Saudi Arabian Mathematical Competitions

Suppose that $b \le c$, we distinguish two cases regarding the value of $p$

1. if $p \ne 2$ implies $p$ is odd, so $p^2 \equiv 1 \pmod 8$. It follows that
$$a^2 + b^2 + c^2 \equiv 9a^2 + 17b^2 + 17c^2 \equiv 73p^2 + 6 \equiv 7 \pmod 8. \quad (2)$$
Otherwise, $a^2 + b^2 + c^2 \equiv 0, 1, 2, 3, 4, 5, 6 \pmod 8$, which is contracts (2).

2. if $p = 2$ then $9a^2 + 17b^2 + 17c^2 = 73p^2 + 6 = 298$. Since $a, b, c \ge 1$, we have
$$298 = 9a^2 + 17b^2 + 17c^2 \ge 26 + 17c^2$$
, which means $c \le 4$. We investigate two cases
* if $b \le c = 4$ then $9a^2 + 17b^2 = 26$, which means $a = b = 1$.
* if $b \le c \le 3$ then it is easy to check that there is no solutions satisfied.
Hence, the only solution satisfied (1) is $(a, b, c, p) = (1, 1, 2, 2)$. $\square$