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: 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
$$(2+14) \mid 2^4 + 14^4\qquad\text{and}\qquad 2+14 \nmid 2^2 + 14^2.$$
A: 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 $32$.

A: 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 $3^4|3^2*629$.
