Prove that if $p$ is a prime number and $p\neq3$, then $3$ divides $p^2+2$

Question

Can someone help me with this? I'm not sure how to approach it.. anything would be helpful!

Prove that if $p$ is a prime number and $p\neq 3$, then $3$ divides $p^2+2$

In my textbook the hint it gives states that: when $p$ is divided by $3$, the remainder is either $0, 1$, or $2$. That is, for some integer $k, \; p=3k$ or $p=3k+1$ or $p=3k+2$.

Answer 1

The next step is to realize from your hint that since $p$ is prime (and is not $3$), it cannot be in the form $3k$. Now what happens when you square a number of the form $3k+1$ or $3k+2$?

Answer 2

Since $\gcd(p,3)=1$, Euler-Fermat tells you that $p^2\equiv1\pmod{3}$. Thus $$p^2+2\equiv1+2\equiv0\pmod{3}.$$

Answer 3

Notice that $\,3\,$ divides one of the three consecutive integers $\,p-1,\, p,\, p+1,\,$
so if $\,3\nmid p\,$ then $\, 3\mid (p\!-\!1)(p\!+\!1) = p^2\!-1,\,$ so $\ 3\mid (p^2\!-1)+3 = p^2+2$

Answer 4

Note that $[p\in\mathbb{P}]\wedge[p\neq3]\implies[p=2]\vee[p\equiv1\pmod6]\vee[p\equiv5\pmod6]$.

---

Therefore, consider the following cases:

• $p=2 \implies p^2+2=6 \implies 3|p^2+2$
• $p\equiv1\pmod6 \implies p^2+2\equiv1^2+2\equiv3\pmod6 \implies 3|p^2+2$
• $p\equiv5\pmod6 \implies p^2+2\equiv5^2+2\equiv27\equiv3\pmod6 \implies 3|p^2+2$

---

Clarification: $n\equiv3\pmod6\implies\exists{k}\in\mathbb{Z}:n=6k+3=3(2k+1)\implies3|n$.

Answer 5

Note $p^2+2 =(p-1)(p+1)+3$. If p is not divisible by 3, then either $(p-1)$ or $(p+1)$ is divisible by 3.