For a Pythagorean triple $a, b, c \in N$, if a is even and a, b, c have no common factors, prove that $c-a$ and $c+a$ are perfect squares.

Question

What property(s) should I examine to approach this problem?

Answer 1

Hint. If $x$ and $y$ are integers and $xy$ is a perfect square, then $x$ and $y$ are both perfect squares. . .

. . . or are they?

Answer 2

Expanding on David's hint: $$a^2+b^2=c^2$$ $$b^2=c^2-a^2$$ $$b^2=(c+a)(c-a)$$ Take it from there?

Answer 3

If a is even and coprime to b then it is represented by $ 2mn $ for $ gcd(n,m) = 1 $ and n , m not both odd in Euclids Formula.

http://en.m.wikipedia.org/wiki/Pythagorean_triple

$ c + a = m^2 + 2mn + n^2 = (m + n)^2 $

$ c - a = m^2 - 2mn + n^2 = (m - n)^2 $