If $A+B \mid A^5$ and $A+B \mid B^5$, what can be concluded? Assume $A+B \mid A^5$  and $A+B \mid B^5$, while all the variables are integers expect zero.
Can we prove that $A=B$?
This is my idea of the proof:

*

*For every prime $p$ that $p\mid A+B$ there is $p \mid A$ and $p \mid B$.


*Let $\gcd(A,B)=P$, so $A=Pw$ and $B=Pv$ such that $w$ and $v$ have no common factors with $P$. Then if $q\ne 1$, such that $q$ and $P$ are coprime, and $q\mid A+B$ then $q\mid w+v$, and hence $q$ divides both $A$ and $B$, and this is not possible.


*Hence $A=P$ and $B=P$.
In this proof, I did not use the exponent $5$ at all. So I feel there is something wrong with the proof.
Is there any thoerem that can be used here?
 A: As mentioned there are easy counterexamples. We can characterize the solutions as follows. By the gcd Universal Property & Freshman's Dream we have
$\begin{align} A+B\mid A^5,B^5\!\iff\! A+B&\mid (A^5,B^5)=(A,B)^5\\[.2em]
\iff a\ +\ b&\mid\, c^4\ \ \text{by dividing prior by }\,c := (A,B), \ {\rm  with }\,\ a,b = \frac{A}c,\frac{B}c
\end{align}$
So let $\,c\,$ and coprime $\,a,b\,$ be solutions of $\,a+b = c^4.\,$ Then $\,A\!+\!B\mid A^5,B^5$ for  $\,A,B = ac,bc$.
e.g. let's view the first counterexample in the comments this way
$$\begin{align} 
1+2&\mid 3^4\\[.2em]
\iff\ 3+6&\mid 3^5 = (3,6)^5 = (3^6,6^5)\\[.2em]
\iff\  3+6&\mid 3^5,6^5\\[.3em]
\text{for a bigger example }\ \ \  
40\ +\ 41\ &\mid\ 3^4\\[.2em]
\iff\  120+123&\mid 120^5, 123^5
\end{align}\qquad\qquad\qquad\qquad\quad $$
A: 
NB : I'll be using $(x,y)$ to denote the values of $A$ and $B$ as $(A,B)$.

"Suppose $A$ and $B$ are integers other than $0$, such that $A + B|A^5$ and $A + B|B^5$, what inferences can be drawn from this ?" - this is what has been asked.
If the above conditions have to be satisfied, $A^5$ and $B^5$ should have $A + B$ as a factor (or some other number divisible by that sum) . This means that $A = 0$ or $B = 0$ , which is invalid as per the question. Therefore, we can consider values like $(3, 6)$ (as @paulinho suggested) , but at the same time, we have exceptions : $(2,9) [\because 11 \nmid2^5 , 11 \nmid 9^5]$ and such values. So, maybe, $A +B$ should be some power of primes to satisfy the condition (as in the example $(3,6) : 3 + 6 = 9 = 3^2$; or even $(2,6):2 + 6 = 8 = 2^3$).
On the whole, if $p$ is a prime, a pair of integers $(p, pn)$ [for some $n$ such that $p \mid (n + 1)$ and $(n + 1) = p^m$ for some $m <  5$] satisfies the conditions.

PS : For a better answer, look for the one written by @BillDubuque. I think that his answer is more sensible and more understandable than mine.

