Mixing
$a + b = c + c$
$begingroup$
So I have the following problem: $a + b = c + c.
I want to prove that the equation has infinitely many relatively prime integer solutions.
What I did first was factor the right side to get:
(
algebra-precalculus number-theory prime-numbers diophantine-equations
asked Feb 2 at 18:51
user604720user604720
558
$endgroup$
add a comment |
$begingroup$
So I have the following problem: $a + b = c + c.
I want to prove that the equation has infinitely many relatively prime integer solutions.
What I did first was factor the right side to get:
(
algebra-precalculus number-theory prime-numbers diophantine-equations
asked Feb 2 at 18:51
user604720user604720
558
$endgroup$
2
$begingroup$
Is the question $a^2+b^2=c^5+c$ or $a^5+b^5=c^5+c$?
$endgroup$
– Peter Foreman
Feb 2 at 18:53
$begingroup$
Sorry I mistyped its a^2 + b^2 = c^5 + c
$endgroup$
– user604720
Feb 2 at 18:54
$begingroup$
If you know that primes congruent to $1$ mod $4$ can be written as sums of squares (and if you know there are infinitely many such primes), you should be good to go.
$endgroup$
– Barry Cipra
Feb 2 at 19:01
$begingroup$
Excuse me, but what is the unknown value?
$endgroup$
– El borito
Feb 2 at 19:05
$begingroup$
Come to think of it, it suffices to know (or show) that $c=5^n$ can always be written as the sum of two relatively prime squares.
$endgroup$
– Barry Cipra
Feb 2 at 19:10
add a comment |
$begingroup$
So I have the following problem: $a + b = c + c.
I want to prove that the equation has infinitely many relatively prime integer solutions.
What I did first was factor the right side to get:
(
algebra-precalculus number-theory prime-numbers diophantine-equations
asked Feb 2 at 18:51
user604720user604720
558
$endgroup$
So I have the following problem: $a + b = c + c.
I want to prove that the equation has infinitely many relatively prime integer solutions.
What I did first was factor the right side to get:
(
algebra-precalculus number-theory prime-numbers diophantine-equations
algebra-precalculus number-theory prime-numbers diophantine-equations
asked Feb 2 at 18:51
user604720user604720
558
asked Feb 2 at 18:51
user604720user604720
558
edited Feb 25 at 4:32
user604720
asked Feb 2 at 18:51
user604720user604720
558
asked Feb 2 at 18:51
user604720user604720
558
asked Feb 2 at 18:51
user604720user604720
558
558
2
$begingroup$
Is the question $a^2+b^2=c^5+c$ or $a^5+b^5=c^5+c$?
$endgroup$
– Peter Foreman
Feb 2 at 18:53
$begingroup$
Sorry I mistyped its a^2 + b^2 = c^5 + c
$endgroup$
– user604720
Feb 2 at 18:54
$begingroup$
If you know that primes congruent to $1$ mod $4$ can be written as sums of squares (and if you know there are infinitely many such primes), you should be good to go.
$endgroup$
– Barry Cipra
Feb 2 at 19:01
$begingroup$
Excuse me, but what is the unknown value?
$endgroup$
– El borito
Feb 2 at 19:05
$begingroup$
Come to think of it, it suffices to know (or show) that $c=5^n$ can always be written as the sum of two relatively prime squares.
$endgroup$
– Barry Cipra
Feb 2 at 19:10
add a comment |
2
$begingroup$
Is the question $a^2+b^2=c^5+c$ or $a^5+b^5=c^5+c$?
$endgroup$
– Peter Foreman
Feb 2 at 18:53
$begingroup$
Sorry I mistyped its a^2 + b^2 = c^5 + c
$endgroup$
– user604720
Feb 2 at 18:54
$begingroup$
If you know that primes congruent to $1$ mod $4$ can be written as sums of squares (and if you know there are infinitely many such primes), you should be good to go.
