derived the equation in $x,y$ and the subsequent claims.
$endgroup$
– user1035795
Jan 20 at 12:51
$begingroup$
Multiply out the first equation (in $r, s$ and $d$) and then divide by $s^6$. The machinery needed to understand Chabauty's method is quite advanced. A book by Cassels and Flynn ("Prolegomena....") is a good place to start, but assumes a solid undergraduate background.
$endgroup$
– Mike Bennett
Jan 20 at 17:11
$begingroup$
@user1035795 From $a^2+b^2=c^2$ you get the primitive pythagorean parameterization $a=r^2-s^2,b=2rs$, then a substitution into $d^2=a^3+b^3$ gives you equation 1. Next since $sneq 0$, let $y=d/s^3$ and $x=r/s$ to get the curve over $mathbb Q$ as in equation 2.
$endgroup$
– Yong Hao Ng
Jan 21 at 4:52
$begingroup$
Thank you Mike, yes plain as day now you (and Yong Hao Ng) pointed out. I'll check out the Chabauty method, thanks.
$endgroup$
– user1035795
Jan 21 at 10:27
$begingroup$
What area of maths do I need to investigate in order to understand how you derived the equation in $x,y$ and the subsequent claims.
$endgroup$
– user1035795
Jan 20 at 12:51
$begingroup$
What area of maths do I need to investigate in order to understand how you derived the equation in $x,y$ and the subsequent claims.
$endgroup$
– user1035795
Jan 20 at 12:51
$begingroup$
Multiply out the first equation (in $r, s$ and $d$) and then divide by $s^6$. The machinery needed to understand Chabauty's method is quite advanced. A book by Cassels and Flynn ("Prolegomena....") is a good place to start, but assumes a solid undergraduate background.
$endgroup$
– Mike Bennett
Jan 20 at 17:11
$begingroup$
Multiply out the first equation (in $r, s$ and $d$) and then divide by $s^6$. The machinery needed to understand Chabauty's method is quite advanced. A book by Cassels and Flynn ("Prolegomena....") is a good place to start, but assumes a solid undergraduate background.
$endgroup$
– Mike Bennett
Jan 20 at 17:11
$begingroup$
@user1035795 From $a^2+b^2=c^2$ you get the primitive pythagorean parameterization $a=r^2-s^2,b=2rs$, then a substitution into $d^2=a^3+b^3$ gives you equation 1. Next since $sneq 0$, let $y=d/s^3$ and $x=r/s$ to get the curve over $mathbb Q$ as in equation 2.
$endgroup$
– Yong Hao Ng
Jan 21 at 4:52
$begingroup$
@user1035795 From $a^2+b^2=c^2$ you get the primitive pythagorean parameterization $a=r^2-s^2,b=2rs$, then a substitution into $d^2=a^3+b^3$ gives you equation 1. Next since $sneq 0$, let $y=d/s^3$ and $x=r/s$ to get the curve over $mathbb Q$ as in equation 2.
$endgroup$
– Yong Hao Ng
Jan 21 at 4:52
$begingroup$
Thank you Mike, yes plain as day now you (and Yong Hao Ng) pointed out. I'll check out the Chabauty method, thanks.
$endgroup$
– user1035795
Jan 21 at 10:27
$begingroup$
Thank you Mike, yes plain as day now you (and Yong Hao Ng) pointed out. I'll check out the Chabauty method, thanks.
$endgroup$
– user1035795
Jan 21 at 10:27
add a comment |
$begingroup$
equations $$X^3+Y^3=Z^2$$
Can be expressed by integers $p,s,a,b,c$ . where the number of $c$ characterizes the degree of primitiveness.
$$X=(3a^2+4ab+b^2)(3a^2+b^2)c^2$$
$$Y=2b(a+b)(3a^2+b^2)c^2$$
$$Z=3(a+b)^2(3a^2+b^2)^2c^3$$
And more.
$$X=2b(b-a)(3a^2+b^2)c^2$$
$$Y=2b(b+a)(3a^2+b^2)c^2$$
$$Z=4b^2(3a^2+b^2)^2c^3$$
If we decide to factor $$X^3+Y^3=qZ^2$$
For a compact notation we replace :
$$a=s(2p-s)$$
$$b=p^2-s^2$$
$$t=p^2-ps+s^2$$
then:
$$X=qb(a+b)c^2$$
$$Y=qa(a+b)c^2$$
$$Z=qt(a+b)^2c^3$$
And the most beautiful solution. If we use the solutions of Pell's equation: $p^2-3a^2s^2=1$
by the way $a$ May appear as a factor in the decision and.
