Mixing
Derivation of Pythagorean Triple General Solution Starting Point:
$begingroup$
I was reading on proof wiki about the derivation of the general solution to the pythagorean triple diophantine equation:
$$
x^2 + y^2 = z^2,
$$
where $x,y,z > 0$ are integers.
I came across the following general solution to the primitive function:
begin{align*}
x &= 2mn\
y &= (m^2 - n^2)\
z &= (m^2 + n^2)\
end{align*}
for coprime $m,n$.
I looked at the proof of it working (if you square $x$ and $y$ and add it it does indeed equal $z^2$)
My one qualm was, how the hell did they start with that? For example, is there a natural way by starting with the original problem that you end up with the expression above?
I noticed that when attempting to derive the general solution myself, from start to finish,
I would begin by noting I can find all pairs such that $z$ and $y$ differ by a constant $k$... but I cannot make that final leap to end up with the equation above so that for any given $k$ you can find a solution.
Thanks ahead of time!
elementary-number-theory diophantine-equations pythagorean-triples
edited Mar 26 '15 at 11:10
Martin Sleziak
44.8k10119272
asked Jun 17 '13 at 23:22
frogeyedpeasfrogeyedpeas
7,55772053
$endgroup$
add a comment |
$begingroup$
I was reading on proof wiki about the derivation of the general solution to the pythagorean triple diophantine equation:
$$
x^2 + y^2 = z^2,
$$
where $x,y,z > 0$ are integers.
I came across the following general solution to the primitive function:
begin{align*}
x &= 2mn\
y &= (m^2 - n^2)\
z &= (m^2 + n^2)\
end{align*}
for coprime $m,n$.
I looked at the proof of it working (if you square $x$ and $y$ and add it it does indeed equal $z^2$)
My one qualm was, how the hell did they start with that? For example, is there a natural way by starting with the original problem that you end up with the expression above?
I noticed that when attempting to derive the general solution myself, from start to finish,
I would begin by noting I can find all pairs such that $z$ and $y$ differ by a constant $k$... but I cannot make that final leap to end up with the equation above so that for any given $k$ you can find a solution.
Thanks ahead of time!
elementary-number-theory diophantine-equations pythagorean-triples
edited Mar 26 '15 at 11:10
Martin Sleziak
44.8k10119272
asked Jun 17 '13 at 23:22
frogeyedpeasfrogeyedpeas
7,55772053
$endgroup$
$begingroup$
That's not a generating formula for all triples. If we make restriction on $m$ and $n$ (relatively prime, opposite parity, $m gt n$) we get all primitive triples. To see don't get all triples, we can't get $9,12,15$.
$endgroup$
– André Nicolas
Jun 17 '13 at 23:31
$begingroup$
find a rational parameterization of the unit circle.
$endgroup$
– yoyo
Jun 17 '13 at 23:41
$begingroup$
@André: That's the result of primitive triplets ${3,4,5 }$, right?
$endgroup$
– Inceptio
Jun 18 '13 at 7:46
$begingroup$
Yes. The details are nicely done in the answer by robjohn.
$endgroup$
– André Nicolas
Jun 18 '13 at 10:56
$begingroup$
The hardest part for me is the values of the variables equaling.
$endgroup$
– Roddy MacPhee
Feb 23 at 13:50
add a comment |
$begingroup$
I was reading on proof wiki about the derivation of the general solution to the pythagorean triple diophantine equation:
$$
x^2 + y^2 = z^2,
$$
where $x,y,z > 0$ are integers.
I came across the following general solution to the primitive function:
begin{align*}
x &= 2mn\
y &= (m^2 - n^2)\
z &= (m^2 + n^2)\
end{align*}
for coprime $m,n$.
I looked at the proof of it working (if you square $x$ and $y$ and add it it does indeed equal $z^2$)
My one qualm was, how the hell did they start with that? For example, is there a natural way by starting with the original problem that you end up with the expression above?
I noticed that when attempting to derive the general solution myself, from start to finish,
I would begin by noting I can find all pairs such that $z$ and $y$ differ by a constant $k$... but I cannot make that final leap to end up with the equation above so that for any given $k$ you can find a solution.
Thanks ahead of time!
elementary-number-theory diophantine-equations pythagorean-triples
edited Mar 26 '15 at 11:10
Martin Sleziak
44.8k10119272
asked Jun 17 '13 at 23:22
frogeyedpeasfrogeyedpeas
7,55772053
$endgroup$
I was reading on proof wiki about the derivation of the general solution to the pythagorean triple diophantine equation:
$$
x^2 + y^2 = z^2,
$$
where $x,y,z > 0$ are integers.
I came across the following general solution to the primitive function:
begin{align*}
x &= 2mn\
y &= (m^2 - n^2)\
z &= (m^2 + n^2)\
end{align*}
for coprime $m,n$.
I looked at the proof of it working (if you square $x$ and $y$ and add it it does indeed equal $z^2$)
My one qualm was, how the hell did they start with that? For example, is there a natural way by starting with the original problem that you end up with the expression above?
I noticed that when attempting to derive the general solution myself, from start to finish,
I would begin by noting I can find all pairs such that $z$ and $y$ differ by a constant $k$... but I cannot make that final leap to end up with the equation above so that for any given $k$ you can find a solution.
Thanks ahead of time!
elementary-number-theory diophantine-equations pythagorean-triples
elementary-number-theory diophantine-equations pythagorean-triples
edited Mar 26 '15 at 11:10
Martin Sleziak
44.8k10119272
asked Jun 17 '13 at 23:22
frogeyedpeasfrogeyedpeas
7,55772053
edited Mar 26 '15 at 11:10
Martin Sleziak
44.8k10119272
asked Jun 17 '13 at 23:22
frogeyedpeasfrogeyedpeas
7,55772053
edited Mar 26 '15 at 11:10
Martin Sleziak
44.8k10119272
edited Mar 26 '15 at 11:10
Martin Sleziak
44.8k10119272
edited Mar 26 '15 at 11:10
Martin Sleziak
44.8k10119272
44.8k10119272
asked Jun 17 '13 at 23:22
frogeyedpeasfrogeyedpeas
7,55772053
asked Jun 17 '13 at 23:22
frogeyedpeasfrogeyedpeas
7,55772053
asked Jun 17 '13 at 23:22
frogeyedpeasfrogeyedpeas
7,55772053
7,55772053
$begingroup$
That's not a generating formula for all triples. If we make restriction on $m$ and $n$ (relatively prime, opposite parity, $m gt n$) we get all primitive triples. To see don't get all triples, we can't get $9,12,15$.
