If $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$; $a,b,$ are positive integers. Is it true: $a+b\mid a^4+b^4$ then $a+b\mid a^2+b^2$?
Somehow I can't find counterexample nor to prove it. I try to write it $a=gx$ and $b=gy$ where $g=\gcd(a,b)$ but didn't help. It seems that there is no $a\ne b$ such that $a+b\mid a^4+b^4$. Of course, if we prove this stronger statement we are done. Any idea?

A little handwavy but enough to get a counter example.
It's easy to verify $a+b|a^{2k} - b^{2k}$ (by noting $\frac {a^m - b^m}{a-b} = a^{m-1}+ a^{m-2}b+ ... + ab^{m-2} + b^{m-1}$ and replacing $b$ with $-b$ and noting $(-b)^{2k} = b^{2k}$.)
$a + b|a^4 - b^4$ and $a+b|a^2 - b^2$
so if $a+b|a^4 + b^4$ then $a+b|(a^4+b^4)\pm (a^4 - b^4)$ so $a+b|2a^4$ and $a+b|2b^4$.
If $a+b|a^2 + b^2$ then by the same argument $a+b|2a^2$ and $a+b|2b^2$
So for a counter example we need $a+b|2a^4,2b^4$ but not $a+b|2a^2,2b^2$. 
One way to do that would be there were a common factor, $p$, of $a$ and $b$ with $a+b|p^4$ but $p^4$ not dividing  $a$ or $b$.  
That's easy enough to find once we understand what we are looking for.
Example: Let $p=3$ we want $a+b|3^4=81$ and $3$ but no higher power  divides $a$ and $b$.  To keep it simple, let $a+b = 81$ and $a=6$ and $b=75$.
That should do it and indeed:
$6^4 + 75^4 = 3^4(2^4 + 25^4) = (6+75)(2^4 + 25^4)$ whereas
$6^2 + 75^2 = 3^2(2^2 + 25^2)=3^2(4 + 625)=3^2*629$.  And $3\not \mid 629$ so $3^4|3^2*629$.
