Is it possible to write any integer $p^n (n\in \mathbb{N})$ in the form $6m\pm 1$.

Question

Any integer $p^n (n\in \mathbb{N})$ is of the form $6m\pm 1$, where $p$ is odd prime.

Any prime $p>2$ is of the form $6m\pm 1, m\in \mathbb{N}$.
Now, $p^n=(6m+ 1)^n=\sum\limits_{r=0}^n {\binom n r} (6m)^{n-r}=\sum\limits_{r=1}^n {\binom n r} (6m)^{n-r}+1=6k+ 1$ $p^n=(6m-1)^n=\sum\limits_{r=0}^n (-1)^r {\binom n r} (6m)^{n-r}=\sum\limits_{r=1}^n {\binom n r} (-1)^r (6m)^{n-r}- 1=6k- 1$

Hence $p^n=6k\pm 1$. Is this procedure correct?

Answer 1

Minor nitpick - $p=3$ does not work.

Your procedure is correct. However, we can greatly simplify this using mods: for all $p > 3$, we have $p \equiv \pm 1 \pmod 6$, so $p^n \equiv (\pm 1)^n$ which is clearly $1$ or $-1$.

Answer 2

Your way of argument is correct, but there are some flaws in the technicalities.

• $p>3$

• starting point of summation

• transition from second step to third step is incorrect (should be from $n$ to $n-1$)

A correct proof would be $$p^n=(6m\pm 1)^n=\sum\limits_{r=0}^n {\binom n r} (6m)^{n-r}(\pm1)^r=(\pm1)^n+\sum\limits_{r=0}^{n-1} {\binom n r} (6m)^{n-r}(\pm1)^r=6k\pm 1$$ where $$k=\frac16\left[\sum\limits_{r=1}^{n-1} {\binom n r} (6m)^{n-r}(\pm1)^r\right]=\sum\limits_{r=1}^{n-1} {\binom n r}6^{n-r-1}\cdot m^{n-r}(\pm1)^r\in\Bbb Z$$ since $n-r-1\ge0$.