<s>[INST] Prove that $\left(A+B\right)^{2}\nmid A^{2n+1}+B^{2n+1}$ Let it be A and B two coprime positive integers. I know how to prove by induction that $A+B\mid A^{2n+1}+B^{2n+1}$, but I am having a bit trouble proving that $\left(A+B\right)^{2}\nmid A^{2n+1}+B^{2n+1}$.
For the case $n=1$ I have managed to prove it in the following way:
$$A^{2n+1}+B^{2n+1}=A^{3}+B^{3}$$
$$A^{3}+B^{3}=A^{2}(A+B)+B\left(B^{2}-A^{2}\right)$$
$$A^{3}+B^{3}=A^{2}(A+B)+B\left(A+B\right)\left(B-A\right)$$
$$A^{3}+B^{3}=\left(A+B\right)\left(A^{2}+B^{2}-AB\right)$$
As $\left(A+B\right)^{2}=A^{2}+B^{2}+2AB$, then we have that $A^{2}+B^{2}-AB=\left(A+B\right)^{2}-3AB$. 
Thus, substituting, we get that
$$A^{3}+B^{3}=\left(A+B\right)\left(\left(A+B\right)^{2}-3AB\right)$$
$\left(A+B\right)^{2}\mid A^{3}+B^{3}$ only if $\left(A+B\right)\mid\left(A+B\right)^{2}-3AB$. As $\left(A+B\right)\mid\left(A+B\right)^{2}$, then it follows that $\left(A+B\right)\mid\left(A+B\right)^{2}-3AB$ only if $\left(A+B\right)\mid3AB$. 
As A and B are coprime, it follows that $A+B\nmid3AB$. Therefore,
$$\left(A+B\right)^{2}\nmid A^{3}+B^{3}$$
The problem is that it seems not easy (or maybe possible) to apply induction on this method.
My questions are: 


*

*Is the above proof correct? 

*How could it be proved (if possible) that $\left(A+B\right)^{2}\nmid A^{2n+1}+B^{2n+1}$?


Thanks in advance!
 [/INST] The statement is false. We show how to find counterexamples.
Since we know that $$A^{2n+1}+B^{2n+1}=(A+B)(A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n}),$$ it is equivalent to finding $A,B$ such that $$A+B\mid A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n},$$ which is the same as $$A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n}\equiv 0\pmod{A+B}.$$
However, we have from $A\equiv -B\pmod{A+B}$ that
\begin{align*}
A^{2n}-A^{2n-1}B+\dots-AB^{2n-1}+B^{2n}&\equiv (-B)^{2n}-(-B)^{2n-1}B+\dots-(-B)B^{2n-1}+B^{2n}\pmod{A+B}\\
&\equiv (2n+1)\cdot B^{2n}\pmod{A+B}.
\end{align*}
As such, we see that counterexamples can be constructed by selecting the value of $n$ such that $A+B\mid 2n+1$. For example, $(A,B,n)=(3,4,3)$ is a counterexample because $$(3+4)^2=49\mid 18571=3^7+4^7.$$
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