$endgroup$
– André Nicolas
Jun 17 '13 at 23:31
$begingroup$
find a rational parameterization of the unit circle.
$endgroup$
– yoyo
Jun 17 '13 at 23:41
$begingroup$
@André: That's the result of primitive triplets ${3,4,5 }$, right?
$endgroup$
– Inceptio
Jun 18 '13 at 7:46
$begingroup$
Yes. The details are nicely done in the answer by robjohn.
$endgroup$
– André Nicolas
Jun 18 '13 at 10:56
$begingroup$
The hardest part for me is the values of the variables equaling.
$endgroup$
– Roddy MacPhee
Feb 23 at 13:50
add a comment |
$begingroup$
That's not a generating formula for all triples. If we make restriction on $m$ and $n$ (relatively prime, opposite parity, $m gt n$) we get all primitive triples. To see don't get all triples, we can't get $9,12,15$.
$endgroup$
– André Nicolas
Jun 17 '13 at 23:31
$begingroup$
find a rational parameterization of the unit circle.
$endgroup$
– yoyo
Jun 17 '13 at 23:41
$begingroup$
@André: That's the result of primitive triplets ${3,4,5 }$, right?
$endgroup$
– Inceptio
Jun 18 '13 at 7:46
$begingroup$
Yes. The details are nicely done in the answer by robjohn.
$endgroup$
– André Nicolas
Jun 18 '13 at 10:56
$begingroup$
The hardest part for me is the values of the variables equaling.
$endgroup$
– Roddy MacPhee
Feb 23 at 13:50
$begingroup$
That's not a generating formula for all triples. If we make restriction on $m$ and $n$ (relatively prime, opposite parity, $m gt n$) we get all primitive triples. To see don't get all triples, we can't get $9,12,15$.
$endgroup$
– André Nicolas
Jun 17 '13 at 23:31
$begingroup$
That's not a generating formula for all triples. If we make restriction on $m$ and $n$ (relatively prime, opposite parity, $m gt n$) we get all primitive triples. To see don't get all triples, we can't get $9,12,15$.
$endgroup$
– André Nicolas
Jun 17 '13 at 23:31
$begingroup$
find a rational parameterization of the unit circle.
$endgroup$
– yoyo
Jun 17 '13 at 23:41
$begingroup$
find a rational parameterization of the unit circle.
$endgroup$
– yoyo
Jun 17 '13 at 23:41
$begingroup$
@André: That's the result of primitive triplets ${3,4,5 }$, right?
$endgroup$
– Inceptio
Jun 18 '13 at 7:46
$begingroup$
@André: That's the result of primitive triplets ${3,4,5 }$, right?
$endgroup$
– Inceptio
Jun 18 '13 at 7:46
$begingroup$
Yes. The details are nicely done in the answer by robjohn.
$endgroup$
– André Nicolas
Jun 18 '13 at 10:56
$begingroup$
Yes. The details are nicely done in the answer by robjohn.
$endgroup$
– André Nicolas
Jun 18 '13 at 10:56
$begingroup$
The hardest part for me is the values of the variables equaling.
$endgroup$
– Roddy MacPhee
Feb 23 at 13:50
$begingroup$
The hardest part for me is the values of the variables equaling.
$endgroup$
– Roddy MacPhee
Feb 23 at 13:50
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
There are a number of ways to find those expressions organically. One way is via complex numbers (and this method can be found in the first chapter of Marcus' Number Fields).
Suppose $x^2 + y^2 = z^2$, with $x,y,zinBbb N$, and also suppose that $x$, $y$, and $z$ have no common factor. Looking mod $4$, we can tell that $z$ must be odd: the only squares mod $4$ are $0$ and $1$, so the only possible equations are $0^2 + 0^2equiv 0^2mod 4$ and $0^2 + 1^2 equiv 1^2mod 4$. But in the first case, everything is divisible by $4$, so we don't have a primitive triple.
Proceeding, we can factor the left hand side and get $(x + iy)(x - iy) = z^2$. So, we now have a problem concerning $Bbb Z[i] = {a + bimid a,binBbb Z}$. One can show that $Bbb Z[i]$ has unique factorization of elements into primes (for a precise definition of all this, look to any abstract algebra text), so suppose some prime $p$ divides $x + iy$. $p$ clearly divides $z^2$ an even number of times, so we want to show that $p$ does not divide $x - iy$. If it did, we would have $pmid (x + yi) + (x - yi)$, or $pmid 2x$. Since $p$ divides $z$ as well, and $x$ and $z$ are relatively prime, we know that we can find $n,minBbb{Z}$ such that $2xm + zn = 1$ (recall that $z$ is odd). However, this implies that $p$ divides $1$, but the only elements dividing $1$ in $Bbb{Z}[i]$ are $pm 1$, $pm i$, none of which are primes - so we have a contradiction. Therefore, by unique factorization, we must have $x + iy = ualpha^2$, where $uin{pm1,pm i}$. Writing $alpha = m + in$, we have $alpha^2 = m^2 - n^2 + i2mn $, so $x = pm left(m^2 - n^2right)$, $y = pm 2mn$, and solving for $z$, we get $z = pmleft(m^2 + n^2right)$.
If $m$ and $n$ aren't coprime, we don't obtain a primitive triple, as each of $x$, $y$, and $z$ will have a factor in common.
There are numerous other ways of deriving these ways of representing $x$, $y$, and $z$ as well, some of which include parameterizing the unit circle or moving $y^2$ to the other side and writing $x^2 = (z + y)(z - y)$.
answered Jun 17 '13 at 23:58
StahlStahl
16.7k43455
$endgroup$
$begingroup$
This answered my follow up before I asked it, thanks!
$endgroup$
– frogeyedpeas
Jun 18 '13 at 0:27
add a comment |
$begingroup$
Consider the intersection of the line $y=t(x+1)$ with the circle $x^2+y^2=1$. We get points on the circle
$$
(x,y)=left(frac{1-t^2}{1+t^2},frac{2t}{1+t^2}right).
$$
Let $t=m/n$ be rational, plug into $x^2+y^2$ and clear denominators to get
$$
(n^2-m^2)^2+(2mn)^2=(n^2+m^2)^2
$$
answered Jun 17 '13 at 23:58
yoyoyoyo
6,5991726
$endgroup$
$begingroup$
magic step is imagining the line $y=t(x+1)$
$endgroup$
– karakfa
Apr 22 '17 at 17:43
add a comment |
$begingroup$
Here is the way to generate all relatively prime pythagorean triples:
Theorem: Let $m$ and $n$ be positive integers so that
$$
begin{align}
&mgt ntag{1}\
&m+ntext{ is odd}tag{2}\
&mtext{ and }ntext{ are relatively prime}tag{3}
end{align}
$$
Then,
$$
begin{align}
a &= m^2 - n^2\
b &= 2mn\
c &= m^2 + n^2
end{align}tag{4}
$$
gives all positive, relatively prime $a$, $b$, and $c$ so that
$$
a^2 + b^2 = c^2tag{5}
$$
Proof: $(5)Rightarrow(4):$
Suppose $a$, $b$, and $c$ are positive, relatively prime, and $a^2 + b^2 = c^2$.
