If $p \geq 11$ is a prime then either $p^3-1$ or $p^3+1$ is divisible by $14$

Question

True or False:

If $p$ is a prime greater than or equal to $ 11$, then either $p^3-1$ or $p^3+1$ is divisible by $14$

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My try

The statement is true. In order to prove this, we prove either $$p^3\equiv1(\;\text{mod}\;14\;)\;\;\text{or}\;\;p^3\equiv -1(\;\text{mod}\;14\;)$$ Since $\gcd(14,p)=1$, by Euler's theorem , $$p^{\large \phi(14)}\equiv1(\;\text{mod}\;14\;)$$ That is $$p^6\equiv1(\;\text{mod}\;14\;)$$ From the above, can I conclude the desired result ? Any help ?

Answer 1

Consider that for $p\not\equiv0\pmod7$ $$ \left(p^3-1\right)\left(p^3+1\right)=p^6-1\equiv0\pmod7 $$ So either $7\mid p^3-1$ or $7\mid p^3+1$. Since $p$ is odd, $2\mid p^3-1$ and $2\mid p^3+1$. Thus, $$ 14\mid p^3-1\quad\text{or}\quad14\mid p^3+1 $$

Answer 2

Hint: $p^6\equiv 1 \mod 14$ means that $p^6-1 = (p^3-1)(p^3+1)$ is divisible by $14$. We have $2\mid p^3+1$ and $2\mid p^3-1$, since the numbers $p^3\pm 1$ are even. Moreover, $7$ is prime and so $7\mid p^3+1$ or $7\mid p^3-1$. Thus $14$ divides $p^3+1$ or $p^3-1$.

Answer 3

Since the other solutions helped you from what you had, I have another "bruteforce" solution:

If $p > 11$ is prime, $2\not \mid p \land 7\not \mid p \implies \bar{p} \in \{\bar{1}, \bar{3}, \bar{5}, \bar{-5}, \bar{-3}, \bar{-1}\}$. $1^3 \equiv 1, 5^3 \equiv -1, 3^3 \equiv -1$ and we're done.