$endgroup$
– Barry Cipra
Feb 2 at 19:01
$begingroup$
Excuse me, but what is the unknown value?
$endgroup$
– El borito
Feb 2 at 19:05
$begingroup$
Come to think of it, it suffices to know (or show) that $c=5^n$ can always be written as the sum of two relatively prime squares.
$endgroup$
– Barry Cipra
Feb 2 at 19:10
2
2
$begingroup$
Is the question $a^2+b^2=c^5+c$ or $a^5+b^5=c^5+c$?
$endgroup$
– Peter Foreman
Feb 2 at 18:53
$begingroup$
Is the question $a^2+b^2=c^5+c$ or $a^5+b^5=c^5+c$?
$endgroup$
– Peter Foreman
Feb 2 at 18:53
$begingroup$
Sorry I mistyped its a^2 + b^2 = c^5 + c
$endgroup$
– user604720
Feb 2 at 18:54
$begingroup$
Sorry I mistyped its a^2 + b^2 = c^5 + c
$endgroup$
– user604720
Feb 2 at 18:54
$begingroup$
If you know that primes congruent to $1$ mod $4$ can be written as sums of squares (and if you know there are infinitely many such primes), you should be good to go.
$endgroup$
– Barry Cipra
Feb 2 at 19:01
$begingroup$
If you know that primes congruent to $1$ mod $4$ can be written as sums of squares (and if you know there are infinitely many such primes), you should be good to go.
$endgroup$
– Barry Cipra
Feb 2 at 19:01
$begingroup$
Excuse me, but what is the unknown value?
$endgroup$
– El borito
Feb 2 at 19:05
$begingroup$
Excuse me, but what is the unknown value?
$endgroup$
– El borito
Feb 2 at 19:05
$begingroup$
Come to think of it, it suffices to know (or show) that $c=5^n$ can always be written as the sum of two relatively prime squares.
$endgroup$
– Barry Cipra
Feb 2 at 19:10
$begingroup$
Come to think of it, it suffices to know (or show) that $c=5^n$ can always be written as the sum of two relatively prime squares.
$endgroup$
– Barry Cipra
Feb 2 at 19:10
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Suppose $c=u^2+v^2$ then you have $$(uc^2+v)^2+(vc^2-u)^2=c(c^4+1)$$ and you want $uc^2+v$ and $vc^2-u$ to be coprime.
Now just choose $u=1$ so that the expressions are $a=c^2+v$ and $b=vc^2-1$ with $c=v^2+1$
Note then that that a common factor of $a$ and $b$ is also a factor of $va-b=v^2+1=c$ and $ac^2-b=c^4+1$, but $c$ and $c^4+1$ are coprime.
So you get a family of solutions with
$c=v^2+1$ and
$a=c^2+v=v^4+2v^2+v+1$ and
$b=vc^2-1=v^5+2v^3+v-1$
[Which, I notice is just Sam's parametric solution with $u=1$. It should be obvious what to do to generalise.]
answered Feb 2 at 21:45
Mark BennetMark Bennet
82k984183
$endgroup$
$begingroup$
Hi, I need some help understanding. How did you generalize this?
$endgroup$
– user604720
Feb 11 at 0:52
$begingroup$
@user604720 Just keeping in the $u$ rather than setting $u=1$ and working through the logic gives the parametric family. Along the way you get $va-ub=c$ and $ac^2-b=u(c^4+1)$ and it is not so easy to prove coprime for the parametric solutions because of the factor $u$, but since this starts with an arbitrary $c=u^2+v^2$ one can pick $u$ and $v$ to make this obvious. $u=1$ is an easy choice.
