Can $7n + 13$ ever equal a square? If not, why not? Can it be proved? And if it can be proved that it does equal a square (which I doubt), what is the smallest value for which this occurs?  
 A: A square can only be $0,1,4,2 \mod(7)$. But $7n+13 \equiv 6 \mod(7)$ which is never possible. So $7n+13$ can never be a square.
A: \begin{align}
1^2 &\equiv 1\pmod 7 \\
2^2 &\equiv 4\pmod 7 \\
3^2 &\equiv 2\pmod 7 \\
4^2 &\equiv 2\pmod 7 \\
5^2 &\equiv 4\pmod 7 \\
6^2 &\equiv 1\pmod 7 \\
7^2 &\equiv 0\pmod 7 \\
8^2 &\equiv 1\pmod 7 \\
9^2 &\equiv 4\pmod 7 \\
10^2 &\equiv 2\pmod 7 \\
11^2 &\equiv 2\pmod7\\
12^2 &\equiv 4\pmod7 \\
& \space\space\space\space\space\space\space\vdots
\end{align}  
It repeats, so a perfect square can only be 0,1,2,4 mod 7    
$7n+13 \equiv 6 \mod 7$  which is clearly not a perfect square.
A: The answer by Prathyush Poduval is entirely correct, but the following perhaps goes a little deeper.
There can be at most $3$ non-zero quadratic residues $\pmod 7$. To see this, note that if $k$ is any non-negative integer and $n$ is any of the $6$ integers such that $0<n<7$, and $n^2 \equiv m \pmod 7$ then:
$$(7k+n)^2 \equiv 7(7k^2+2kn)+n^2 \equiv n^2 \equiv m \pmod 7$$
$$(7-n)^2 \equiv 7(7-2n)+n^2 \equiv n^2 \equiv m \pmod 7$$
Hence the number of non-zero quadratic residues is at most $6/2 = 3$.  To find all the non-zero quadratic residues it is therefore only necessary to consider:
$$1^2 \equiv 1 \pmod 7$$
$$2^2 \equiv 4 \pmod 7$$
$$3^2 \equiv 2 \pmod 7$$ 
