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?

Completing the square;
$$4x^2+8x+5=(2x+2)^2+1,$$
 so solving the congruence $\;4x^2+8x+5\equiv 0\mod 13$ amounts to solving
$$\bigl(2(x+1)\bigr)^2\equiv -1\mod 13,$$
and ultimately to finding the square roots $y$ of $-1\bmod 13$ (we know this is possible because $13\equiv 1\mod 4$).
Note that $5^2\equiv -1\mod 13$, hence the other square root  is $-5\equiv 8$, and since $2^{-1}\equiv 7\mod 13$, the solutions in $x$ are
$$ x\equiv 7\cdot\pm 5 -1\equiv 8,3\mod 13.$$