$endgroup$
– Mark Bennet
Feb 11 at 6:51
add a comment |
$begingroup$
Above equation shown below has parametric solution:
$(a^2+b^2)=c(c^4+1)$
$a=u^5+uv^4+2u^3v^2+v$
$b=v^5+u^4v+2u^2v^3-u$
$c=u^2+v^2$
For $(u,v)=(3,2)$ we get:
$(509^2+335^2)=13(13^4+1)$
answered Feb 2 at 20:45
SamSam
311
$endgroup$
$begingroup$
Welcome to Math.SE. I see you've used the OP's idea of letting $c=u^2+v^2$. And it is clear that the RHS's outer terms are $u^{10}$ and $v^{10}$, hence your $u^5$ and $v^5$ terms. But how did you find out the other terms?
$endgroup$
– Rosie F
Feb 2 at 21:12
$begingroup$
@RosieF I have put a solution which begins in a different place, and shows an infinite number of coprime pairs - it turns out to be essentially the same as this and shows where the numbers come from.
$endgroup$
– Mark Bennet
Feb 2 at 21:47
$begingroup$
Hi, how did you go about putting the equation in parametric form?
$endgroup$
– user604720
Feb 11 at 0:44
add a comment |
$begingroup$
In the equation:
$$X^2+Y^2=Z^5+Z$$
I think this formula should be written in a more general form:
$$Z=a^2+b^2$$
$$X=a(a^2+b^2)^2+b$$
$$Y=b(a^2+b^2)^2-a$$
And yet another formula:
$$Z=frac{a^2+b^2}{2}$$
$$X=frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$$
$$Y=frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$$Z=frac{(a^2+b^2)^2}{2}$$
$$X=frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$$
$$Y=frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$$
where $a,b$ - any integers asked us.
Well, a simple solution:
$$Z=(a^2+b^2)^2$$
$$X=a^2+2(a^2+b^2)^4ab-b^2$$
$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$
answered Feb 3 at 4:38
individindivid
3,2521916
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $c=u^2+v^2$ then you have $$(uc^2+v)^2+(vc^2-u)^2=c(c^4+1)$$ and you want $uc^2+v$ and $vc^2-u$ to be coprime.
Now just choose $u=1$ so that the expressions are $a=c^2+v$ and $b=vc^2-1$ with $c=v^2+1$
Note then that that a common factor of $a$ and $b$ is also a factor of $va-b=v^2+1=c$ and $ac^2-b=c^4+1$, but $c$ and $c^4+1$ are coprime.
So you get a family of solutions with
$c=v^2+1$ and
$a=c^2+v=v^4+2v^2+v+1$ and
$b=vc^2-1=v^5+2v^3+v-1$
[Which, I notice is just Sam's parametric solution with $u=1$. It should be obvious what to do to generalise.]
answered Feb 2 at 21:45
Mark BennetMark Bennet
82k984183
$endgroup$
$begingroup$
Hi, I need some help understanding. How did you generalize this?
$endgroup$
– user604720
Feb 11 at 0:52
$begingroup$
@user604720 Just keeping in the $u$ rather than setting $u=1$ and working through the logic gives the parametric family. Along the way you get $va-ub=c$ and $ac^2-b=u(c^4+1)$ and it is not so easy to prove coprime for the parametric solutions because of the factor $u$, but since this starts with an arbitrary $c=u^2+v^2$ one can pick $u$ and $v$ to make this obvious. $u=1$ is an easy choice.
$endgroup$
– Mark Bennet
Feb 11 at 6:51
add a comment |
$begingroup$
Suppose $c=u^2+v^2$ then you have $$(uc^2+v)^2+(vc^2-u)^2=c(c^4+1)$$ and you want $uc^2+v$ and $vc^2-u$ to be coprime.
Now just choose $u=1$ so that the expressions are $a=c^2+v$ and $b=vc^2-1$ with $c=v^2+1$
Note then that that a common factor of $a$ and $b$ is also a factor of $va-b=v^2+1=c$ and $ac^2-b=c^4+1$, but $c$ and $c^4+1$ are coprime.
