Lemma Let $n >1$ be a positive integer relatively prime to $6$. Then there exists $f( x,y) \in \mathbb{Z}[ x,y]$ such that $( x+y)^{n} -\left( x^{n} +y^{n}\right) =f( x,y)\left( x^{2} +xy+y^{2}\right)$. proof of the lemmaLet $g( x) =\left( x^{n} +1\right) -( x+1)^{n}$. It is sufficient to show that $g( \omega ) =0$ where $\omega $ is a primitive third root of unity. Note that $g( \omega ) =\left( \omega ^{n} +1\right) -( \omega +1)^{n} =\omega ^{n} +1-\left( -\omega ^{2}\right)^{n} =\omega ^{2n} +\omega ^{n} +1$. Since $\left\{\omega ^{2n} ,\omega ^{n}\right\} =\left\{\omega ^{2} ,\omega \right\}$, we have $\omega ^{2n} +\omega ^{n} +1=\omega ^{2} +\omega +1=0$. So we must have $g( \omega ) =0$. $p\mid a^{n} +b^{n} +c^{n} \cdots ( \star )$ $p\mid a^{2} +ab+b^{2} \cdots ( \triangledown )$ Assume that $p\mid \gcd( a,b)$. From $( \star )$ and the assumption, we have $p\mid c$, which implies $p\mid a+b+c$. So we must have $p\nmid \gcd( a,b) \ \cdots ( \clubsuit )$. From $( \triangledown )$ and $( \clubsuit )$, we have $6\mid p-1\ \cdots ( \spadesuit )$ If $\gcd( n,6) >1$, then $\gcd( p-1,n) >1$ due to $( \spadesuit )$. Suppose, henceforth, $n$ is relatively prime to $6$. From the lemma, there exists $f( x,y) \in \mathbb{Z}[ x,y]$ such that $( a+b)^{n} -\left( a^{n} +b^{n}\right) =f( a,b)\left( a^{2} +ab+b^{2}\right) \cdots ( \neptune )$. From $( \neptune )$ and $( \triangledown )$, we have $( a+b)^{n} \equiv a^{n} +b^{n}\pmod p\ \cdots ( \heartsuit )$. From $( \heartsuit )$ and $( \star )$, we have $p\mid ( a+b)^{n} +c^{n} \cdots ( \diamondsuit )$. From $( \diamondsuit )$, we have $p\mid ( a+b)^{2n} -c^{2n} \cdots ( \twonotes )$. From $( \diamondsuit )$ and $p\nmid a+b+c$, we have $p\nmid c$ and $p\nmid a+b\ \cdots ( \leftmoon )$. Assume, for the sake of contradiction, $\gcd( p-1,n) =1$. Since $\gcd( p-1,n) =1$, we have $\gcd( p-1,2n) =2\ \cdots ( \uranus )$ From $( \leftmoon )$, we have $p\mid ( a+b)^{p-1} -c^{p-1} \cdots ( \earth )$. From $( \twonotes )$ and $( \earth )$, we have $p\mid \gcd\left(( a+b)^{p-1} -c^{p-1} ,( a+b)^{2n} -c^{2n}\right)$ Note that $\gcd\left(( a+b)^{p-1} -c^{p-1} ,( a+b)^{2n} -c^{2n}\right) =( a+b)^{2} -c^{2}$ due to $( \uranus )$. So we must have $p\mid ( a+b-c)( a+b+c)$, which implies $a+b\equiv c\pmod p\ \cdots ( \eighthnote )$. From $( \eighthnote )$ and $( \heartsuit )$, we have $p\mid a^{n} +b^{n} -c^{n} \ \cdots ( \flat )$. From $( \flat )$ and $( \star )$, we have $p\mid 2c^{n}$, which implise $p\mid c$. This is clearly impossible due to $( \leftmoon )$. $\blacksquare $