# 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 $b^2$.
$$a+b|(a+b)(a-b)=a^2-b^2$$