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?

We have that $$a^4+b^4=(a+b)(a^3-a^2b+ab^2-b^3)+2b^4$$ so if $a+b$ is a factor of $a^4+b^4$ it is also a factor of $2b^4$ and (by symmetry) $2a^4$
If the highest common factor of $a$ and $b$ is $y$ so that $a=py$ and $b=qy$ we find that $(p+q)y$ is a factor of $2q^4y^4$. Now $p+q$ can have no factor in common with $q$ by construction, so $p+q|2y^4$. We find an easy solution by setting $p+q=y$.
You might want to think about constructing a counterexample before going further, by tightening things up a bit.

 If we want to be tight against a constraint we might try $y^4=\frac {p+q}2$. With $y=2$ this would give $p+q=32$. Then $a=2p$ and $b=2q$ and $a^4+b^4=16 (p^4+q^4)$ and this is divisible by $p+q=32$ because $p$ and $q$ have the same parity. But now put $p=1, q=31$ with $a=2, b=62$ and $a^2+b^2=4+3844=3848$ is not divisible by $32$.