So you get a family of solutions with
$c=v^2+1$ and
$a=c^2+v=v^4+2v^2+v+1$ and
$b=vc^2-1=v^5+2v^3+v-1$
[Which, I notice is just Sam's parametric solution with $u=1$. It should be obvious what to do to generalise.]
answered Feb 2 at 21:45
Mark BennetMark Bennet
82k984183
$endgroup$
$begingroup$
Hi, I need some help understanding. How did you generalize this?
$endgroup$
– user604720
Feb 11 at 0:52
$begingroup$
@user604720 Just keeping in the $u$ rather than setting $u=1$ and working through the logic gives the parametric family. Along the way you get $va-ub=c$ and $ac^2-b=u(c^4+1)$ and it is not so easy to prove coprime for the parametric solutions because of the factor $u$, but since this starts with an arbitrary $c=u^2+v^2$ one can pick $u$ and $v$ to make this obvious. $u=1$ is an easy choice.
$endgroup$
– Mark Bennet
Feb 11 at 6:51
add a comment |
$begingroup$
Suppose $c=u^2+v^2$ then you have $$(uc^2+v)^2+(vc^2-u)^2=c(c^4+1)$$ and you want $uc^2+v$ and $vc^2-u$ to be coprime.
Now just choose $u=1$ so that the expressions are $a=c^2+v$ and $b=vc^2-1$ with $c=v^2+1$
Note then that that a common factor of $a$ and $b$ is also a factor of $va-b=v^2+1=c$ and $ac^2-b=c^4+1$, but $c$ and $c^4+1$ are coprime.
So you get a family of solutions with
$c=v^2+1$ and
$a=c^2+v=v^4+2v^2+v+1$ and
$b=vc^2-1=v^5+2v^3+v-1$
[Which, I notice is just Sam's parametric solution with $u=1$. It should be obvious what to do to generalise.]
answered Feb 2 at 21:45
Mark BennetMark Bennet
82k984183
$endgroup$
Suppose $c=u^2+v^2$ then you have $$(uc^2+v)^2+(vc^2-u)^2=c(c^4+1)$$ and you want $uc^2+v$ and $vc^2-u$ to be coprime.
Now just choose $u=1$ so that the expressions are $a=c^2+v$ and $b=vc^2-1$ with $c=v^2+1$
Note then that that a common factor of $a$ and $b$ is also a factor of $va-b=v^2+1=c$ and $ac^2-b=c^4+1$, but $c$ and $c^4+1$ are coprime.
So you get a family of solutions with
$c=v^2+1$ and
$a=c^2+v=v^4+2v^2+v+1$ and
$b=vc^2-1=v^5+2v^3+v-1$
[Which, I notice is just Sam's parametric solution with $u=1$. It should be obvious what to do to generalise.]
answered Feb 2 at 21:45
Mark BennetMark Bennet
82k984183
edited Feb 11 at 6:41
answered Feb 2 at 21:45
Mark BennetMark Bennet
82k984183
answered Feb 2 at 21:45
Mark BennetMark Bennet
82k984183
answered Feb 2 at 21:45
Mark BennetMark Bennet
82k984183
82k984183
$begingroup$
Hi, I need some help understanding. How did you generalize this?
$endgroup$
– user604720
Feb 11 at 0:52
$begingroup$
@user604720 Just keeping in the $u$ rather than setting $u=1$ and working through the logic gives the parametric family. Along the way you get $va-ub=c$ and $ac^2-b=u(c^4+1)$ and it is not so easy to prove coprime for the parametric solutions because of the factor $u$, but since this starts with an arbitrary $c=u^2+v^2$ one can pick $u$ and $v$ to make this obvious. $u=1$ is an easy choice.
$endgroup$
– Mark Bennet
Feb 11 at 6:51
add a comment |
$begingroup$
Hi, I need some help understanding. How did you generalize this?