Because $(2k)^2 = 4k^2$ and $(2k+1)^2 = 4(k+1)k + 1$, the square of an even
integer must be $0 bmod{4}$ and the square of an odd integer must be $1 bmod{4}$.
At least one of $a$ and $b$ must be odd; otherwise $a$, $b$, and $c$ would share a
common factor of $2$. If both are odd, then $c^2$ would need to be $2 bmod{4}$,
which is impossible. Thus, one must be even and the other must be odd. This means that $c$ must be odd. Without loss of generality, let $b$ be even.
Let $M = (c+a)/2$ and $N = (c-a)/2$. Then
$$
begin{align}
a &= M - Ntag{6}\
c &= M + Ntag{7}\
b^2 &= 4MNtag{8}
end{align}
$$
Thus, we have that $M gt N gt 0$ and one of $M$ and $N$ must be even and the othermust be odd. Furthermore, $gcd(M,N)$ divides $a$, $b$, and $c$; thus, $gcd(M,N) = 1$. Since $b^2 = 4MN$ and $gcd(M,N) = 1$, both $M$ and $N$ must be perfect squares. Let $M = m^2$ and $N = n^2$, where $m$ and $n$ are positive; then, $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied.
$(4)Rightarrow(5):$
Suppose $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. Then $(5)$ is satisfied:
$$
begin{align}
a^2 + b^2
&= (m^2 - n^2)^2 + (2mn)^2\
&= m^4 - 2 m^2 n^2 + n^4 + 4 m^2 n^2\
&= m^4 + 2 m^2 n^2 + n^4\
&= (m^2 + n^2)^2\
&= c^2tag{9}
end{align}
$$
Furthermore, $a$ and $b$ are relatively prime since
$$
begin{align}
gcd(a,b)
&= gcd(m^2-n^2,2mn)\
&:mid:gcd(m-n,2) gcd(m-n,m) gcd(m-n,n)\
×gcd(m+n,2) gcd(m+n,m) gcd(m+n,n)\
&=gcd(m+n,2)^2 gcd(n,m)^4\
&= 1tag{10}
end{align}\
$$
$square$
answered Jun 18 '13 at 0:34
robjohn♦robjohn
268k27309634
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add a comment |
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The difficulty seems to be that if you specify integer lengths for the two legs, or integer lengths for the hypotenuse and one of the legs, the remaining side will most often have irrational length. If one starts in this way, it is perhaps not immediately obvious how to do the number-theoretic analysis to determine the conditions under which the quantity under the square-root sign is a perfect square.
Interestingly, there is a slightly modified geometry problem in which specifying rational lengths for two of three parameters automatically leads to the third also being rational. This sort of problem has a long history, going back to ancient Mesopotamia and Egypt. See Poles and walls in Mesopotamia and Egypt by Duncan J. Melville and page 307 of Methods and traditions of Babylonian mathematics: Plimpton 322, Pythagorean triples, and the Babylonian triangle parameter equations by Jöran Friberg.
Here's one version: a vertical cane of unknown length sits against a wall. The cane slips so that the top of the cane moves down the wall a distance $u$ and the bottom of the cane moves out (perpendicularly) from the wall a distance $a$. How long is the cane?
Let the length of the cane be $c$. In its new position, the cane, the wall, and the ground form a right triangle with legs $a$ and $b=c-u$ and hypotenuse $c$. Now
$$
c^2=a^2+(c-u)^2=a^2+c^2-2cu+u^2
$$
so that $2cu=a^2+u^2$ and
$$
c=frac{a^2+u^2}{2u}.
$$
Notice that we have computed $c$ without taking any square roots. This means that if we choose the lengths $a$ and $u$ to be rational, then so are $c$ and $c-u$. Therefore we get rational Pythagorean triples this way:
$$
(a,b,c)=(a,c-u,c)=left(a,frac{a^2-u^2}{2u},frac{a^2+u^2}{2u}right).
$$
If we want integer triples, we can multiply through by $2u$ and require $a$ and $u$ to be integer. Now that we have a form that gives integer triples, we can, if we wish, investigate whether integer triples always have this form and go on to analyze the conditions under which such triples are primitive.
answered Jan 20 at 18:51
Will OrrickWill Orrick
13.7k13461
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are a number of ways to find those expressions organically. One way is via complex numbers (and this method can be found in the first chapter of Marcus' Number Fields).
Suppose $x^2 + y^2 = z^2$, with $x,y,zinBbb N$, and also suppose that $x$, $y$, and $z$ have no common factor. Looking mod $4$, we can tell that $z$ must be odd: the only squares mod $4$ are $0$ and $1$, so the only possible equations are $0^2 + 0^2equiv 0^2mod 4$ and $0^2 + 1^2 equiv 1^2mod 4$. But in the first case, everything is divisible by $4$, so we don't have a primitive triple.
Proceeding, we can factor the left hand side and get $(x + iy)(x - iy) = z^2$. So, we now have a problem concerning $Bbb Z[i] = {a + bimid a,binBbb Z}$. One can show that $Bbb Z[i]$ has unique factorization of elements into primes (for a precise definition of all this, look to any abstract algebra text), so suppose some prime $p$ divides $x + iy$. $p$ clearly divides $z^2$ an even number of times, so we want to show that $p$ does not divide $x - iy$. If it did, we would have $pmid (x + yi) + (x - yi)$, or $pmid 2x$. Since $p$ divides $z$ as well, and $x$ and $z$ are relatively prime, we know that we can find $n,minBbb{Z}$ such that $2xm + zn = 1$ (recall that $z$ is odd). However, this implies that $p$ divides $1$, but the only elements dividing $1$ in $Bbb{Z}[i]$ are $pm 1$, $pm i$, none of which are primes - so we have a contradiction. Therefore, by unique factorization, we must have $x + iy = ualpha^2$, where $uin{pm1,pm i}$. Writing $alpha = m + in$, we have $alpha^2 = m^2 - n^2 + i2mn $, so $x = pm left(m^2 - n^2right)$, $y = pm 2mn$, and solving for $z$, we get $z = pmleft(m^2 + n^2right)$.
