Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.

Question

Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.

I have tried to solve this question and did pretty well until I reached the end, so I was wondering if I could get help on that part. Here is what I did. $$a^2+b^2=c^2$$ $$b^2=(c-a)(c+a)$$ Since $a, b, c > 0 \therefore (c+a) \ne 0$ $$\therefore c-a={b^2\over c+a}$$ Similarly we get, $$c-b={a^2\over c+b}$$ $$\therefore {1\over2}(c-a)(c-b)={1\over2}({b^2\over c+a})({a^2\over c+b})$$ $$={(ab)^2\over 2c^2+2ab+2bc+2ca}$$ $$={(ab^2)\over a^2+b^2+c^2+2ab+2bc+2ca}$$ $$={(ab)^2\over (a+b+c)^2}$$ $$=({ab\over a+b+c})^2$$ However, I was unable to prove that ${ab\over a+b+c} \in \mathbb{Z}$ Is there a way to prove it? Thank you

Answer 1

Alternatively, use Formulas for generating Pythagorean triples

WLOG $a=2pqk, b=(p^2-q^2)k,c=(p^2+q^2)k$

$c-a=k(p-q)^2$

$c-b=2kq^2$

Answer 2

Multiply by conjugate: $${ab\over a+b+c}={ab\over a+b+c}\cdot \frac{a+b-c}{a+b-c}=\frac{ab(a+b-c)}{2ab}=\frac{a+b-c}{2}\in \mathbb Z^+,$$ because: $$a+b>c$$ and there are two cases for $a^2+b^2=c^2$: 1) $a,b,c$ are even; 2) one is even, the other two are odd. And for each case, $a+b-c$ is even.

Answer 3

$$(a+b-c)^2 = 2(c-a)(c-b)$$ is an equivalent statement of the Pythagorean formula. It is a well known fact that either a or b must be even. So from there, identify that $b^2=(c-a)(c+a) \text{ and } a^2=(c-b)(c+b)$. So it is unavoidable that either $(c-a)$ or $(c-b)$ must be even. This covers off the right side of the equation. On the left side, you have 2 odd numbers and an even number being added/subtracted together, resulting in an guaranteed even number. The final step is to simply divide both sides by 4! $$\bigg(\frac{a+b-c}{2}\bigg)^2 = \frac{(c-a)(c-b)}{2}$$