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?

Well,
\begin{align}
&& a+b &\mid a^4 + b^4 \\
&\iff & a+b &\mid a^4 + b^4 - (a+b)^4 + 4ab(a+b)^2 \\
&\iff & a+b &\mid 2a^2b^2 \\
&\iff & a+b &\mid ab\bigl((a+b)^2 - (a^2+b^2)\bigr) \\
&\iff & a+b &\mid ab(a^2+b^2)\,
\end{align}
so for coprime $a,b$ it follows that $a+b \mid a^4+b^4 \iff a+b \mid a^2+b^2$.
Writing $a = gx,\, b = gy$ with $g = \gcd(a,b)$, we see that $a+b \mid a^4 + b^4$ if and only if $x+y \mid g^3(x^2+y^2)$, and $a+b \mid a^2+b^2$ if and only if $x+y \mid g(x^2+y^2)$. So if we find coprime $x,y$ and a $g$ such that $x+y \nmid g(x^2+y^2)$ but $x+y \mid g^3(x^2+y^2)$, we have a counterexample.
Choosing $x = 1,\, y = 7$ and $g = 2$ provides one, $1+7 = 8 \nmid 100 = 2(1+7^2)$, but $8 \mid 400$. So
$$(2+14) \mid 2^4 + 14^4\qquad\text{and}\qquad 2+14 \nmid 2^2 + 14^2.$$