Then the solutions are of the form::
$$X=qa(2p-3as)sc^2$$
$$Y=q(p-2as)pc^2$$
$$Z=q(p^2-3aps+3a^2s^2)c^3$$
If we change the sign : $$Y^3-X^3=qZ^2$$
Then the solutions are of the form:
$$X=qa(2p+3as)sc^2$$
$$Y=q(p+2as)pc^2$$
$$Z=q(p^2+3aps+3a^2s^2)c^3$$
Another solution of the equation: $$X^3+Y^3=qZ^2$$
$p,s$ - integers asked us.
To facilitate the calculations we make the change. $a,b,c$
If the ratio is as follows : $q=3t^2+1$
$$b=2q(q+2mp{6t})p^2+6p(tmp1)ps+(q-1mp{3t})s^2$$
$$c=6q(q-2(1pm{t}))p^2+6q(tmp1)ps+3(1mp{t})s^2$$
$$a=12q(1mp{t})p^2+6(4tmp{q})ps+3(1mp{t})s^2$$
If the ratio is as follows: $q=t^2+3$
$$b=3(q-1)(1pm{t})s^2+2(3pm{(q-1)t})ps+(1pm{t})p^2$$
$$c=3(6-(q-1)(q-3mp{t}))s^2+6(1pm{t})ps+(q-3pm{t})p^2$$
$$a=3(6-(q-1)(1mp{t}))s^2+6(1pm{t})ps+(1pm{t})p^2$$
Then the solutions are of the form:
$$X=2c(c-b)$$
$$Y=(c-3b)(c-b)$$
$$Z=3a(c-b)^2$$
Then the solutions are of the form:
$$X=2(c-b)c$$
$$Y=2(c+b)c$$
$$Z=4ac^2$$
If the ratio is as follows : $q=t^2+3$
$$c=6(q-4)(2pm{t})p^2+4(6pm{(q-4)t})ps+2(2pm{t})s^2$$
$$b=3(24-(q-4)(q-3mp{2t}))p^2+12(2pm{t})ps+(q-3pm{2t})s^2$$
$$a=3(24-(q-4)(4mp{2t}))p^2+12(2pm{t})ps+2(2pm{t})s^2$$
If the ratio is as follows $q=3t^2+4$
$$c=q(-q+7(4mp{3t}))p^2+6q(tmp{1})ps+(q-4mp{3t})s^2$$
$$b=3q(2q-7(1pm{t}))p^2+6q(tmp{1})ps+3(1mp{t})s^2$$
$$a=21q(1mp{t})p^2+6(7tmp{q})ps+3(1mp{t})s^2$$
Then the solutions are of the form :
$$X=2(3c-2b)c$$
$$Y=2(3c+2b)c$$
$$Z=12ac^2$$
Then the solutions are of the form :
$$X=(2b-c)b$$
$$Y=(2b+c)b$$
$$Z=2ab^2$$
answered Jan 18 at 16:01
individindivid
3,2621916
$endgroup$
$begingroup$
So with all of that can you give an example solution?
$endgroup$
– user1035795
Jan 20 at 12:53
add a comment |
$begingroup$
equations $$X^3+Y^3=Z^2$$
Can be expressed by integers $p,s,a,b,c$ . where the number of $c$ characterizes the degree of primitiveness.
$$X=(3a^2+4ab+b^2)(3a^2+b^2)c^2$$
$$Y=2b(a+b)(3a^2+b^2)c^2$$
$$Z=3(a+b)^2(3a^2+b^2)^2c^3$$
And more.
$$X=2b(b-a)(3a^2+b^2)c^2$$
$$Y=2b(b+a)(3a^2+b^2)c^2$$
$$Z=4b^2(3a^2+b^2)^2c^3$$
If we decide to factor $$X^3+Y^3=qZ^2$$
For a compact notation we replace :
$$a=s(2p-s)$$
$$b=p^2-s^2$$
$$t=p^2-ps+s^2$$
then:
$$X=qb(a+b)c^2$$
$$Y=qa(a+b)c^2$$
$$Z=qt(a+b)^2c^3$$
And the most beautiful solution. If we use the solutions of Pell's equation: $p^2-3a^2s^2=1$
by the way $a$ May appear as a factor in the decision and.
