$ab$ 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: Hint $\ n = \dfrac{a^2\!+b^2}{ab} = \dfrac{a}b + \dfrac{b}a =\, x+x^{-1}\,\overset{\large {\times\, x}}\Longrightarrow\,x^2-n\,x + 1 = 0$
By RRT = Rational Root Test $\ a/b\, =\, x\, = \pm 1.\,$  It is special case $\, j = 1 = k,\, c_1 = 0\,$ of below.  
Generally applying RRT as above yields the degree $\,j+k\,$ homogeneous generalization
$$a,b,c_i\in\Bbb Z,\,\  a^{\large j}b^{\large k}\mid \color{#c00}{\bf 1}\:\! a^{\large j+k}\! + c_1 a^{\large j+k-1} b + \cdots + c_{\large j+k-1} a b^{\large j+k-1}\! + \color{#c00}{\bf 1}\:\!b^{\large j+k}\Rightarrow\, a = \pm b \qquad $$
$\qquad\qquad\ \ \ \ \ \  $ e.g. $\ a^2b \mid a^3 + c_1 a^2b + c_2 ab^2 + b^3\,\Rightarrow\, a = \pm b,\ $  e.g. here  (see also here). 
Alternatively the statement is homogeneous in $\,a,b\,$ so we can cancel $\,\gcd(a,b)^{\large j+k}$ to reduce to the case  $\,a,b\,$  coprime. The dividend $\,c\,$  has form $\,a^{\large n}\!+b^{\large n}\! + abm\,$ so by Euclid it is coprime to $a,b$ thus $\,a,b\mid c\,\Rightarrow\, a,b = \pm1$.
A: 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. 