If $m$ and $n$ aren't coprime, we don't obtain a primitive triple, as each of $x$, $y$, and $z$ will have a factor in common.
There are numerous other ways of deriving these ways of representing $x$, $y$, and $z$ as well, some of which include parameterizing the unit circle or moving $y^2$ to the other side and writing $x^2 = (z + y)(z - y)$.
answered Jun 17 '13 at 23:58
StahlStahl
16.7k43455
$endgroup$
$begingroup$
This answered my follow up before I asked it, thanks!
$endgroup$
– frogeyedpeas
Jun 18 '13 at 0:27
add a comment |
$begingroup$
There are a number of ways to find those expressions organically. One way is via complex numbers (and this method can be found in the first chapter of Marcus' Number Fields).
Suppose $x^2 + y^2 = z^2$, with $x,y,zinBbb N$, and also suppose that $x$, $y$, and $z$ have no common factor. Looking mod $4$, we can tell that $z$ must be odd: the only squares mod $4$ are $0$ and $1$, so the only possible equations are $0^2 + 0^2equiv 0^2mod 4$ and $0^2 + 1^2 equiv 1^2mod 4$. But in the first case, everything is divisible by $4$, so we don't have a primitive triple.
Proceeding, we can factor the left hand side and get $(x + iy)(x - iy) = z^2$. So, we now have a problem concerning $Bbb Z[i] = {a + bimid a,binBbb Z}$. One can show that $Bbb Z[i]$ has unique factorization of elements into primes (for a precise definition of all this, look to any abstract algebra text), so suppose some prime $p$ divides $x + iy$. $p$ clearly divides $z^2$ an even number of times, so we want to show that $p$ does not divide $x - iy$. If it did, we would have $pmid (x + yi) + (x - yi)$, or $pmid 2x$. Since $p$ divides $z$ as well, and $x$ and $z$ are relatively prime, we know that we can find $n,minBbb{Z}$ such that $2xm + zn = 1$ (recall that $z$ is odd). However, this implies that $p$ divides $1$, but the only elements dividing $1$ in $Bbb{Z}[i]$ are $pm 1$, $pm i$, none of which are primes - so we have a contradiction. Therefore, by unique factorization, we must have $x + iy = ualpha^2$, where $uin{pm1,pm i}$. Writing $alpha = m + in$, we have $alpha^2 = m^2 - n^2 + i2mn $, so $x = pm left(m^2 - n^2right)$, $y = pm 2mn$, and solving for $z$, we get $z = pmleft(m^2 + n^2right)$.
If $m$ and $n$ aren't coprime, we don't obtain a primitive triple, as each of $x$, $y$, and $z$ will have a factor in common.
There are numerous other ways of deriving these ways of representing $x$, $y$, and $z$ as well, some of which include parameterizing the unit circle or moving $y^2$ to the other side and writing $x^2 = (z + y)(z - y)$.
answered Jun 17 '13 at 23:58
StahlStahl
16.7k43455
$endgroup$
$begingroup$
This answered my follow up before I asked it, thanks!
$endgroup$
– frogeyedpeas
Jun 18 '13 at 0:27
add a comment |
$begingroup$
There are a number of ways to find those expressions organically. One way is via complex numbers (and this method can be found in the first chapter of Marcus' Number Fields).
Suppose $x^2 + y^2 = z^2$, with $x,y,zinBbb N$, and also suppose that $x$, $y$, and $z$ have no common factor. Looking mod $4$, we can tell that $z$ must be odd: the only squares mod $4$ are $0$ and $1$, so the only possible equations are $0^2 + 0^2equiv 0^2mod 4$ and $0^2 + 1^2 equiv 1^2mod 4$. But in the first case, everything is divisible by $4$, so we don't have a primitive triple.
Proceeding, we can factor the left hand side and get $(x + iy)(x - iy) = z^2$. So, we now have a problem concerning $Bbb Z[i] = {a + bimid a,binBbb Z}$. One can show that $Bbb Z[i]$ has unique factorization of elements into primes (for a precise definition of all this, look to any abstract algebra text), so suppose some prime $p$ divides $x + iy$. $p$ clearly divides $z^2$ an even number of times, so we want to show that $p$ does not divide $x - iy$. If it did, we would have $pmid (x + yi) + (x - yi)$, or $pmid 2x$. Since $p$ divides $z$ as well, and $x$ and $z$ are relatively prime, we know that we can find $n,minBbb{Z}$ such that $2xm + zn = 1$ (recall that $z$ is odd). However, this implies that $p$ divides $1$, but the only elements dividing $1$ in $Bbb{Z}[i]$ are $pm 1$, $pm i$, none of which are primes - so we have a contradiction. Therefore, by unique factorization, we must have $x + iy = ualpha^2$, where $uin{pm1,pm i}$. Writing $alpha = m + in$, we have $alpha^2 = m^2 - n^2 + i2mn $, so $x = pm left(m^2 - n^2right)$, $y = pm 2mn$, and solving for $z$, we get $z = pmleft(m^2 + n^2right)$.
If $m$ and $n$ aren't coprime, we don't obtain a primitive triple, as each of $x$, $y$, and $z$ will have a factor in common.
There are numerous other ways of deriving these ways of representing $x$, $y$, and $z$ as well, some of which include parameterizing the unit circle or moving $y^2$ to the other side and writing $x^2 = (z + y)(z - y)$.
answered Jun 17 '13 at 23:58
StahlStahl
16.7k43455
$endgroup$
There are a number of ways to find those expressions organically. One way is via complex numbers (and this method can be found in the first chapter of Marcus' Number Fields).
Suppose $x^2 + y^2 = z^2$, with $x,y,zinBbb N$, and also suppose that $x$, $y$, and $z$ have no common factor. Looking mod $4$, we can tell that $z$ must be odd: the only squares mod $4$ are $0$ and $1$, so the only possible equations are $0^2 + 0^2equiv 0^2mod 4$ and $0^2 + 1^2 equiv 1^2mod 4$. But in the first case, everything is divisible by $4$, so we don't have a primitive triple.
