Can $7n + 13$ ever equal a square?

Question

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?

Answer 1

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.

Answer 2

\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.

Answer 3

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:

$$1^2 \equiv 1 \pmod 7$$ $$2^2 \equiv 4 \pmod 7$$ $$3^2 \equiv 2 \pmod 7$$