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

For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13

if x = 8, y = 325 which is divisible by 13

if x = 16, y = 1157 which is divisible by 13

if x = 21, y = 1937 which is divisible by 13

I am guessing that values of x = 13i + 3 or x = 13i + 8 where i is an integer will result in a value of y that is evenly divisible by 13.

How do you prove that x = 13i + 3 or x = 13i + 8 will result in a value of y that is evenly divisible by 13?

Is there a general proof to find values of x that will result in a value of y that is evenly divisible by an odd integer p?
Let us try with this approach:

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

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:

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

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: $x \in \{13m + 3, 13m +8, \text{where } m\geq 0\}$.
Solution

Check that image, I have tried solving for x = 13i + 3. You can try similarly for x = 13i + 8