Proceeding, we can factor the left hand side and get $(x + iy)(x - iy) = z^2$. So, we now have a problem concerning $Bbb Z[i] = {a + bimid a,binBbb Z}$. One can show that $Bbb Z[i]$ has unique factorization of elements into primes (for a precise definition of all this, look to any abstract algebra text), so suppose some prime $p$ divides $x + iy$. $p$ clearly divides $z^2$ an even number of times, so we want to show that $p$ does not divide $x - iy$. If it did, we would have $pmid (x + yi) + (x - yi)$, or $pmid 2x$. Since $p$ divides $z$ as well, and $x$ and $z$ are relatively prime, we know that we can find $n,minBbb{Z}$ such that $2xm + zn = 1$ (recall that $z$ is odd). However, this implies that $p$ divides $1$, but the only elements dividing $1$ in $Bbb{Z}[i]$ are $pm 1$, $pm i$, none of which are primes - so we have a contradiction. Therefore, by unique factorization, we must have $x + iy = ualpha^2$, where $uin{pm1,pm i}$. Writing $alpha = m + in$, we have $alpha^2 = m^2 - n^2 + i2mn $, so $x = pm left(m^2 - n^2right)$, $y = pm 2mn$, and solving for $z$, we get $z = pmleft(m^2 + n^2right)$.
If $m$ and $n$ aren't coprime, we don't obtain a primitive triple, as each of $x$, $y$, and $z$ will have a factor in common.
There are numerous other ways of deriving these ways of representing $x$, $y$, and $z$ as well, some of which include parameterizing the unit circle or moving $y^2$ to the other side and writing $x^2 = (z + y)(z - y)$.
answered Jun 17 '13 at 23:58
StahlStahl
16.7k43455
answered Jun 17 '13 at 23:58
StahlStahl
16.7k43455
answered Jun 17 '13 at 23:58
StahlStahl
16.7k43455
answered Jun 17 '13 at 23:58
StahlStahl
16.7k43455
16.7k43455
$begingroup$
This answered my follow up before I asked it, thanks!
$endgroup$
– frogeyedpeas
Jun 18 '13 at 0:27
add a comment |
$begingroup$
This answered my follow up before I asked it, thanks!
$endgroup$
– frogeyedpeas
Jun 18 '13 at 0:27
$begingroup$
This answered my follow up before I asked it, thanks!
$endgroup$
– frogeyedpeas
Jun 18 '13 at 0:27
$begingroup$
This answered my follow up before I asked it, thanks!
$endgroup$
– frogeyedpeas
Jun 18 '13 at 0:27
add a comment |
$begingroup$
Consider the intersection of the line $y=t(x+1)$ with the circle $x^2+y^2=1$. We get points on the circle
$$
(x,y)=left(frac{1-t^2}{1+t^2},frac{2t}{1+t^2}right).
$$
Let $t=m/n$ be rational, plug into $x^2+y^2$ and clear denominators to get
$$
(n^2-m^2)^2+(2mn)^2=(n^2+m^2)^2
$$
answered Jun 17 '13 at 23:58
yoyoyoyo
6,5991726
$endgroup$
$begingroup$
magic step is imagining the line $y=t(x+1)$
$endgroup$
– karakfa
Apr 22 '17 at 17:43
add a comment |
$begingroup$
Consider the intersection of the line $y=t(x+1)$ with the circle $x^2+y^2=1$. We get points on the circle
$$
(x,y)=left(frac{1-t^2}{1+t^2},frac{2t}{1+t^2}right).
$$
Let $t=m/n$ be rational, plug into $x^2+y^2$ and clear denominators to get
$$
(n^2-m^2)^2+(2mn)^2=(n^2+m^2)^2
$$
answered Jun 17 '13 at 23:58
yoyoyoyo
6,5991726
$endgroup$
$begingroup$
magic step is imagining the line $y=t(x+1)$
$endgroup$
– karakfa
Apr 22 '17 at 17:43
add a comment |
$begingroup$
Consider the intersection of the line $y=t(x+1)$ with the circle $x^2+y^2=1$. We get points on the circle
$$
(x,y)=left(frac{1-t^2}{1+t^2},frac{2t}{1+t^2}right).
$$
Let $t=m/n$ be rational, plug into $x^2+y^2$ and clear denominators to get
$$
(n^2-m^2)^2+(2mn)^2=(n^2+m^2)^2
$$
answered Jun 17 '13 at 23:58
yoyoyoyo
6,5991726
$endgroup$
Consider the intersection of the line $y=t(x+1)$ with the circle $x^2+y^2=1$. We get points on the circle
$$
(x,y)=left(frac{1-t^2}{1+t^2},frac{2t}{1+t^2}right).
$$
Let $t=m/n$ be rational, plug into $x^2+y^2$ and clear denominators to get
$$
(n^2-m^2)^2+(2mn)^2=(n^2+m^2)^2
$$
answered Jun 17 '13 at 23:58
yoyoyoyo
6,5991726
edited Apr 24 '17 at 16:57
answered Jun 17 '13 at 23:58
yoyoyoyo
6,5991726
answered Jun 17 '13 at 23:58
yoyoyoyo
6,5991726
answered Jun 17 '13 at 23:58
yoyoyoyo
6,5991726
6,5991726
$begingroup$
magic step is imagining the line $y=t(x+1)$
$endgroup$
– karakfa
Apr 22 '17 at 17:43
add a comment |
$begingroup$
magic step is imagining the line $y=t(x+1)$
$endgroup$
– karakfa
Apr 22 '17 at 17:43
$begingroup$
magic step is imagining the line $y=t(x+1)$
$endgroup$
– karakfa
Apr 22 '17 at 17:43
$begingroup$
magic step is imagining the line $y=t(x+1)$
$endgroup$
– karakfa
Apr 22 '17 at 17:43
add a comment |
$begingroup$
Here is the way to generate all relatively prime pythagorean triples:
Theorem: Let $m$ and $n$ be positive integers so that
$$
begin{align}
&mgt ntag{1}\
&m+ntext{ is odd}tag{2}\
&mtext{ and }ntext{ are relatively prime}tag{3}
end{align}
$$
Then,
$$
begin{align}
a &= m^2 - n^2\
b &= 2mn\
c &= m^2 + n^2
end{align}tag{4}
$$
gives all positive, relatively prime $a$, $b$, and $c$ so that
$$
a^2 + b^2 = c^2tag{5}
$$
Proof: $(5)Rightarrow(4):$
Suppose $a$, $b$, and $c$ are positive, relatively prime, and $a^2 + b^2 = c^2$.
Because $(2k)^2 = 4k^2$ and $(2k+1)^2 = 4(k+1)k + 1$, the square of an even
integer must be $0 bmod{4}$ and the square of an odd integer must be $1 bmod{4}$.
At least one of $a$ and $b$ must be odd; otherwise $a$, $b$, and $c$ would share a
common factor of $2$. If both are odd, then $c^2$ would need to be $2 bmod{4}$,
which is impossible. Thus, one must be even and the other must be odd. This means that $c$ must be odd. Without loss of generality, let $b$ be even.
