Algebraic manipulation from $a^2+b^2 = abm$ with all variables in Z to a|b?

Question

I know that $a^2+b^2=a\,b\,m$, with $a,b,m$ integers ($a,b$ positive integers)

How can I show that a|b from this?

I know it's true intuitively. I can recall the definition of divides to be $ak=b$ such that $k$ is in $\mathbb{Z}$.

Answer 1

Note: From symmetry (exchanging $a$ and $b$ gives the same equation) $a\mid b$ would also mean $b\mid a$ and together with $a,b >0$ lead to $a=b=1$ and $m=2$.

The equation can written as $$ (a + b)^2 = (2+m) a b $$ Setting $b = a + k$ we get $$ 4a^2 + k^2 + 4ak = (2+m) a (a+k) = (2+m) a^2 + (2+m)ak \iff \\ k^2 = (m - 2) a^2 + (m-2) ak = (m-2) a (a+k) $$