$p^2 + 5pq + 4q^2 = x^2$ $p^2 + 4pq + 4q^2 + pq = x^2$ $(p + 2q)^2 + pq = x^2$ $pq = x^2 - (p + 2q)^2 = (x + p + 2q)(x - p - 2q)$ Since $x + p + 2q > x - p - 2q$, and $p, q$ are prime: Case 1: $x + p + 2q = p, x - p - 2q = q$ $\implies x + 2q = 0$ which is impossible since $x, q > 0$ Case 2: $x + p + 2q = q, x - p - 2q = p$ $\implies x + p + q = 0$ which is impossible since $x, p, q > 0$ Case 3: $x + p + 2q = pq, x - p - 2q = 1$ $\implies 2x = pq + 1$ $\implies x = (pq + 1)/2$ $\implies (pq + 1)/2 - p - 2q = 1$ $\implies pq + 1 - 2p - 4q = 2$ $\implies pq - 2p - 4q - 1 = 0$ $\implies p(q - 2) - 4q + 8 = 9$ $\implies p(q - 2) - 4(q - 2) = 9$ $\implies(p - 4)(q - 2) = 9$ Case 1: $p - 4 = 1, q - 2 = 9$ $\implies p = 5, q = 11$ Case 2: $p - 4 = 3, q - 2 = 3$ $\implies p = 7, q = 5$ Case 3: $p - 4 = 9, q - 2 = 1$ $\implies p = 13, q = 3$ $\implies \boxed{(p, q) = (5, 11), (7, 5), (13, 3)}$