Hi! Can anyone provide a proof for the following:

The following formula

$\displaystyle a=k(m^2-n^2)$

$\displaystyle b=k(2mn)$

$\displaystyle c=k(m^2+n^2)$

generates all the positive integers that satisfy $\displaystyle a^2+b^2=c^2$.

Thanks!

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- Aug 8th 2010, 07:12 PMTerenceCSAbout Pythagorean triple...
Hi! Can anyone provide a proof for the following:

The following formula

$\displaystyle a=k(m^2-n^2)$

$\displaystyle b=k(2mn)$

$\displaystyle c=k(m^2+n^2)$

generates all the positive integers that satisfy $\displaystyle a^2+b^2=c^2$.

Thanks! - Aug 8th 2010, 07:26 PMundefined
I believe we can use an argument similar to the one in this thread post #22.

- Aug 8th 2010, 09:15 PMBruno J.
What undefined suggests is good! Here is another way.

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