Let $M = (c+a)/2$ and $N = (c-a)/2$. Then
$$
begin{align}
a &= M - Ntag{6}\
c &= M + Ntag{7}\
b^2 &= 4MNtag{8}
end{align}
$$
Thus, we have that $M gt N gt 0$ and one of $M$ and $N$ must be even and the othermust be odd. Furthermore, $gcd(M,N)$ divides $a$, $b$, and $c$; thus, $gcd(M,N) = 1$. Since $b^2 = 4MN$ and $gcd(M,N) = 1$, both $M$ and $N$ must be perfect squares. Let $M = m^2$ and $N = n^2$, where $m$ and $n$ are positive; then, $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied.
$(4)Rightarrow(5):$
Suppose $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. Then $(5)$ is satisfied:
$$
begin{align}
a^2 + b^2
&= (m^2 - n^2)^2 + (2mn)^2\
&= m^4 - 2 m^2 n^2 + n^4 + 4 m^2 n^2\
&= m^4 + 2 m^2 n^2 + n^4\
&= (m^2 + n^2)^2\
&= c^2tag{9}
end{align}
$$
Furthermore, $a$ and $b$ are relatively prime since
$$
begin{align}
gcd(a,b)
&= gcd(m^2-n^2,2mn)\
&:mid:gcd(m-n,2) gcd(m-n,m) gcd(m-n,n)\
×gcd(m+n,2) gcd(m+n,m) gcd(m+n,n)\
&=gcd(m+n,2)^2 gcd(n,m)^4\
&= 1tag{10}
end{align}\
$$
$square$
answered Jun 18 '13 at 0:34
robjohn♦robjohn
268k27309634
$endgroup$
add a comment |
$begingroup$
Here is the way to generate all relatively prime pythagorean triples:
Theorem: Let $m$ and $n$ be positive integers so that
$$
begin{align}
&mgt ntag{1}\
&m+ntext{ is odd}tag{2}\
&mtext{ and }ntext{ are relatively prime}tag{3}
end{align}
$$
Then,
$$
begin{align}
a &= m^2 - n^2\
b &= 2mn\
c &= m^2 + n^2
end{align}tag{4}
$$
gives all positive, relatively prime $a$, $b$, and $c$ so that
$$
a^2 + b^2 = c^2tag{5}
$$
Proof: $(5)Rightarrow(4):$
Suppose $a$, $b$, and $c$ are positive, relatively prime, and $a^2 + b^2 = c^2$.
Because $(2k)^2 = 4k^2$ and $(2k+1)^2 = 4(k+1)k + 1$, the square of an even
integer must be $0 bmod{4}$ and the square of an odd integer must be $1 bmod{4}$.
At least one of $a$ and $b$ must be odd; otherwise $a$, $b$, and $c$ would share a
common factor of $2$. If both are odd, then $c^2$ would need to be $2 bmod{4}$,
which is impossible. Thus, one must be even and the other must be odd. This means that $c$ must be odd. Without loss of generality, let $b$ be even.
Let $M = (c+a)/2$ and $N = (c-a)/2$. Then
$$
begin{align}
a &= M - Ntag{6}\
c &= M + Ntag{7}\
b^2 &= 4MNtag{8}
end{align}
$$
Thus, we have that $M gt N gt 0$ and one of $M$ and $N$ must be even and the othermust be odd. Furthermore, $gcd(M,N)$ divides $a$, $b$, and $c$; thus, $gcd(M,N) = 1$. Since $b^2 = 4MN$ and $gcd(M,N) = 1$, both $M$ and $N$ must be perfect squares. Let $M = m^2$ and $N = n^2$, where $m$ and $n$ are positive; then, $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied.
$(4)Rightarrow(5):$
Suppose $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. Then $(5)$ is satisfied:
$$
begin{align}
a^2 + b^2
&= (m^2 - n^2)^2 + (2mn)^2\
&= m^4 - 2 m^2 n^2 + n^4 + 4 m^2 n^2\
&= m^4 + 2 m^2 n^2 + n^4\
&= (m^2 + n^2)^2\
&= c^2tag{9}
end{align}
$$
Furthermore, $a$ and $b$ are relatively prime since
$$
begin{align}
gcd(a,b)
&= gcd(m^2-n^2,2mn)\
&:mid:gcd(m-n,2) gcd(m-n,m) gcd(m-n,n)\
×gcd(m+n,2) gcd(m+n,m) gcd(m+n,n)\
&=gcd(m+n,2)^2 gcd(n,m)^4\
&= 1tag{10}
end{align}\
$$
$square$
answered Jun 18 '13 at 0:34
robjohn♦robjohn
268k27309634
$endgroup$
add a comment |
$begingroup$
Here is the way to generate all relatively prime pythagorean triples:
Theorem: Let $m$ and $n$ be positive integers so that
$$
begin{align}
&mgt ntag{1}\
&m+ntext{ is odd}tag{2}\
&mtext{ and }ntext{ are relatively prime}tag{3}
end{align}
$$
Then,
$$
begin{align}
a &= m^2 - n^2\
b &= 2mn\
c &= m^2 + n^2
end{align}tag{4}
$$
gives all positive, relatively prime $a$, $b$, and $c$ so that
$$
a^2 + b^2 = c^2tag{5}
$$
Proof: $(5)Rightarrow(4):$
Suppose $a$, $b$, and $c$ are positive, relatively prime, and $a^2 + b^2 = c^2$.
Because $(2k)^2 = 4k^2$ and $(2k+1)^2 = 4(k+1)k + 1$, the square of an even
integer must be $0 bmod{4}$ and the square of an odd integer must be $1 bmod{4}$.
At least one of $a$ and $b$ must be odd; otherwise $a$, $b$, and $c$ would share a
common factor of $2$. If both are odd, then $c^2$ would need to be $2 bmod{4}$,
which is impossible. Thus, one must be even and the other must be odd. This means that $c$ must be odd. Without loss of generality, let $b$ be even.
Let $M = (c+a)/2$ and $N = (c-a)/2$. Then
$$
begin{align}
a &= M - Ntag{6}\
c &= M + Ntag{7}\
b^2 &= 4MNtag{8}
end{align}
$$
Thus, we have that $M gt N gt 0$ and one of $M$ and $N$ must be even and the othermust be odd. Furthermore, $gcd(M,N)$ divides $a$, $b$, and $c$; thus, $gcd(M,N) = 1$. Since $b^2 = 4MN$ and $gcd(M,N) = 1$, both $M$ and $N$ must be perfect squares. Let $M = m^2$ and $N = n^2$, where $m$ and $n$ are positive; then, $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied.
