Let $p \geq 5$ and prime. Show $p^2 + 2$ is divisible by three.

Question

I know I have to use the division algorithm to put into the form $p^2 + 2 = 3q + r$ but everything I've tried after that has lead me to a dead end. I've mainly been trying to show $r=0$ or to make the right side a product. Any help or hints would be appreciated.

Answer 1

If $p\geq 5$ is prime then $p=3n+k$ where $k\in\{1,2\}$ so that $$ 2+p^2=2+(3n+k)^2=2+9n^2+6nk+k^2=9n^2+6nk+3+(k-1)(k+1). $$ With $k\in\{1,2\}$, every term on the rightmost expression above is divisible by $3$ so $2+p^2$ is divisible by $3$.

Answer 2

If you have already covered Fermat's little theorem, you could use $p^2\equiv 1\pmod 3$, so $p^2+2\equiv1+2\equiv0\pmod 3$ and therefore $p^2+2$ is divisible by 3.

Answer 3

Do you know about congruences?

If so, then since $p > 3$ but $3 \nmid p$, you have $p^2 \equiv 1 \bmod 3$. Hence $p^2 + 2 \equiv 0 \bmod 3$, which is what you want.

Answer 4

Every prime larger than $3$ can be expressed in the form $6k \pm 1$. If $p = 6k \pm 1$, then

\begin{align*} p^2 + 2 & = (6k \pm 1)^2 + 2\\ & = 36k^2 \pm 12k + 1 + 2\\ & = 36k^2 \pm 12k + 3\\ & = 3(12k^2 \pm 4k + 1) \end{align*} so $3|p^2 + 2$.