$ab$ divides $a^2+b^2 \implies a=b$ Let $a$ and $b$ be two positive integers. 

If $ab$ divides $a^2+b^2$ then $a=b$.  

I can show that $a$ divides $b^2$ and $b$ divides $a^2$ but then I get stuck. Any ideas?

A standard argument: Let $a=da_1$ and $b=db_1$ with $d={\rm gcd}(a,b)$ (that is, ${\rm gcd}(a_1,b_1)=1$). Now, $ab\mid a^2+b^2\iff a_1b_1\mid a_1^2+b_1^2$. Now, we have $a_1\mid a_1^2+b_1^2$ and $b_1\mid a_1^2+b_1^2$. Since $a_1\mid a_1^2$, the first is possible iff $a_1\mid b_1^2$. But as $a_1,b_1$ are coprime, this implies $a_1=1$. Similarly, we get $b_1=1$, from where the conclusion follows. 
