Find all primes $p$ such that there exist integers $a, b, c$, and $k$ satisfying the equations

$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}=p \\
& a^{4}+b^{4}+c^{4}=k p .
\end{aligned}
$$
The answers are 2 and 3. It is clear that for them we can use the values $a=b=k=1$ and $c=0$ or 1 respectively. Assume $p \geq 5$. Subtracting twice the second equation from the square of the first, we find that $p$ divides

$$
2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4},
$$

an expression which factors as

$$
(a+b+c)(a+b-c)(a-b+c)(-a+b+c) .
$$

One of the four factors must be divisible by $p$. By flipping the signs on $a, b$, and $c$, we can assume it is the first one. Using the inequality $x^{2}-x \geq 0$ (for integers $x$, equality holding at $x=0$ and $x=1$ ), we have

$$
p \leq a+b+c \leq a^{2}+b^{2}+c^{2}=p .
$$

This equality condition can only hold if $a, b$, and $c$ are each equal to their squares, implying that they are at most 1 and $p \leq 1+1+1=3$.