How to prove properties about a specific modular arithmetic equivalence

Question

Ever since I was introduced to modular arithmetic, I've had some trouble with it. I think it uses a part of my brain that I haven't used often. Anyways, I've been thinking about this specific equivalence: $$a^3 \equiv 5 \, (\text{mod } 7)$$ And I have a hunch that no $a$ exists s.t. this equivalence is true. Simulating it, it's clear that there is a pattern: 6, 1, 6, 6, 0, 1, 1, 6, 1, 6, 6, 0, 1, 1, 6, 1, 6, 6, 0...

But I can't figure out how to formally prove 1. that this pattern is the actual pattern and as an extension, 2. that the equivalence above doesn't hold (it should be trivial if I can prove 1).

Can anyone help? Thanks so much.

Answer 1

You can prove it by calculating the value of $a^3 \bmod 7$ for $a=0,1,2,3,4,5,6$; if none of those yield 5, then you have proven the claim.

Why is this sufficient? Well, if $a \equiv b \pmod 7$, then $a^3 \equiv b^3 \pmod 7$. So, if there was any solution to $a^3 \equiv 5 \pmod 7$, then you could take $b = a \bmod 7$, and that would be another solution. Now $b$ is one of $0,1,2,3,4,5,6$, so we've proven that if there is any solution, then one of $0,1,2,3,4,5,6$ must be a solution. Conversely, if none of $0,1,2,3,4,5,6$ is a solution, then there is no solution whatsoever.

For the special case of squaring (rather than cubing), you might be interested in quadratic reciprocity, which is a more advanced technique that allows to check for the existence of solutions to such an equation. There is also cubic reciprocity, though I'm not sure whether it leads to an efficient algorithm to check for solutions when you have a cube instead of a square.