Here is the official solution: https://cms.math.ca/Competitions/CMO/solutions/sol_2015.pdf

The first two primes $p$ are $p=5,7$. Now if $p\ge 11$ we easily see that $p$ must be of the form $6k+5\equiv 2\pmod 3$. Therefore, we can apply the lemma that $p\mid x^2+xy+y^2\equiv 0\pmod p,$ $x,y\in \mathbb{Z},$ $p\equiv 2\pmod 3$ $\Rightarrow p\mid x,y$ to get that if $p\mid a^n+b^n+c^n$ then $p\nmid a^{2n}+b^{2n}+c^{2n}$. To see why, observe that $p\mid a^n+b^n+c^n$ $\Rightarrow c^n\equiv -(a^n+b^n)\pmod p$; thus $a^{2n}+b^{2n}+c^{2n}$ $\equiv a^{2n}+b^{2n}+(a^n+b^n)^2$ $\equiv 2(a^{2n}+a^nb^n+b^{2n})\pmod p$, so if this is $\equiv 0\pmod p$, then by the lemma we would have $p\mid a^n,b^n$ and thus $p\mid a,b$, contradicting $p\nmid a,b,c$. I was wondering if this approach could be carried out to solve the problem?

Let $s$ be the number of $n$ that satisfy $a^n+b^n\equiv -1$. Then for a constant $1 \le \ell \le p-2$, if $a^n+b^n=a^{n+\ell}+b^{n+\ell}$ then $a^n(a^{\ell}-1)+b^n(b^{\ell}-1)=0$. If $ \ell \neq (p-1)/2$ this has at most two solutions $n$. Therefore in the set of differences between $n$ that satisfy, each number appears at most twice except $(p-1)/2$, which appears at most $s$ times. This is equivalent to $s^2-2s \le 2(p-2)$, and by the general formula this implies $s \le \left( \frac{1}{2} \right)(2+\sqrt{4+8(p-2)}) < 1+\sqrt{2p}$, so we are done.

JuanOrtiz wrote: Let $s$ be the number of $n$ that satisfy $a^n+b^n\equiv -1$. Then for a constant $1 \le \ell \le p-2$, if $a^n+b^n=a^{n+\ell}+b^{n+\ell}$ then $a^n(a^{\ell}-1)+b^n(b^{\ell}-1)=0$. If $ \ell \neq (p-1)/2$ this has at most two solutions $n$. Therefore in the set of differences between $n$ that satisfy, each number appears at most twice except $(p-1)/2$, which appears at most $s$ times. This is equivalent to $s^2-2s \le 2(p-2)$, and by the general formula this implies $s \le \left( \frac{1}{2} \right)(2+\sqrt{4+8(p-2)}) < 1+\sqrt{2p}$, so we are done. (I am not good at English)I have a problem, in s^2-2s every unit doublet is counted twice ,but 2(p-2) count once,what's wrong?

Amathlover wrote: JuanOrtiz wrote: Let $s$ be the number of $n$ that satisfy $a^n+b^n\equiv -1$. Then for a constant $1 \le \ell \le p-2$, if $a^n+b^n=a^{n+\ell}+b^{n+\ell}$ then $a^n(a^{\ell}-1)+b^n(b^{\ell}-1)=0$. If $ \ell \neq (p-1)/2$ this has at most two solutions $n$. Therefore in the set of differences between $n$ that satisfy, each number appears at most twice except $(p-1)/2$, which appears at most $s$ times. This is equivalent to $s^2-2s \le 2(p-2)$, and by the general formula this implies $s \le \left( \frac{1}{2} \right)(2+\sqrt{4+8(p-2)}) < 1+\sqrt{2p}$, so we are done. (I am not good at English)I have a problem, in s^2-2s every unit doublet is counted twice ,but 2(p-2) count once,what's wrong? ok，this question is solved.according to Fermat's little threom,every |（i-j)|equal p-1-|i-j|,so we only need to consider |i-j|<p-1/2

Amathlover wrote: Amathlover wrote: JuanOrtiz wrote: Let $s$ be the number of $n$ that satisfy $a^n+b^n\equiv -1$. Then for a constant $1 \le \ell \le p-2$, if $a^n+b^n=a^{n+\ell}+b^{n+\ell}$ then $a^n(a^{\ell}-1)+b^n(b^{\ell}-1)=0$. If $ \ell \neq (p-1)/2$ this has at most two solutions $n$. Therefore in the set of differences between $n$ that satisfy, each number appears at most twice except $(p-1)/2$, which appears at most $s$ times. This is equivalent to $s^2-2s \le 2(p-2)$, and by the general formula this implies $s \le \left( \frac{1}{2} \right)(2+\sqrt{4+8(p-2)}) < 1+\sqrt{2p}$, so we are done. (I am not good at English)I have a problem, in s^2-2s every unit doublet is counted twice ,but 2(p-2) count once,what's wrong? ok，this question is solved.according to Fermat's little theorm,every |（i-j)|equal p-1-|i-j|,so we only need to consider |i-j|<p-1/2

Any other ideas?

Let $q=\tfrac{p-1}{2}$. By scaling WLOG assume $c=1$. Let $S \in \mathbb{Z}/(p-1)\mathbb{Z}$ be the set of all $n$ such that $a^n+b^n \equiv -1 \pmod{p}$. We first do some "auxillary" work. If $\mathrm{ord}_p(a/b)=1 \implies a=b$, then we are looking for $n$ with $a^n \equiv -1/2 \pmod{p}$. Since $p \neq 2,3$, $-1/2 \not \in \{-1,1\} \pmod{p}$, hence at most two $n$ work. If $\mathrm{ord}_p(a/b)=2 \implies a=-b$, then we actually need $n$ even which yields the same equation. Hence $\mathrm{ord}_p(a/b) \in \{q,2q\}$. The main part of the problem is the following claim. Claim: Each difference between distinct elements of $S$ shows up at most twice, except for $q$. Proof: Suppose we had $k \neq l$ (in $\mathbb{Z}/(p-1)\mathbb{Z}$) with $k,l,k+d,l+d \in S$ for some $d \neq 0$. Upon multiplication and cancellation this implies $$a^{l+d}b^k+a^kb^{l+d}\equiv a^{k+d}b^l+a^lb^{k+d} \pmod{p} \implies a^kb^l(a^d-b^d) \equiv a^lb^k(a^d-b^d) \pmod{p}.$$Hence either $(a/b)^d \equiv 1 \pmod{p}$ or $(a/b)^{k-l} \equiv 1 \pmod{p}$. If $d \neq q$, then this means $k-l=q$, which implies the desired claim. $\blacksquare$ For any element of $S$, there are at least $|S|-2$ other elements which don't differ from it by $q$, hence we need $|S|(|S|-2) \leq 2(p-2)$. Solving for $|S|$ yields the desired result. $\blacksquare$