Integer Roots of Quadratic Equation

For $x^2 - 5x + k = 0$ to have integer roots, the discriminant $D = (-5)^2 - 4(1)(k) = 25 - 4k$ must be a perfect square, say $m^2$. Since $k \ge 0$, $25 - 4k \le 25$. $25 - 4k$ can be equal to $5^2, 3^2, 1^2$ (since the parity of $m^2$ must match $25$, so $m$ must be odd).

• If $25 - 4k = 25 \implies 4k = 0 \implies k = 0$. Roots are $0, 5$.
• If $25 - 4k = 9 \implies 4k = 16 \implies k = 4$. Roots are $1, 4$.
• If $25 - 4k = 1 \implies 4k = 24 \implies k = 6$. Roots are $2, 3$. Thus, there are $3$ non-negative integer values of $k$.