$(4)Rightarrow(5):$
Suppose $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. Then $(5)$ is satisfied:
$$
begin{align}
a^2 + b^2
&= (m^2 - n^2)^2 + (2mn)^2\
&= m^4 - 2 m^2 n^2 + n^4 + 4 m^2 n^2\
&= m^4 + 2 m^2 n^2 + n^4\
&= (m^2 + n^2)^2\
&= c^2tag{9}
end{align}
$$
Furthermore, $a$ and $b$ are relatively prime since
$$
begin{align}
gcd(a,b)
&= gcd(m^2-n^2,2mn)\
&:mid:gcd(m-n,2) gcd(m-n,m) gcd(m-n,n)\
×gcd(m+n,2) gcd(m+n,m) gcd(m+n,n)\
&=gcd(m+n,2)^2 gcd(n,m)^4\
&= 1tag{10}
end{align}\
$$
$square$
answered Jun 18 '13 at 0:34
robjohn♦robjohn
268k27309634
$endgroup$
Here is the way to generate all relatively prime pythagorean triples:
Theorem: Let $m$ and $n$ be positive integers so that
$$
begin{align}
&mgt ntag{1}\
&m+ntext{ is odd}tag{2}\
&mtext{ and }ntext{ are relatively prime}tag{3}
end{align}
$$
Then,
$$
begin{align}
a &= m^2 - n^2\
b &= 2mn\
c &= m^2 + n^2
end{align}tag{4}
$$
gives all positive, relatively prime $a$, $b$, and $c$ so that
$$
a^2 + b^2 = c^2tag{5}
$$
Proof: $(5)Rightarrow(4):$
Suppose $a$, $b$, and $c$ are positive, relatively prime, and $a^2 + b^2 = c^2$.
Because $(2k)^2 = 4k^2$ and $(2k+1)^2 = 4(k+1)k + 1$, the square of an even
integer must be $0 bmod{4}$ and the square of an odd integer must be $1 bmod{4}$.
At least one of $a$ and $b$ must be odd; otherwise $a$, $b$, and $c$ would share a
common factor of $2$. If both are odd, then $c^2$ would need to be $2 bmod{4}$,
which is impossible. Thus, one must be even and the other must be odd. This means that $c$ must be odd. Without loss of generality, let $b$ be even.
Let $M = (c+a)/2$ and $N = (c-a)/2$. Then
$$
begin{align}
a &= M - Ntag{6}\
c &= M + Ntag{7}\
b^2 &= 4MNtag{8}
end{align}
$$
Thus, we have that $M gt N gt 0$ and one of $M$ and $N$ must be even and the othermust be odd. Furthermore, $gcd(M,N)$ divides $a$, $b$, and $c$; thus, $gcd(M,N) = 1$. Since $b^2 = 4MN$ and $gcd(M,N) = 1$, both $M$ and $N$ must be perfect squares. Let $M = m^2$ and $N = n^2$, where $m$ and $n$ are positive; then, $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied.
$(4)Rightarrow(5):$
Suppose $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. Then $(5)$ is satisfied:
$$
begin{align}
a^2 + b^2
&= (m^2 - n^2)^2 + (2mn)^2\
&= m^4 - 2 m^2 n^2 + n^4 + 4 m^2 n^2\
&= m^4 + 2 m^2 n^2 + n^4\
&= (m^2 + n^2)^2\
&= c^2tag{9}
end{align}
$$
Furthermore, $a$ and $b$ are relatively prime since
$$
begin{align}
gcd(a,b)
&= gcd(m^2-n^2,2mn)\
&:mid:gcd(m-n,2) gcd(m-n,m) gcd(m-n,n)\
×gcd(m+n,2) gcd(m+n,m) gcd(m+n,n)\
&=gcd(m+n,2)^2 gcd(n,m)^4\
&= 1tag{10}
end{align}\
$$
$square$
answered Jun 18 '13 at 0:34
robjohn♦robjohn
268k27309634
answered Jun 18 '13 at 0:34
robjohn♦robjohn
268k27309634
answered Jun 18 '13 at 0:34
robjohn♦robjohn
268k27309634
answered Jun 18 '13 at 0:34
robjohn♦robjohn
268k27309634
268k27309634
add a comment |
add a comment |
$begingroup$
The difficulty seems to be that if you specify integer lengths for the two legs, or integer lengths for the hypotenuse and one of the legs, the remaining side will most often have irrational length. If one starts in this way, it is perhaps not immediately obvious how to do the number-theoretic analysis to determine the conditions under which the quantity under the square-root sign is a perfect square.
Interestingly, there is a slightly modified geometry problem in which specifying rational lengths for two of three parameters automatically leads to the third also being rational. This sort of problem has a long history, going back to ancient Mesopotamia and Egypt. See Poles and walls in Mesopotamia and Egypt by Duncan J. Melville and page 307 of Methods and traditions of Babylonian mathematics: Plimpton 322, Pythagorean triples, and the Babylonian triangle parameter equations by Jöran Friberg.
Here's one version: a vertical cane of unknown length sits against a wall. The cane slips so that the top of the cane moves down the wall a distance $u$ and the bottom of the cane moves out (perpendicularly) from the wall a distance $a$. How long is the cane?
Let the length of the cane be $c$. In its new position, the cane, the wall, and the ground form a right triangle with legs $a$ and $b=c-u$ and hypotenuse $c$. Now
$$
c^2=a^2+(c-u)^2=a^2+c^2-2cu+u^2
$$
so that $2cu=a^2+u^2$ and
$$
c=frac{a^2+u^2}{2u}.
$$
Notice that we have computed $c$ without taking any square roots. This means that if we choose the lengths $a$ and $u$ to be rational, then so are $c$ and $c-u$. Therefore we get rational Pythagorean triples this way:
$$
(a,b,c)=(a,c-u,c)=left(a,frac{a^2-u^2}{2u},frac{a^2+u^2}{2u}right).
$$
If we want integer triples, we can multiply through by $2u$ and require $a$ and $u$ to be integer. Now that we have a form that gives integer triples, we can, if we wish, investigate whether integer triples always have this form and go on to analyze the conditions under which such triples are primitive.
answered Jan 20 at 18:51
Will OrrickWill Orrick
13.7k13461
$endgroup$
add a comment |
$begingroup$
The difficulty seems to be that if you specify integer lengths for the two legs, or integer lengths for the hypotenuse and one of the legs, the remaining side will most often have irrational length. If one starts in this way, it is perhaps not immediately obvious how to do the number-theoretic analysis to determine the conditions under which the quantity under the square-root sign is a perfect square.
