Prove that if $(a+b)$ divides $a^2$ then $(a+b)$ divides $b^2$ I'm trying to solve the following exercise: "Prove that if $(a+b)$ divides $a^2$ then $(a+b)$ divides $b^2$".
It's quite obvious how to prove divisibility for a product, but how to do it for a sum?

If $a+b$ divides $a^2$, then you can write $a^2=(a+b)k$.
But then $$b^2= a^2-a^2+b^2 = a^2 -(a^2-b^2)$$
$$=\underbrace{(a+b)k}_{a^2} - (a+b)(a-b)$$
$$=(a+b)(k-(a-b))$$
$$=(a+b)(k-a+b)$$
so $(a+b)$ divides $b^2$.
