Mastering quadratic

For $x^2+7x-14(q^2+1)=0$ to have integer solutions, its discriminant $7^2-(-4)14(q^2+1) = 7(7+8(q^2+1))$ must be a perfect square. This means that $7+8(q^2+1)$ must be a multiple of 7 or $7+8(q^2+1) \bmod 7 = 0$ . But $7+8(q^2+1) \equiv 0 + q^2 + 1 \text{ (mod 7)}$ , $\implies 7+8(q^2+1) \bmod 7 = 1, 2, 3, 5$ , since the possible values of $q^2 \bmod 7 = 0, 1, 2, 4$ . Therefore, $x^2+7x-14(q^2+1)=0$ has $\boxed 0$ integer solution.