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Let $n$ be a positive integer with $n\equiv 4\mod 6$ and define $p:=\frac{n^2+n+1}{3}$ (which is in this case a positive integer as well).

Conjecture : $$p^2\mid n^n+(n+1)^{n+1}$$ for every $n$ of the given form.

The conjecture is true upto $n=10^9$

Trial : We have to show $(n^2+n+1)^2 \mid 9(n^n+(n+1)^{n+1})$ and with $x^6\equiv (x+1)^6\equiv 1\mod (x^2+x+1)$ I could show $x^2+x+1\mid x^n+(x+1)^{n+1}$ , but I did not manage to find the general remainder of $9(x^n+(x+1)^{n+1})$ modulo $(x^2+x+1)^2$ to finish the proof. Can "lifting the exponent" help here ?

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    $\begingroup$ Does it help that $n^2+n+1\mid n^3-1$? $\endgroup$ Mar 26 at 12:31
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    $\begingroup$ For small $n$, I have $\color{red}3p^2\mid n^n+(n+1)^{n+1}$. I can neither prove it nor find a counterexample. $\endgroup$
    – mathlove
    Apr 1 at 8:32

1 Answer 1

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This can be proven using methods from Squarefree values of trinomial discriminants by David W. Boyd, Greg Martin and Mark Thom. Specifically, we can establish the following:

Proposition. Let $n\equiv 5\pmod 6$, then $$ \left(\frac{n^2-n+1}{3}\right)^2\text{ divides }n^n+(n-1)^{n-1}.\tag{*} $$

Proof. This follows the proof strategy of Proposition 4.3 in the cited paper. We set $m=\varepsilon=1$, $\varepsilon'=-1$ to show that $|\operatorname{Disc}(x^n+x-1)|=n^n+(n-1)^{n-1}$ and $x^n+x-1=(x^2-x+1)h(x)$.

This proposition aligns with your statement by shifting $n\mapsto n+1$.

Note: An even stronger assertion noted by @mathlove in the comments can be derived using the same approach. Just note that Lemma 2.1 implies a stronger divisibility: $$\operatorname{Disc}(x^2-x+1)\operatorname{Res}(x^2-x+1,h(x))^2 \text{ divides } |\operatorname{Disc}(x^n+x-1)|.$$ Now just apply $\operatorname{Disc}(x^2-x+1)=-3$.

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