# Asymptotics on the largest prime for which $x^n+1\equiv y^n$ has no nonzero solution

It $\let\epsilon\varepsilon\let\leq\leqslant\let\geq\geqslant$is a well known result that for every $n\in\mathbb N$, $x^n+1\equiv y^n\pmod p$ is non-trivially solvable for sufficiently large primes $p$. Let $P_n$ denote the largest prime for which it is unsolvable. Ramsey theory gives (I think the proof was first given by Schur, if I'm not mistaking) $$P_n<R(\underset{n\;3\text{'s}}{\underbrace{3,\ldots,3}}),$$ where $R(\ldots)$ denotes the Ramsey number. Since $$R(\underset{n\;3\text{'s}}{\underbrace{3,\ldots,3}})\leq2-n+n\cdot R(\underset{n-1\;3\text{'s}}{\underbrace{3,\ldots,3}})\leq n\cdot R(\underset{n-1\;3\text{'s}}{\underbrace{3,\ldots,3}})$$ for $n\geq2$ and $R(3,3,3)=17$, the best explicit bound I can get from this is $P_n<\frac{17}6\cdot n!$.

A heuristic approach goes as follows: there are at least $\frac{p-1}n$ nonzero $n$th powers modulo primes $p$. The 'probability' that none of these yields a solution to the congruence is thus at most $\left(1-\frac{\frac{p-1}n}p\right)^{\frac{p-1}n}$, if we assume the $n$th powers to be distributed uniformly. If we want to bound this from above by a fixed $\epsilon>0$, we have \begin{align*}\left(1-\frac{\frac{p-1}n}p\right)^{\frac{p-1}n}&\leq\epsilon\\ \frac{p-1}n\log\left(1-\frac1n\right)\approx\frac{p-1}n\log\left(1-\frac{p-1}{np}\right)&\leq\log\epsilon\\ p-1&\lessapprox\frac{n\log\epsilon}{\log\left(1-\frac1n\right)}=n^2\log\epsilon+O(n).\end{align*} I'm not sure if these calculations make any sense at all, but from this I would conjecture that $P_n=O(n^2)$.

Finally, my question:

Is it known that $P_n=O(n^2)$? If not, what is the best known bound, or what do heuristics tell us?

From the Hasse Weil Bound, it follows that the number of solutions $N_p$ to $x^n + y^n = z^n$ in $\mathbb{F}_p$ satisfies $$|N_p - (p + 1)|\le 2g\sqrt{p}$$
where $g$ is the genus. By the genus degree formula, $g\le\frac{(n - 1)(n - 2)}{2}$
Therefore, for all $p> \left((n - 1)(n - 2)\right)^2$, $N_p>0$. This doesn't give your bound of $O(n^2)$ and instead gives $O(n^4)$