Then the solutions are of the form::
$$X=qa(2p-3as)sc^2$$
$$Y=q(p-2as)pc^2$$
$$Z=q(p^2-3aps+3a^2s^2)c^3$$
If we change the sign : $$Y^3-X^3=qZ^2$$
Then the solutions are of the form:
$$X=qa(2p+3as)sc^2$$
$$Y=q(p+2as)pc^2$$
$$Z=q(p^2+3aps+3a^2s^2)c^3$$
Another solution of the equation: $$X^3+Y^3=qZ^2$$
$p,s$ - integers asked us.
To facilitate the calculations we make the change. $a,b,c$
If the ratio is as follows : $q=3t^2+1$
$$b=2q(q+2mp{6t})p^2+6p(tmp1)ps+(q-1mp{3t})s^2$$
$$c=6q(q-2(1pm{t}))p^2+6q(tmp1)ps+3(1mp{t})s^2$$
$$a=12q(1mp{t})p^2+6(4tmp{q})ps+3(1mp{t})s^2$$
If the ratio is as follows: $q=t^2+3$
$$b=3(q-1)(1pm{t})s^2+2(3pm{(q-1)t})ps+(1pm{t})p^2$$
$$c=3(6-(q-1)(q-3mp{t}))s^2+6(1pm{t})ps+(q-3pm{t})p^2$$
$$a=3(6-(q-1)(1mp{t}))s^2+6(1pm{t})ps+(1pm{t})p^2$$
Then the solutions are of the form:
$$X=2c(c-b)$$
$$Y=(c-3b)(c-b)$$
$$Z=3a(c-b)^2$$
Then the solutions are of the form:
$$X=2(c-b)c$$
$$Y=2(c+b)c$$
$$Z=4ac^2$$
If the ratio is as follows : $q=t^2+3$
$$c=6(q-4)(2pm{t})p^2+4(6pm{(q-4)t})ps+2(2pm{t})s^2$$
$$b=3(24-(q-4)(q-3mp{2t}))p^2+12(2pm{t})ps+(q-3pm{2t})s^2$$
$$a=3(24-(q-4)(4mp{2t}))p^2+12(2pm{t})ps+2(2pm{t})s^2$$
If the ratio is as follows $q=3t^2+4$
$$c=q(-q+7(4mp{3t}))p^2+6q(tmp{1})ps+(q-4mp{3t})s^2$$
$$b=3q(2q-7(1pm{t}))p^2+6q(tmp{1})ps+3(1mp{t})s^2$$
$$a=21q(1mp{t})p^2+6(7tmp{q})ps+3(1mp{t})s^2$$
Then the solutions are of the form :
$$X=2(3c-2b)c$$
$$Y=2(3c+2b)c$$
$$Z=12ac^2$$
Then the solutions are of the form :
$$X=(2b-c)b$$
$$Y=(2b+c)b$$
$$Z=2ab^2$$
answered Jan 18 at 16:01
individindivid
3,2621916
$endgroup$
$begingroup$
So with all of that can you give an example solution?
$endgroup$
– user1035795
Jan 20 at 12:53
add a comment |
$begingroup$
equations $$X^3+Y^3=Z^2$$
Can be expressed by integers $p,s,a,b,c$ . where the number of $c$ characterizes the degree of primitiveness.
$$X=(3a^2+4ab+b^2)(3a^2+b^2)c^2$$
$$Y=2b(a+b)(3a^2+b^2)c^2$$
$$Z=3(a+b)^2(3a^2+b^2)^2c^3$$
And more.
$$X=2b(b-a)(3a^2+b^2)c^2$$
$$Y=2b(b+a)(3a^2+b^2)c^2$$
$$Z=4b^2(3a^2+b^2)^2c^3$$
If we decide to factor $$X^3+Y^3=qZ^2$$
For a compact notation we replace :
$$a=s(2p-s)$$
$$b=p^2-s^2$$
$$t=p^2-ps+s^2$$
then:
$$X=qb(a+b)c^2$$
$$Y=qa(a+b)c^2$$
$$Z=qt(a+b)^2c^3$$
And the most beautiful solution. If we use the solutions of Pell's equation: $p^2-3a^2s^2=1$
by the way $a$ May appear as a factor in the decision and.
