Thailand Mathematical Olympiad

Suppose for the sake of contradiction that an integer pair $(x, y)$ satisfies $2560x^2 + 5x + 6 = y^5$. Thus, $6 \equiv y^5 \equiv y \pmod{5}$, and so $y \equiv 1 \pmod{5}$. Let $k$ be an integer such that $y = 5k + 1$. We now have
$$2560x^2 + 5x + 6 = (5k + 1)^5 = 1 + \sum_{j=1}^{5} \binom{5}{j} (5k)^j.$$
Dividing the above equation by $5$ yields
$$512x^2 + x + 1 = 5^4k^5 + 5^4k^4 + 10 \cdot 5^2k^3 + 10 \cdot 5k^2 + 5k.$$
Taking the above equation modulo $5$, we obtain $2x^2 + x + 1 \equiv 0 \pmod{5}$. This makes $1 \equiv 2x^2 - 4x + 2 \equiv 2(x - 1)^2 \pmod{5}$, and so $3 \equiv (x - 1)^2 \pmod{5}$. This is a contradiction since $3$ is not a square modulo $5$.