Berkeley Math Circle Monthly Contest 5

Solution:
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 $p \leq 1 + 1 + 1 = 3$.