For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?

Let us try with this approach:

$\begin{align} 4x^2+8x+5 &= 4(x+1)^2+1 &\equiv 0 &\quad(\text{mod} 13)\\ &\Rightarrow 4(x+1)^2 &\equiv 12 &\quad(\text{mod} 13)\\ &\Rightarrow (x+1)^2 &\equiv 3 &\quad(\text{mod} 13)\end{align}\\$

Substituting for $x+1 = f$ we are looking to take the square-root of 3 modulo 13. The following holds true: $$n^2 \equiv (13-n)^2 \quad (\text{mod} 13)$$

It suffices to look at numbers from 0 to 6:

$\begin{align} 0^2 &\equiv 0 (\text{mod} 13),\\ 1^2 &\equiv 1 (\text{mod} 13),\\ 2^2 &\equiv 4 (\text{mod} 13),\\ 3^2 &\equiv 9 (\text{mod} 13),\\ 4^2 &\equiv 3 (\text{mod} 13),\\ 5^2 &\equiv 12 (\text{mod} 13),\\ 6^2 &\equiv 10 (\text{mod} 13). \end{align}$

Thus numbers of the form $f = 13m + 4$ solve the problem. Then also $f = 13m + 9$ will solve the problem. Since $x = f-1$ our solutions are numbers of the form: $x \in \{13m + 3, 13m +8, \text{where } m\geq 0\}$.