Then the solutions are of the form::
$$X=qa(2p-3as)sc^2$$
$$Y=q(p-2as)pc^2$$
$$Z=q(p^2-3aps+3a^2s^2)c^3$$
If we change the sign : $$Y^3-X^3=qZ^2$$
Then the solutions are of the form:
$$X=qa(2p+3as)sc^2$$
$$Y=q(p+2as)pc^2$$
$$Z=q(p^2+3aps+3a^2s^2)c^3$$
Another solution of the equation: $$X^3+Y^3=qZ^2$$
$p,s$ - integers asked us.
To facilitate the calculations we make the change. $a,b,c$
If the ratio is as follows : $q=3t^2+1$
$$b=2q(q+2mp{6t})p^2+6p(tmp1)ps+(q-1mp{3t})s^2$$
$$c=6q(q-2(1pm{t}))p^2+6q(tmp1)ps+3(1mp{t})s^2$$
$$a=12q(1mp{t})p^2+6(4tmp{q})ps+3(1mp{t})s^2$$
If the ratio is as follows: $q=t^2+3$
$$b=3(q-1)(1pm{t})s^2+2(3pm{(q-1)t})ps+(1pm{t})p^2$$
$$c=3(6-(q-1)(q-3mp{t}))s^2+6(1pm{t})ps+(q-3pm{t})p^2$$
$$a=3(6-(q-1)(1mp{t}))s^2+6(1pm{t})ps+(1pm{t})p^2$$
Then the solutions are of the form:
$$X=2c(c-b)$$
$$Y=(c-3b)(c-b)$$
$$Z=3a(c-b)^2$$
Then the solutions are of the form:
$$X=2(c-b)c$$
$$Y=2(c+b)c$$
$$Z=4ac^2$$
If the ratio is as follows : $q=t^2+3$
$$c=6(q-4)(2pm{t})p^2+4(6pm{(q-4)t})ps+2(2pm{t})s^2$$
$$b=3(24-(q-4)(q-3mp{2t}))p^2+12(2pm{t})ps+(q-3pm{2t})s^2$$
$$a=3(24-(q-4)(4mp{2t}))p^2+12(2pm{t})ps+2(2pm{t})s^2$$
If the ratio is as follows $q=3t^2+4$
$$c=q(-q+7(4mp{3t}))p^2+6q(tmp{1})ps+(q-4mp{3t})s^2$$
$$b=3q(2q-7(1pm{t}))p^2+6q(tmp{1})ps+3(1mp{t})s^2$$
$$a=21q(1mp{t})p^2+6(7tmp{q})ps+3(1mp{t})s^2$$
Then the solutions are of the form :
$$X=2(3c-2b)c$$
$$Y=2(3c+2b)c$$
$$Z=12ac^2$$
Then the solutions are of the form :
$$X=(2b-c)b$$
$$Y=(2b+c)b$$
$$Z=2ab^2$$
answered Jan 18 at 16:01
individindivid
3,2621916
$endgroup$
equations $$X^3+Y^3=Z^2$$
Can be expressed by integers $p,s,a,b,c$ . where the number of $c$ characterizes the degree of primitiveness.
$$X=(3a^2+4ab+b^2)(3a^2+b^2)c^2$$
$$Y=2b(a+b)(3a^2+b^2)c^2$$
$$Z=3(a+b)^2(3a^2+b^2)^2c^3$$
And more.
$$X=2b(b-a)(3a^2+b^2)c^2$$
$$Y=2b(b+a)(3a^2+b^2)c^2$$
$$Z=4b^2(3a^2+b^2)^2c^3$$
If we decide to factor $$X^3+Y^3=qZ^2$$
For a compact notation we replace :
$$a=s(2p-s)$$
$$b=p^2-s^2$$
$$t=p^2-ps+s^2$$
then:
$$X=qb(a+b)c^2$$
$$Y=qa(a+b)c^2$$
$$Z=qt(a+b)^2c^3$$
And the most beautiful solution. If we use the solutions of Pell's equation: $p^2-3a^2s^2=1$
by the way $a$ May appear as a factor in the decision and.
Then the solutions are of the form::
$$X=qa(2p-3as)sc^2$$
$$Y=q(p-2as)pc^2$$
$$Z=q(p^2-3aps+3a^2s^2)c^3$$
If we change the sign : $$Y^3-X^3=qZ^2$$
Then the solutions are of the form:
$$X=qa(2p+3as)sc^2$$
$$Y=q(p+2as)pc^2$$
$$Z=q(p^2+3aps+3a^2s^2)c^3$$
Another solution of the equation: $$X^3+Y^3=qZ^2$$
$p,s$ - integers asked us.
