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.


*

*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.


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

Thank you.
 A: We only need to check the cubes $1^3,2^3,\ldots ,6^3$ modulo $7$. The result is
$$
1^3=2^3\equiv 1\bmod 7, 3^3\equiv -1 \bmod 7, 4^3\equiv 1\bmod 7,5^3\equiv -1 \bmod 7, 6^3\equiv -1 \bmod 7.
$$
A: Just to write the complete steps of the proof.
Well, first as you said, we have $$(7k + a)^3 \equiv (0*k + a)^3 \equiv a^3 \bmod 7.$$
Then, we need to show that the statement is true for each $a=1,\ 2,\ 3,\ 4,\ 5,\ 6$. For this, as Joffan stated, you may compute the cubes that leads to 
$1^3\equiv 1\bmod 7,$
$2^3\equiv 1\bmod 7,$
$3^3\equiv -1 \bmod 7,$
$4^3\equiv 1\bmod 7,$
$5^3\equiv -1 \bmod 7,$ 
$6^3\equiv -1 \bmod 7.$
This completes the proof.
A: Say $a$ is a number not a multiple of $7$.
Then we have that $$a \equiv \pm 1 ,\pm 2,\pm 3 \pmod7 $$
$$\implies a^3 \equiv (\pm 1)^3 ,(\pm 2)^3,(\pm 3)^3\equiv \pm 1 ,\pm 8,\pm 27 \pmod7 $$
$$\implies a^3 \equiv \pm 1 \pmod7 $$
Hope this is the shortest proof possible.
A: You can first show this for $\{1,2,3,4,5,6\}$ and then show that applies to all non-multiples of $7$.
\begin{align}
1^3 &= 1\\
2^3 &= 8 = 7+1\\
3^3 &= 27 = 4\cdot 7 -1\\
4^3 &= 64 = 9\cdot 7+1\\
5^3 &= 125 = 18\cdot 7 -1\\
6^3 &= 216 = 31\cdot 7 -1 \\
\end{align}
Then use your construction to complete the proof for all numbers not multiples of $7$,
A: You only have to consider: $1^3 = 1, 2^3 = 1, 3^3 = -1, 4^3 = 1, 5^3 = -1, 6^3 = -1 \pmod 7$ . 
A: 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.
A: 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.
A: That's part of a proof.  For a full proof, you should check the cases $a=1,2,3,4,5,6$.  Actually, you only need three of these since $(7-a)^3 \equiv (-a)^3 \equiv -a^3 \mod 7$.
