Prove 7 is not a factor of any $113_n$ where the base $n>3$

Question

Actually, I was trying to prove that $131_n$ and $311_n$ were not factorable by 7 as well. However, I would believe that the proofs would be similar.

I tried working with $x^2+x+3=\frac{c}{7}$ where x represents the base; however, I could not get anywhere with it. I then jotted down the evaluations for the left side for n from 1 through about 15 and found a pattern in the differences between the results and the nearest multiple of 7.

Is there a way to prove this? or, better yet, is there a way to mathematically represent the pattern described above?

Answer 1

The equation $x^2+x+3=0$ in the field $\mathbb F_7$ has exactly two solutions (maybe in a quadratic extension of $\mathbb F_7$). These solutions are given by $$x=\frac{-1\pm\sqrt{1-12}}{2}=\frac{-1\pm\sqrt{-11}}{2}=\frac{-1\pm\sqrt{3}}{2}$$ In order to have $x\in\mathbb F_7$ we need that $3$ be a quadratic residue modulo $7$ which is not the case by either the law of quadratic reciprocity or, simply, because the squares in $\mathbb F_7$ are $\{1,4,2,0\}$.

Answer 2

If $b$ is the base you are asking if it can be that $b^2+b+3 \equiv 0 \pmod 7$ For a small number like this you can just try all the possibilities and find that none of them work.