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Triple Pythagorean with $a^2+b^2=c^4$

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0

$begingroup$

It is well known that there exist integer solutions to the equation $a^2+b^2=c^2$.
For example, an explicit formula for integer values of $a$ , $b$ , and $c$ is

beginaligna&=2mn \ b&=m^2-n^2 \ c&=m^2+n^2.endalign

I want to find solutions to the equation $a^2+b^2=c^4$. However, I got $a,b,c$ in a general form that substitutes $x$ in integer and get $a,b,c$ in the integer. What should I do first? I'm very stuck!

elementary-number-theory diophantine-equations pythagorean-triples

share|cite|improve this question

edited Mar 22 at 5:48

Martin Sleziak

45k10122277

asked Mar 17 at 16:26

HeartHeart

30519

$endgroup$

add a comment | 

0

$begingroup$

It is well known that there exist integer solutions to the equation $a^2+b^2=c^2$.
For example, an explicit formula for integer values of $a$ , $b$ , and $c$ is

beginaligna&=2mn \ b&=m^2-n^2 \ c&=m^2+n^2.endalign

I want to find solutions to the equation $a^2+b^2=c^4$. However, I got $a,b,c$ in a general form that substitutes $x$ in integer and get $a,b,c$ in the integer. What should I do first? I'm very stuck!

elementary-number-theory diophantine-equations pythagorean-triples

share|cite|improve this question

edited Mar 22 at 5:48

Martin Sleziak

45k10122277

asked Mar 17 at 16:26

HeartHeart

30519

$endgroup$

add a comment | 

$begingroup$

It is well known that there exist integer solutions to the equation $a^2+b^2=c^2$.
For example, an explicit formula for integer values of $a$ , $b$ , and $c$ is

beginaligna&=2mn \ b&=m^2-n^2 \ c&=m^2+n^2.endalign

I want to find solutions to the equation $a^2+b^2=c^4$. However, I got $a,b,c$ in a general form that substitutes $x$ in integer and get $a,b,c$ in the integer. What should I do first? I'm very stuck!

elementary-number-theory diophantine-equations pythagorean-triples

share|cite|improve this question

edited Mar 22 at 5:48

Martin Sleziak

45k10122277

asked Mar 17 at 16:26

HeartHeart

30519

$endgroup$

share|cite|improve this question

edited Mar 22 at 5:48

Martin Sleziak

45k10122277

asked Mar 17 at 16:26

HeartHeart

30519

share|cite|improve this question

edited Mar 22 at 5:48

Martin Sleziak

45k10122277

asked Mar 17 at 16:26

HeartHeart

30519

edited Mar 22 at 5:48

Martin Sleziak

45k10122277

edited Mar 22 at 5:48

Martin Sleziak

45k10122277

asked Mar 17 at 16:26

HeartHeart

30519

asked Mar 17 at 16:26

HeartHeart

30519

2 Answers
2

active

oldest

votes

4

$begingroup$

Your equation, $a^2 + b^2 = c^4$, is the same as $a^2 + b^2 = c^2$, except that $c$ has been replaced with $c^2$. Thus you can find solutions by choosing $m$ and $n$ so that
$$
beginalign
a&=2mn \ b&=m^2-n^2 \ c^2&=m^2+n^2.
endalign
$$
But how can you make sure that $c$ is an integer? Easy: that last equation, $c^2 = m^2 + n^2$ says that $m,n$ and $c$ also form a Pythagorean triple. Thus you can find solutions with two new parameters, $r$ and $s$ such that
$$
beginalign
m&=2rs \ n&=r^2-s^2 \ c&=r^2+s^2.
endalign
$$
Substituting these in the first set of equations to find $a$ and $b$, you have
$$
beginalign
a&=4rs(r^2-s^2) \ b&=6r^2 s^2-r^4-s^4 \ c&=r^2 + s^2.
endalign
$$
For example, choosing $r=2$ and $s=1$ gives you $a = 24$, $b=7$, $c=5$ and, indeed, $24^2 + 7^2 = 5^4$.

share|cite|improve this answer

answered Mar 17 at 17:15

FredHFredH

3,7051023

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Finding any triplet where C is an odd square will satisfy the equation because the square root of C taken to the 4th power does is.
  $endgroup$
  – poetasis
  Mar 31 at 16:50
  

  
  
  
  

add a comment | 

4

$begingroup$

Whenever you have $a^2+b^2=c^2$ you also get $(ca)^2+(cb)^2=c^4$ -- does that count for you?

share|cite|improve this answer

answered Mar 17 at 16:51

Henning MakholmHenning Makholm

243k17309554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1, but are those all solutions?
  $endgroup$
  – tarit goswami
  Mar 22 at 3:10
  

  
  