To facilitate the calculations we make the change. $a,b,c$
If the ratio is as follows : $q=3t^2+1$
$$b=2q(q+2mp{6t})p^2+6p(tmp1)ps+(q-1mp{3t})s^2$$
$$c=6q(q-2(1pm{t}))p^2+6q(tmp1)ps+3(1mp{t})s^2$$
$$a=12q(1mp{t})p^2+6(4tmp{q})ps+3(1mp{t})s^2$$
If the ratio is as follows: $q=t^2+3$
$$b=3(q-1)(1pm{t})s^2+2(3pm{(q-1)t})ps+(1pm{t})p^2$$
$$c=3(6-(q-1)(q-3mp{t}))s^2+6(1pm{t})ps+(q-3pm{t})p^2$$
$$a=3(6-(q-1)(1mp{t}))s^2+6(1pm{t})ps+(1pm{t})p^2$$
Then the solutions are of the form:
$$X=2c(c-b)$$
$$Y=(c-3b)(c-b)$$
$$Z=3a(c-b)^2$$
Then the solutions are of the form:
$$X=2(c-b)c$$
$$Y=2(c+b)c$$
$$Z=4ac^2$$
If the ratio is as follows : $q=t^2+3$
$$c=6(q-4)(2pm{t})p^2+4(6pm{(q-4)t})ps+2(2pm{t})s^2$$
$$b=3(24-(q-4)(q-3mp{2t}))p^2+12(2pm{t})ps+(q-3pm{2t})s^2$$
$$a=3(24-(q-4)(4mp{2t}))p^2+12(2pm{t})ps+2(2pm{t})s^2$$
If the ratio is as follows $q=3t^2+4$
$$c=q(-q+7(4mp{3t}))p^2+6q(tmp{1})ps+(q-4mp{3t})s^2$$
$$b=3q(2q-7(1pm{t}))p^2+6q(tmp{1})ps+3(1mp{t})s^2$$
$$a=21q(1mp{t})p^2+6(7tmp{q})ps+3(1mp{t})s^2$$
Then the solutions are of the form :
$$X=2(3c-2b)c$$
$$Y=2(3c+2b)c$$
$$Z=12ac^2$$
Then the solutions are of the form :
$$X=(2b-c)b$$
$$Y=(2b+c)b$$
$$Z=2ab^2$$
answered Jan 18 at 16:01
individindivid
3,2621916
answered Jan 18 at 16:01
individindivid
3,2621916
answered Jan 18 at 16:01
individindivid
3,2621916
answered Jan 18 at 16:01
individindivid
3,2621916
3,2621916
$begingroup$
So with all of that can you give an example solution?
$endgroup$
– user1035795
Jan 20 at 12:53
add a comment |
$begingroup$
So with all of that can you give an example solution?
$endgroup$
– user1035795
Jan 20 at 12:53
$begingroup$
So with all of that can you give an example solution?
$endgroup$
– user1035795
Jan 20 at 12:53
$begingroup$
So with all of that can you give an example solution?
$endgroup$
– user1035795
Jan 20 at 12:53
add a comment |
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$begingroup$
Is there a particular reason that you're requiring primitivity? The answer might well be different without it, since your last equation is no longer homogeneous.
$endgroup$
– user3482749
Jan 17 at 15:20
$begingroup$
You can just multiply $a$, $b$ and $c$ by any $n^2$ and $d$ by $n^3$ (for any $ninmathbb N$) to form a new solution if you have one, so it makes some kind of sense to rule that out, but that's not the same as requiring primitivity, which I also don't see a reason for.
$endgroup$
– Henrik
Jan 17 at 15:30
2
$begingroup$
Because we can always find a trivial solution i.e. $3^3+4^3=91$, if use that as a common factor $a=3.91, b=4.91$ then $a^3+b^3 = 91^3(3^3+4^3)=91^4$ which is square
$endgroup$
– user1035795
Jan 17 at 15:34
1
$begingroup$
But doesn't that just mean that we can find solutions where the triple isn't primitive, not that all solutions where the triple isn't primitive is trivial (in that sense)? I.e. might solutions where the triple isn't primitive not still be interesting?
$endgroup$
– Henrik
Jan 17 at 16:27
1
$begingroup$
Yes, they might, I guess "Interest" is subjective. My interest is the existence of a coprime solution
$endgroup$
– user1035795
Jan 17 at 16:36