For a primitive pythagorean triple $(a, b, c)$ (with $a$ odd, $b$ even), prove $(c+b), (c-b)$ are both squares.

My thoughts

Having $a$ be odd, gives $a^2$ as odd also.

From $a^2 + b^2 = c^2$ we have $a^2 = (c-b)(c+b)$.

I have figured out that $\gcd(c-b, c+b) = 1$, which seems relevant.

From the fundamental theorem of arithmetic, we now know that the prime factorizations of $c-b, c+b$ will have no common factors. This also seems relevant.

But I'm having a hard time tying all this together.

Any help would be appreciated!

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    $\begingroup$ Any prime factor of $a^2$ appears an even number of times therein, but it cannot appear in both of $c-b$ and $c+b$. $\endgroup$ – Lord Shark the Unknown Jan 3 '18 at 4:41
  • $\begingroup$ @LordSharktheUnknown - Oooh, right! So all factors in each of c-b and c+b must appear an even number of times. Yep, that'll do it, thanks! $\endgroup$ – Alec Jan 3 '18 at 4:43

Lord Shark's answer in the comments is probably the most elementary one. An alternative answer is to use the classification of primitive Pythagorean triples.

There are coprime, natural numbers $m,n$ (one of them even, the other odd) such that $$a=m^2-n^2\\b=2mn\\c=m^2+n^2$$ It's a bit of work to prove this, but once we have this result, the answer is immediate: $$c-b=(m-n)^2\\c+b=(m+n)^2$$


Given $A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$ $$C-B=m^2+n^2-2mn=(m-n)^2$$ $$C+B=m^2+n^2+2mn=(m+n)^2$$


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