The cube of any number not a multiple of $7$, will equal one more or one less than a multiple of $7$

Yeah so I'm kind of stuck on this problem, and I have two questions.

1. Is there a way to define a number mathematically so that it cannot be a multiple of $7$? I know $7k+1,\ 7k+2,\ 7k+3,\ \cdots$, but that will take ages to prove for each case.

2. Is this a proof? $$(7k + a)^3 \equiv (0\cdot k + a)^3 \equiv a^3 \bmod 7$$

Thank you.

Actually there is a more general theorem.

THEOREM. Let $p = 2a+1$ be a prime number. Then $n^a \equiv \pm 1 \pmod p$ for all $n \not \equiv 0 \pmod p$

PROOF. Because $p$ is a prime number, $\mathbb Z_p$ is a field and $\mathbf U_p$ is a cyclic multiplicative group. Hence there exists an integer, $g$ such that $\mathbf U_p = \{g, g^2, g^3, \dots g^{p-1}\}$. Hence the polynomial $x^{p-1} - 1$ has $p-1$ distinct roots; namely the members of $\mathbf U_p$.

Since $\dfrac{p-1}{2} = a$, $x^{p-1} - 1 \equiv (x^a-1)(x^a+1) \pmod p$. Hence $a$ of the members of $\mathbf U_p$ are roots of $x^a-1$ and the other $a$ members of $\mathbf U_p$ are roots of $x^a+1$. The theorem follows.

By Fermat's little theorem, if $p$ is prime, $$x\not\equiv 0\pmod p \Leftrightarrow x^{p-1}\equiv 1\pmod p$$

Further,$$y^2\equiv 1\pmod p \Leftrightarrow y\equiv \pm 1\pmod p$$

As 7 is prime, we can enter the value into the first expression and the result follows from the second.

$$x^6\equiv 1\pmod 7 \Rightarrow x^3\equiv \pm 1\pmod 7$$

Calculation of cases is unnecessary.