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