About Pythagorean triple...

**TerenceCS** · August 8th 2010, 07:12 PM

Hi! Can anyone provide a proof for the following:
The following formula
$a=k(m^2-n^2)$
$b=k(2mn)$
$c=k(m^2+n^2)$
generates all the positive integers that satisfy $a^2+b^2=c^2$ .
Thanks!

**undefined** · August 8th 2010, 07:26 PM

Originally Posted by TerenceCS

Hi! Can anyone provide a proof for the following:
The following formula
$a=k(m^2-n^2)$
$b=k(2mn)$
$c=k(m^2+n^2)$
generates all the positive integers that satisfy $a^2+b^2=c^2$ .
Thanks!

I believe we can use an argument similar to the one in this thread post #22.

**Bruno J.** · August 8th 2010, 09:15 PM

Originally Posted by TerenceCS

Hi! Can anyone provide a proof for the following:
The following formula
$a=k(m^2-n^2)$
$b=k(2mn)$
$c=k(m^2+n^2)$
generates all the positive integers that satisfy $a^2+b^2=c^2$ .
Thanks!

What undefined suggests is good! Here is another way.

You can suppose that $a,b,c$ are relatively prime. Then it is impossible for both $a,b$ to be odd (because then $c^2\equiv 2 \mod 4$ , which is impossible), and impossible for both of them to be even, which would contradict the hypothesis that $a,b,c$ are relatively prime. Hence, say $a$ is even and $b,c$ are odd. Write $(c-b)(c+b)=a^2$ . Now both $c-b,c+b$ are even, say $c-b=2u,c+b=2v$ ; moreover $u,v$ are relatively prime. Since $a^2=4uv$ is a square, both $u,v$ must be squares, so we have $c-b=2m^2. c+b=2n^2$ , i.e. $c=m^2+n^2, b=n^2-n^2, a=2mn$ .

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