<s>[INST] If $a$ is not divisible by $7$, then $a^3 - 1$ or $a^3 + 1$ is divisible by $7$ Determine is, in general, true or false. Recall that a universal statement is true if it is true for all possible cases while it is false if there is even one counterexample. Be prepared to prove that your answer is correct by supplying a proof or counterexample, whichever is appropriate.
If $a$ is not divisible by $7$, then $a^3 - 1$ or $a^3 + 1$ is divisible by $7$.
Solution:
According to Fermat's Little Theorem, we know that if $p$ is prime and $a$ is an integer and $p \nmid a$, then $a^{p - 1} \equiv 1 \mbox{ (mod $p$)}$.
Hence, using the above theorem we get that
$$
\begin{array}{rcl}
a^{7-1} & \equiv & 1 \mbox{ (mod 7)}\\
a^6 & \equiv & 1 \mbox{ (mod 7)}\\
 a^6  -1 & = & 7k \mbox{ where $k$ is an integer}\\
(a^3-1)(a^3+1) & = & 7k
 \end{array}
$$
Since $k$ is integer
$(a^3-1)$ or $(a^3+1)$ is divisible by $7$.
Could you check it for me please is it correct or not?
 [/INST] It's correct. An alternative can be:
You could realize that 
$$x^3 \equiv 0, 1, -1 \pmod 7$$
So, if $7 \nmid x$ then $x^3 \equiv 1, -1 \pmod 7$.
By the definition of congruences we have, either $x^3 + 1 = 7k$ or $x^3 - 1 = 7k$.
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