<s>[INST] 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?
 [/INST] You are looking for solutions to $4x^2+8x+5\equiv 0 \pmod {13}$.  You can do the usual quadratic formula and find $x=\frac {-8 \pm \sqrt{64-80}}8$.  Don't let the negative number under the square root bother you because $\pmod {13}$  we have $-16 \equiv 10$.  You need to find whether $10$ is a square $\bmod 13$.  One way is to just try them.  Here we find $6^2 \equiv 7^2 \equiv 10 \pmod {13}$.  Now noting that $8\cdot 5 \equiv 1 \pmod {13}$ we can say that $x=5(-8\pm 6)\equiv 3,8 \pmod {13}$ 
There is a whole theory of quadratic reciprocity that says when things are square roots in rings of integers.  I have not studied it.
</s>