Pythagorean triple division proof

Question

Show that in a Pythagorean triple, if the largest term is divisible by 4, then so are the other two terms.

My approach: Given c as the largest term, I know that pythagorean theorem holds. $$ c^2 = a^2 + b ^2$$

Since c is divisible by 4, I let c = 4k, where k is some positive integer. And by letting c = 4k, I know that $c^2 = 16k^2$. But I don't know how this proves $ a= 4l$ and $b = 4m$ where l, m are some positive integer.

Answer 1

Hint:

First of all, both $a,b$ should have same parity to keep $c$ even

Now if $a,b$ both are odd, let $a=2d+1,b=2e+1$

$$(2d+1)^2+(2e+1)^2=8\cdot\dfrac{c(c+1)+e(e+1)}2+2\equiv2\pmod8\equiv2\pmod4$$

But for any integer $n, n^2\equiv0,1\pmod4$

So, $a,b$ must be even

Can you take it from here?

Answer 2

Show that if $a,b$ are coprime, then $c$ is odd. Conclude.

Answer 3

Hint: $$a^2\equiv_4 0,1\\ b^2\equiv_4 0,1 $$

Answer 4

In all primitive triples, C is odd. Therefore, if C is divisible by $4$, the triple must be a $4k$ multiple of a primitive and all terms are divisible by $4$.