$endgroup$
– user604720
Feb 11 at 0:52
$begingroup$
@user604720 Just keeping in the $u$ rather than setting $u=1$ and working through the logic gives the parametric family. Along the way you get $va-ub=c$ and $ac^2-b=u(c^4+1)$ and it is not so easy to prove coprime for the parametric solutions because of the factor $u$, but since this starts with an arbitrary $c=u^2+v^2$ one can pick $u$ and $v$ to make this obvious. $u=1$ is an easy choice.
$endgroup$
– Mark Bennet
Feb 11 at 6:51
$begingroup$
Hi, I need some help understanding. How did you generalize this?
$endgroup$
– user604720
Feb 11 at 0:52
$begingroup$
Hi, I need some help understanding. How did you generalize this?
$endgroup$
– user604720
Feb 11 at 0:52
$begingroup$
@user604720 Just keeping in the $u$ rather than setting $u=1$ and working through the logic gives the parametric family. Along the way you get $va-ub=c$ and $ac^2-b=u(c^4+1)$ and it is not so easy to prove coprime for the parametric solutions because of the factor $u$, but since this starts with an arbitrary $c=u^2+v^2$ one can pick $u$ and $v$ to make this obvious. $u=1$ is an easy choice.
$endgroup$
– Mark Bennet
Feb 11 at 6:51
$begingroup$
@user604720 Just keeping in the $u$ rather than setting $u=1$ and working through the logic gives the parametric family. Along the way you get $va-ub=c$ and $ac^2-b=u(c^4+1)$ and it is not so easy to prove coprime for the parametric solutions because of the factor $u$, but since this starts with an arbitrary $c=u^2+v^2$ one can pick $u$ and $v$ to make this obvious. $u=1$ is an easy choice.
$endgroup$
– Mark Bennet
Feb 11 at 6:51
add a comment |
$begingroup$
Above equation shown below has parametric solution:
$(a^2+b^2)=c(c^4+1)$
$a=u^5+uv^4+2u^3v^2+v$
$b=v^5+u^4v+2u^2v^3-u$
$c=u^2+v^2$
For $(u,v)=(3,2)$ we get:
$(509^2+335^2)=13(13^4+1)$
answered Feb 2 at 20:45
SamSam
311
$endgroup$
$begingroup$
Welcome to Math.SE. I see you've used the OP's idea of letting $c=u^2+v^2$. And it is clear that the RHS's outer terms are $u^{10}$ and $v^{10}$, hence your $u^5$ and $v^5$ terms. But how did you find out the other terms?
$endgroup$
– Rosie F
Feb 2 at 21:12
$begingroup$
@RosieF I have put a solution which begins in a different place, and shows an infinite number of coprime pairs - it turns out to be essentially the same as this and shows where the numbers come from.
$endgroup$
– Mark Bennet
Feb 2 at 21:47
$begingroup$
Hi, how did you go about putting the equation in parametric form?
$endgroup$
– user604720
Feb 11 at 0:44
add a comment |
$begingroup$
Above equation shown below has parametric solution:
$(a^2+b^2)=c(c^4+1)$
$a=u^5+uv^4+2u^3v^2+v$
$b=v^5+u^4v+2u^2v^3-u$
$c=u^2+v^2$
For $(u,v)=(3,2)$ we get:
$(509^2+335^2)=13(13^4+1)$
answered Feb 2 at 20:45
SamSam
311
$endgroup$
$begingroup$
Welcome to Math.SE. I see you've used the OP's idea of letting $c=u^2+v^2$. And it is clear that the RHS's outer terms are $u^{10}$ and $v^{10}$, hence your $u^5$ and $v^5$ terms. But how did you find out the other terms?
$endgroup$
– Rosie F
Feb 2 at 21:12
$begingroup$
@RosieF I have put a solution which begins in a different place, and shows an infinite number of coprime pairs - it turns out to be essentially the same as this and shows where the numbers come from.
$endgroup$
– Mark Bennet
Feb 2 at 21:47
$begingroup$
Hi, how did you go about putting the equation in parametric form?