  
  

add a comment | 

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4

$begingroup$

Your equation, $a^2 + b^2 = c^4$, is the same as $a^2 + b^2 = c^2$, except that $c$ has been replaced with $c^2$. Thus you can find solutions by choosing $m$ and $n$ so that
$$
beginalign
a&=2mn \ b&=m^2-n^2 \ c^2&=m^2+n^2.
endalign
$$
But how can you make sure that $c$ is an integer? Easy: that last equation, $c^2 = m^2 + n^2$ says that $m,n$ and $c$ also form a Pythagorean triple. Thus you can find solutions with two new parameters, $r$ and $s$ such that
$$
beginalign
m&=2rs \ n&=r^2-s^2 \ c&=r^2+s^2.
endalign
$$
Substituting these in the first set of equations to find $a$ and $b$, you have
$$
beginalign
a&=4rs(r^2-s^2) \ b&=6r^2 s^2-r^4-s^4 \ c&=r^2 + s^2.
endalign
$$
For example, choosing $r=2$ and $s=1$ gives you $a = 24$, $b=7$, $c=5$ and, indeed, $24^2 + 7^2 = 5^4$.

share|cite|improve this answer

answered Mar 17 at 17:15

FredHFredH

3,7051023

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Finding any triplet where C is an odd square will satisfy the equation because the square root of C taken to the 4th power does is.
  $endgroup$
  – poetasis
  Mar 31 at 16:50
  

  
  
  
  

add a comment | 

4

$begingroup$

Your equation, $a^2 + b^2 = c^4$, is the same as $a^2 + b^2 = c^2$, except that $c$ has been replaced with $c^2$. Thus you can find solutions by choosing $m$ and $n$ so that
$$
beginalign
a&=2mn \ b&=m^2-n^2 \ c^2&=m^2+n^2.
endalign
$$
But how can you make sure that $c$ is an integer? Easy: that last equation, $c^2 = m^2 + n^2$ says that $m,n$ and $c$ also form a Pythagorean triple. Thus you can find solutions with two new parameters, $r$ and $s$ such that
$$
beginalign
m&=2rs \ n&=r^2-s^2 \ c&=r^2+s^2.
endalign
$$
Substituting these in the first set of equations to find $a$ and $b$, you have
$$
beginalign
a&=4rs(r^2-s^2) \ b&=6r^2 s^2-r^4-s^4 \ c&=r^2 + s^2.
endalign
$$
For example, choosing $r=2$ and $s=1$ gives you $a = 24$, $b=7$, $c=5$ and, indeed, $24^2 + 7^2 = 5^4$.

share|cite|improve this answer

answered Mar 17 at 17:15

FredHFredH

3,7051023

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Finding any triplet where C is an odd square will satisfy the equation because the square root of C taken to the 4th power does is.
  $endgroup$
  – poetasis
  Mar 31 at 16:50
  

  
  
  
  

add a comment | 

$begingroup$

Your equation, $a^2 + b^2 = c^4$, is the same as $a^2 + b^2 = c^2$, except that $c$ has been replaced with $c^2$. Thus you can find solutions by choosing $m$ and $n$ so that
$$
beginalign
a&=2mn \ b&=m^2-n^2 \ c^2&=m^2+n^2.
endalign
$$
But how can you make sure that $c$ is an integer? Easy: that last equation, $c^2 = m^2 + n^2$ says that $m,n$ and $c$ also form a Pythagorean triple. Thus you can find solutions with two new parameters, $r$ and $s$ such that
$$
beginalign
m&=2rs \ n&=r^2-s^2 \ c&=r^2+s^2.
endalign
$$
Substituting these in the first set of equations to find $a$ and $b$, you have
$$
beginalign
a&=4rs(r^2-s^2) \ b&=6r^2 s^2-r^4-s^4 \ c&=r^2 + s^2.
endalign
$$
For example, choosing $r=2$ and $s=1$ gives you $a = 24$, $b=7$, $c=5$ and, indeed, $24^2 + 7^2 = 5^4$.

share|cite|improve this answer

answered Mar 17 at 17:15

FredHFredH

3,7051023

$endgroup$

share|cite|improve this answer

answered Mar 17 at 17:15

FredHFredH

3,7051023

answered Mar 17 at 17:15

FredHFredH

3,7051023

answered Mar 17 at 17:15

FredHFredH

3,7051023

4

$begingroup$

Whenever you have $a^2+b^2=c^2$ you also get $(ca)^2+(cb)^2=c^4$ -- does that count for you?

share|cite|improve this answer

answered Mar 17 at 16:51

Henning MakholmHenning Makholm

243k17309554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1, but are those all solutions?
  $endgroup$
  – tarit goswami
  Mar 22 at 3:10
  

  
  
  
  

add a comment | 

4

$begingroup$

Whenever you have $a^2+b^2=c^2$ you also get $(ca)^2+(cb)^2=c^4$ -- does that count for you?

share|cite|improve this answer

answered Mar 17 at 16:51

Henning MakholmHenning Makholm

243k17309554

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1, but are those all solutions?
  $endgroup$
  – tarit goswami
  Mar 22 at 3:10
  

  
  
  
  

add a comment | 

$begingroup$

Whenever you have $a^2+b^2=c^2$ you also get $(ca)^2+(cb)^2=c^4$ -- does that count for you?

share|cite|improve this answer

answered Mar 17 at 16:51

Henning MakholmHenning Makholm

243k17309554

$endgroup$

share|cite|improve this answer

answered Mar 17 at 16:51

Henning MakholmHenning Makholm

243k17309554

answered Mar 17 at 16:51

Henning MakholmHenning Makholm

243k17309554

answered Mar 17 at 16:51

Henning MakholmHenning Makholm

243k17309554