$a^4+b^4+c^4\equiv a^2(b+c)^2+b^2(c+a)^2+c^2(a+b)^2 \equiv 2(a^2b^2+b^2c^2+c^2a^2)$$+2abc(a+b+c)\equiv 2(a^2b^2+b^2c^2+c^2a^2)(modp)$ Also, $a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$ Hence, $p|a^2+b^2+c^2$ and we get $(2b+c)^2\equiv-3c^2(modp) $ $-3$ isn't a quadratic residue $mod p$ since $p\equiv-1(mod3)$ and hence $p|c$ and similarly, $p$ divides the others

lajanugen wrote: and we get $(2b+c)^2\equiv-3c^2(modp) $ Umm. I don't see how do you get that.

MeKnowsNothing wrote: A prime $p$ and integers $a,b,c$ satisfies \[6\mid p+1,\quad p\mid a+b+c,\quad p\mid a^4+b^4+c^4.\] Prove that $p\mid a,b,c$. $a \equiv -b-c \bmod p \Rightarrow a^4+b^4+c^4 \equiv (b+c)^4 + b^4+c^4 $ $= 2(b^4+c^4+2b^3c+2bc^3+3b^2c^2) = 2(b^2+bc+c^2)^2$, so $p \mid b^2+bc+c^2 \mid b^3-c^3$ Then $b^3 \equiv c^3 \Rightarrow \left ( b^3 \right )^{(p+1)/3} \equiv \left ( c^3 \right )^{(p+1)/3} \iff b^2 \equiv c^2$, and hence $bc^2 \equiv c^3, b^3 \equiv b^2c \bmod p$, and therefore $b \equiv c \bmod p$, or $p \mid b,c$. $p \mid b,c$ gives $p \mid a$, and we are done. $b \equiv c \bmod p$ gives: $a \equiv -2b, a^4 \equiv -2b^4 \bmod p \Rightarrow (-2b)^4 \equiv a^4 \equiv -2b^4 \bmod p \iff$ $p \mid 18b^4 \iff p \mid b$, and therefore $p \mid b,c$, and therefore $p \mid a$, so we are done.

I found this solution on the Internet which was given by Matti K. Sinisalo: As $p|a^2+ab+b^2$ then for some integer $k$ we have $a^2 + ab + b^2 -kp = 0$ which is equivalent to $b^2 + ab + (a^2-kp) = 0$ Its discriminant must be a square so there is an integer $d$ satisfying $4kp- 3a^2 = d^2$. Therefore $d^2 \equiv -3a^2 \pmod{p}$ Let $e=a^{-1}\pmod p$. Then $(de)^2 \equiv -3\pmod{p}$. Thus $ -3\equiv p-3 \pmod p$ must be a quadratic residue $\pmod{p}$. But $(-3/p) = (p/3) = (-1/3) = -1$ as $6|p+1$ gives $3|p+1$ or $p\equiv -1 \pmod{3}$. But then $-3$ is a quadratic non-residue $\pmod{p}$ which is a contradiction.

Mathias_DK wrote: Then $b^3 \equiv c^3 \Rightarrow \left ( b^3 \right )^{(p+1)/3} \equiv \left ( c^3 \right )^{(p+1)/3} \iff b^2 \equiv c^2$, and hence $bc^2 \equiv c^3, b^3 \equiv b^2c \bmod p$, and therefore $b \equiv c \bmod p$, or $p \mid b,c$. I don't see how do you argue that $\left ( b^3 \right )^{(p+1)/3} \equiv \left ( c^3 \right )^{(p+1)/3} \iff b^2 \equiv c^2$ and the part $b^2 \equiv c^2$, and hence $bc^2 \equiv c^3, b^3 \equiv b^2c \bmod p$. And if $b\equiv c\pmod p$ then I think $p\mid b-c$, not $p\mid b,c$. Mathias_DK wrote: $a^4 \equiv -2b^4 \bmod p$ Why?

MeKnowsNothing wrote: I don't see how do you argue that $\left ( b^3 \right )^{(p+1)/3} \equiv \left ( c^3 \right )^{(p+1)/3} \iff b^2 \equiv c^2$ $3 \mid p+1$, so $\frac{p+1}{3}$ is a natural number. And $x^{p+1} \equiv x^2$ by Fermat's Little Theorem. It should be $\Rightarrow$ instead of $\iff$ of course. MeKnowsNothing wrote: $b^2 \equiv c^2$, and hence $bc^2 \equiv c^3, b^3 \equiv b^2c \bmod p$. And if $b\equiv c\pmod p$ then I think $p\mid b-c$, not $p\mid b,c$. If $b^3 \equiv c^3$ then $b^2 \cdot b \equiv c^3$. Since $b^2 \equiv c^2$ we get $bc^2 \equiv c^3$. This means $p \mid bc^2-c^3 \iff p \mid c$ or $b \equiv c \bmod p$. So we get either $p \mid b,c$ or $b \equiv c \bmod p$. I handle the two cases seperatly and come to the conclusion that $p \mid a,b,c$ in both cases. MeKnowsNothing wrote: Mathias_DK wrote: $a^4 \equiv -2b^4 \bmod p$ Why? I took the case $b \equiv c \bmod p$. Hence $p \mid a^4+b^4+c^4 \equiv a^4+2b^4$, and therefore $a^4 \equiv -2b^4$.

MeKnowsNothing wrote: I found this solution on the Internet which was given by Matti K. Sinisalo: As $p|a^2+ab+b^2$ then for some integer $k$ we have $a^2 + ab + b^2 -kp = 0$ which is equivalent to $b^2 + ab + (a^2-kp) = 0$ Its discriminant must be a square so there is an integer $d$ satisfying $4kp- 3a^2 = d^2$. Therefore $d^2 \equiv -3a^2 \pmod{p}$ Let $e=a^{-1}\pmod p$. Then $(de)^2 \equiv -3\pmod{p}$. Thus $ -3\equiv p-3 \pmod p$ must be a quadratic residue $\pmod{p}$. But $(-3/p) = (p/3) = (-1/3) = -1$ as $6|p+1$ gives $3|p+1$ or $p\equiv -1 \pmod{3}$. But then $-3$ is a quadratic non-residue $\pmod{p}$ which is a contradiction. An easier way to see that $-3$ must be a quadratic residue if $p \nmid a,b$: $p \mid 4(a^2+ab+b^2) = (2a+b)^2+3b^2$