$endgroup$
– user604720
Feb 11 at 0:44
add a comment |
$begingroup$
Above equation shown below has parametric solution:
$(a^2+b^2)=c(c^4+1)$
$a=u^5+uv^4+2u^3v^2+v$
$b=v^5+u^4v+2u^2v^3-u$
$c=u^2+v^2$
For $(u,v)=(3,2)$ we get:
$(509^2+335^2)=13(13^4+1)$
answered Feb 2 at 20:45
SamSam
311
$endgroup$
Above equation shown below has parametric solution:
$(a^2+b^2)=c(c^4+1)$
$a=u^5+uv^4+2u^3v^2+v$
$b=v^5+u^4v+2u^2v^3-u$
$c=u^2+v^2$
For $(u,v)=(3,2)$ we get:
$(509^2+335^2)=13(13^4+1)$
answered Feb 2 at 20:45
SamSam
311
answered Feb 2 at 20:45
SamSam
311
answered Feb 2 at 20:45
SamSam
311
answered Feb 2 at 20:45
SamSam
311
311
$begingroup$
Welcome to Math.SE. I see you've used the OP's idea of letting $c=u^2+v^2$. And it is clear that the RHS's outer terms are $u^{10}$ and $v^{10}$, hence your $u^5$ and $v^5$ terms. But how did you find out the other terms?
$endgroup$
– Rosie F
Feb 2 at 21:12
$begingroup$
@RosieF I have put a solution which begins in a different place, and shows an infinite number of coprime pairs - it turns out to be essentially the same as this and shows where the numbers come from.
$endgroup$
– Mark Bennet
Feb 2 at 21:47
$begingroup$
Hi, how did you go about putting the equation in parametric form?
$endgroup$
– user604720
Feb 11 at 0:44
add a comment |
$begingroup$
Welcome to Math.SE. I see you've used the OP's idea of letting $c=u^2+v^2$. And it is clear that the RHS's outer terms are $u^{10}$ and $v^{10}$, hence your $u^5$ and $v^5$ terms. But how did you find out the other terms?
$endgroup$
– Rosie F
Feb 2 at 21:12
$begingroup$
@RosieF I have put a solution which begins in a different place, and shows an infinite number of coprime pairs - it turns out to be essentially the same as this and shows where the numbers come from.
$endgroup$
– Mark Bennet
Feb 2 at 21:47
$begingroup$
Hi, how did you go about putting the equation in parametric form?
$endgroup$
– user604720
Feb 11 at 0:44
$begingroup$
Welcome to Math.SE. I see you've used the OP's idea of letting $c=u^2+v^2$. And it is clear that the RHS's outer terms are $u^{10}$ and $v^{10}$, hence your $u^5$ and $v^5$ terms. But how did you find out the other terms?
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– Rosie F
Feb 2 at 21:12
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Welcome to Math.SE. I see you've used the OP's idea of letting $c=u^2+v^2$. And it is clear that the RHS's outer terms are $u^{10}$ and $v^{10}$, hence your $u^5$ and $v^5$ terms. But how did you find out the other terms?
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– Rosie F
Feb 2 at 21:12
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@RosieF I have put a solution which begins in a different place, and shows an infinite number of coprime pairs - it turns out to be essentially the same as this and shows where the numbers come from.
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– Mark Bennet
Feb 2 at 21:47
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@RosieF I have put a solution which begins in a different place, and shows an infinite number of coprime pairs - it turns out to be essentially the same as this and shows where the numbers come from.
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– Mark Bennet
Feb 2 at 21:47
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Hi, how did you go about putting the equation in parametric form?
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– user604720
Feb 11 at 0:44
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Hi, how did you go about putting the equation in parametric form?
