The 14th Thailand Mathematical Olympiad

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