2

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?

John Glenn
  • 2,323
  • 11
  • 25
Dolphin
  • 79
  • Above and beyond the answers you've gotten, the question of 'is $a$ a square mod $b$?' (and note that this is equivalent to the question you've posed; can you see why?) is a surprisingly deep one touching several areas of number theory; the magic phrase to look for is quadratic residue. (https://en.wikipedia.org/wiki/Quadratic_residue) – Steven Stadnicki Mar 28 '18 at 17:44

3 Answers3

4

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.

Jim Haddocc
  • 2,485
  • Literally just figured that out myself about 10 seconds ago. Never thought to look at the answers modulo 7. But this notwithstanding, is there a deeper reason even beyond this, or an alternative explanation. – Dolphin Mar 28 '18 at 17:43
  • @Dolphin well, it depends on what you mean by "deeper reason". When you are given things like $7n+13$, it's usually meant to be solved the modulo way, which is satisfactory for me. I'll edit my post if anything else comes to mind – Jim Haddocc Mar 28 '18 at 17:44
0

\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.

kayush
  • 2,441
0

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:

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

Adam Bailey
  • 4,197