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– user604720
Feb 11 at 0:44
add a comment |
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In the equation:
$$X^2+Y^2=Z^5+Z$$
I think this formula should be written in a more general form:
$$Z=a^2+b^2$$
$$X=a(a^2+b^2)^2+b$$
$$Y=b(a^2+b^2)^2-a$$
And yet another formula:
$$Z=frac{a^2+b^2}{2}$$
$$X=frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$$
$$Y=frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$$Z=frac{(a^2+b^2)^2}{2}$$
$$X=frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$$
$$Y=frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$$
where $a,b$ - any integers asked us.
Well, a simple solution:
$$Z=(a^2+b^2)^2$$
$$X=a^2+2(a^2+b^2)^4ab-b^2$$
$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$
answered Feb 3 at 4:38
individindivid
3,2521916
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add a comment |
$begingroup$
In the equation:
$$X^2+Y^2=Z^5+Z$$
I think this formula should be written in a more general form:
$$Z=a^2+b^2$$
$$X=a(a^2+b^2)^2+b$$
$$Y=b(a^2+b^2)^2-a$$
And yet another formula:
$$Z=frac{a^2+b^2}{2}$$
$$X=frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$$
$$Y=frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$$Z=frac{(a^2+b^2)^2}{2}$$
$$X=frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$$
$$Y=frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$$
where $a,b$ - any integers asked us.
Well, a simple solution:
$$Z=(a^2+b^2)^2$$
$$X=a^2+2(a^2+b^2)^4ab-b^2$$
$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$
answered Feb 3 at 4:38
individindivid
3,2521916
$endgroup$
add a comment |
$begingroup$
In the equation:
$$X^2+Y^2=Z^5+Z$$
I think this formula should be written in a more general form:
$$Z=a^2+b^2$$
$$X=a(a^2+b^2)^2+b$$
$$Y=b(a^2+b^2)^2-a$$
And yet another formula:
$$Z=frac{a^2+b^2}{2}$$
$$X=frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$$
$$Y=frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$$Z=frac{(a^2+b^2)^2}{2}$$
$$X=frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$$
$$Y=frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$$
where $a,b$ - any integers asked us.
Well, a simple solution:
$$Z=(a^2+b^2)^2$$
$$X=a^2+2(a^2+b^2)^4ab-b^2$$
$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$
answered Feb 3 at 4:38
individindivid
3,2521916
$endgroup$
In the equation:
$$X^2+Y^2=Z^5+Z$$
I think this formula should be written in a more general form:
$$Z=a^2+b^2$$
$$X=a(a^2+b^2)^2+b$$
$$Y=b(a^2+b^2)^2-a$$
And yet another formula:
$$Z=frac{a^2+b^2}{2}$$
$$X=frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$$
$$Y=frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$$Z=frac{(a^2+b^2)^2}{2}$$
$$X=frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$$
$$Y=frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$$
where $a,b$ - any integers asked us.
Well, a simple solution:
$$Z=(a^2+b^2)^2$$
$$X=a^2+2(a^2+b^2)^4ab-b^2$$
$$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$$
answered Feb 3 at 4:38
individindivid
3,2521916
answered Feb 3 at 4:38
individindivid
3,2521916
answered Feb 3 at 4:38
individindivid
3,2521916
answered Feb 3 at 4:38
individindivid
3,2521916
3,2521916
add a comment |
add a comment |
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2
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Is the question $a^2+b^2=c^5+c$ or $a^5+b^5=c^5+c$?
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– Peter Foreman
Feb 2 at 18:53
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Sorry I mistyped its a^2 + b^2 = c^5 + c
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– user604720
Feb 2 at 18:54
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If you know that primes congruent to $1$ mod $4$ can be written as sums of squares (and if you know there are infinitely many such primes), you should be good to go.
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– Barry Cipra
Feb 2 at 19:01
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Excuse me, but what is the unknown value?
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– El borito
Feb 2 at 19:05
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Come to think of it, it suffices to know (or show) that $c=5^n$ can always be written as the sum of two relatively prime squares.
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– Barry Cipra
Feb 2 at 19:10