<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're on the right track.  
When $x=13i+3$, $y=4(13i+3)^2+8(13i+3)+5= 4(169i^2+78i+9)+8(13i+3)+5$.
Leaving out some multiples of $13$, this is $4\times 9+8\times3+5=36+24+5=65=5 \times 13$.
This shows that, when $x=13i+3$, $y$ is a multiple of $13.$
(I'll leave $x=13i+8$ as an exercise.)
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