Interestingly, there is a slightly modified geometry problem in which specifying rational lengths for two of three parameters automatically leads to the third also being rational. This sort of problem has a long history, going back to ancient Mesopotamia and Egypt. See Poles and walls in Mesopotamia and Egypt by Duncan J. Melville and page 307 of Methods and traditions of Babylonian mathematics: Plimpton 322, Pythagorean triples, and the Babylonian triangle parameter equations by Jöran Friberg.
Here's one version: a vertical cane of unknown length sits against a wall. The cane slips so that the top of the cane moves down the wall a distance $u$ and the bottom of the cane moves out (perpendicularly) from the wall a distance $a$. How long is the cane?
Let the length of the cane be $c$. In its new position, the cane, the wall, and the ground form a right triangle with legs $a$ and $b=c-u$ and hypotenuse $c$. Now
$$
c^2=a^2+(c-u)^2=a^2+c^2-2cu+u^2
$$
so that $2cu=a^2+u^2$ and
$$
c=frac{a^2+u^2}{2u}.
$$
Notice that we have computed $c$ without taking any square roots. This means that if we choose the lengths $a$ and $u$ to be rational, then so are $c$ and $c-u$. Therefore we get rational Pythagorean triples this way:
$$
(a,b,c)=(a,c-u,c)=left(a,frac{a^2-u^2}{2u},frac{a^2+u^2}{2u}right).
$$
If we want integer triples, we can multiply through by $2u$ and require $a$ and $u$ to be integer. Now that we have a form that gives integer triples, we can, if we wish, investigate whether integer triples always have this form and go on to analyze the conditions under which such triples are primitive.
answered Jan 20 at 18:51
Will OrrickWill Orrick
13.7k13461
$endgroup$
add a comment |
$begingroup$
The difficulty seems to be that if you specify integer lengths for the two legs, or integer lengths for the hypotenuse and one of the legs, the remaining side will most often have irrational length. If one starts in this way, it is perhaps not immediately obvious how to do the number-theoretic analysis to determine the conditions under which the quantity under the square-root sign is a perfect square.
Interestingly, there is a slightly modified geometry problem in which specifying rational lengths for two of three parameters automatically leads to the third also being rational. This sort of problem has a long history, going back to ancient Mesopotamia and Egypt. See Poles and walls in Mesopotamia and Egypt by Duncan J. Melville and page 307 of Methods and traditions of Babylonian mathematics: Plimpton 322, Pythagorean triples, and the Babylonian triangle parameter equations by Jöran Friberg.
Here's one version: a vertical cane of unknown length sits against a wall. The cane slips so that the top of the cane moves down the wall a distance $u$ and the bottom of the cane moves out (perpendicularly) from the wall a distance $a$. How long is the cane?
Let the length of the cane be $c$. In its new position, the cane, the wall, and the ground form a right triangle with legs $a$ and $b=c-u$ and hypotenuse $c$. Now
$$
c^2=a^2+(c-u)^2=a^2+c^2-2cu+u^2
$$
so that $2cu=a^2+u^2$ and
$$
c=frac{a^2+u^2}{2u}.
$$
Notice that we have computed $c$ without taking any square roots. This means that if we choose the lengths $a$ and $u$ to be rational, then so are $c$ and $c-u$. Therefore we get rational Pythagorean triples this way:
$$
(a,b,c)=(a,c-u,c)=left(a,frac{a^2-u^2}{2u},frac{a^2+u^2}{2u}right).
$$
If we want integer triples, we can multiply through by $2u$ and require $a$ and $u$ to be integer. Now that we have a form that gives integer triples, we can, if we wish, investigate whether integer triples always have this form and go on to analyze the conditions under which such triples are primitive.
answered Jan 20 at 18:51
Will OrrickWill Orrick
13.7k13461
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The difficulty seems to be that if you specify integer lengths for the two legs, or integer lengths for the hypotenuse and one of the legs, the remaining side will most often have irrational length. If one starts in this way, it is perhaps not immediately obvious how to do the number-theoretic analysis to determine the conditions under which the quantity under the square-root sign is a perfect square.
Interestingly, there is a slightly modified geometry problem in which specifying rational lengths for two of three parameters automatically leads to the third also being rational. This sort of problem has a long history, going back to ancient Mesopotamia and Egypt. See Poles and walls in Mesopotamia and Egypt by Duncan J. Melville and page 307 of Methods and traditions of Babylonian mathematics: Plimpton 322, Pythagorean triples, and the Babylonian triangle parameter equations by Jöran Friberg.
Here's one version: a vertical cane of unknown length sits against a wall. The cane slips so that the top of the cane moves down the wall a distance $u$ and the bottom of the cane moves out (perpendicularly) from the wall a distance $a$. How long is the cane?
Let the length of the cane be $c$. In its new position, the cane, the wall, and the ground form a right triangle with legs $a$ and $b=c-u$ and hypotenuse $c$. Now
$$
c^2=a^2+(c-u)^2=a^2+c^2-2cu+u^2
$$
so that $2cu=a^2+u^2$ and
$$
c=frac{a^2+u^2}{2u}.
$$
Notice that we have computed $c$ without taking any square roots. This means that if we choose the lengths $a$ and $u$ to be rational, then so are $c$ and $c-u$. Therefore we get rational Pythagorean triples this way:
$$
(a,b,c)=(a,c-u,c)=left(a,frac{a^2-u^2}{2u},frac{a^2+u^2}{2u}right).
$$
If we want integer triples, we can multiply through by $2u$ and require $a$ and $u$ to be integer. Now that we have a form that gives integer triples, we can, if we wish, investigate whether integer triples always have this form and go on to analyze the conditions under which such triples are primitive.
answered Jan 20 at 18:51
Will OrrickWill Orrick
13.7k13461
answered Jan 20 at 18:51
Will OrrickWill Orrick
13.7k13461
answered Jan 20 at 18:51
Will OrrickWill Orrick
13.7k13461
answered Jan 20 at 18:51
Will OrrickWill Orrick
13.7k13461
13.7k13461
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$begingroup$
That's not a generating formula for all triples. If we make restriction on $m$ and $n$ (relatively prime, opposite parity, $m gt n$) we get all primitive triples. To see don't get all triples, we can't get $9,12,15$.
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– André Nicolas
Jun 17 '13 at 23:31
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find a rational parameterization of the unit circle.
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– yoyo
Jun 17 '13 at 23:41
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@André: That's the result of primitive triplets ${3,4,5 }$, right?
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– Inceptio
Jun 18 '13 at 7:46
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Yes. The details are nicely done in the answer by robjohn.
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– André Nicolas
Jun 18 '13 at 10:56
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The hardest part for me is the values of the variables equaling.
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– Roddy MacPhee
Feb 23 at 13:50