Finding a divisor $d^n$ of $a^n+b^n$ such that $d\mid a+b$ but $d^n\nmid a+b$

Question

Let us suppose that $d\mid (a+b)$, where $\mid$ means "divides", and such that $d>1$, $n>1$ and $\gcd(a,b)=1$

Could it be proved that $d^n\mid a^n+b^n$ only if $d^n\mid a+b$? If false, could you show some counterexample?

The question comes from the observation that $$a+b\mid a^{2k+1}+b^{2k+1} \forall k \in \mathbb N$$ However, I have not been able to find any power $d^n$ dividing $a^n+b^n$ and not dividing also $a+b$.

Thanks in advance!

EDIT

From the counterexample provided, I have noticed that for $a+b=p^{k}$, $p$ being some odd prime number, it holds that $p^{k+1}\mid a^p+b^p$. This result can be derived both from the observation provided in the OP, and Fermat's Little Theorem, and provides a family of counterexamples setting $d=n=p$ and $a+b=p^{p-1}$. Thus, I think it would be interesting to show some counterexample such that $d\neq n$

Answer 1

Have you tried $$a=8,b=1,n=3,d=3$$