For any prime $p>3$ show that 3 divides $2p^2+1$ Does anyone know how to show this preferable without using modular
For any prime $p>3$ show that 3 divides $2p^2+1$ 
 A: Without using mod: note $\, 2p^2+1 = 2(p^2-1)+3\,$ so it suffices to show that $\,3\mid p^2-1.\,$ One of the $\,3\,$ consecutive integers $\,p-1,\,p,\,p+1$ must be a multiple of $\,3$. It cannot be the prime $\,p> 3,\,$ therefore either $\,3\mid p-1\,$ or $\,3\mid p+1,\,$ hence $\,3\mid(p-1)(p+1) = p^2-1\ \ $ QED 
Using mod, it is easier:  $\,\ 3\nmid p\,\Rightarrow\ {\rm mod}\ 3\!:\ p\equiv \pm 1\,\Rightarrow\, p^2\equiv 1\,\Rightarrow\ 2p^2+1\equiv 3\equiv 0 \ \ $ QED
Remark $\ $ So the result is true for any integer $p$ not divisible by $3$. It's equivalent to the special case $\,q=3\,$ of Fermat's little Theorem: $\,q\nmid p\,\Rightarrow\,p^{q-1}\equiv 1\pmod q$
A: If you want to avoid modular arithmetic, you can easily do the explicit calculations, distinguishing the two cases


*

*$p = 1 + 3 k$, then $$2 p^{2} + 1 = 2 (1 + 6k + 9 k^{2}) + 1 = 3 (1 + 4k + 6 k^{2})$$

*$p = 2 + 3 k$, then $$2 p^{2} + 1 = 2 (4 + 12k + 9 k^{2}) + 1 = 3 (3 + 8k + 6 k^{2})$$

A: For any prime $p>3$ we have $p\equiv 1\mod 3$ or $p\equiv 2\mod 3$ and in the two cases we have
$$2p^2+1\equiv 0\mod 3$$
A: Since $p>3$ it holds that $p\equiv 1,2\pmod{3}$. Then $p^2\equiv 1\pmod{3}$ and then $2p^2\equiv 2\pmod{3}$. Adding $1$ yields the result.
A: Generalization:
If $q$ is prime not dividing integer  $\displaystyle m,(m,q)=1\implies m^{q-1}\equiv1\pmod q$ (by Fermat's Little Theorem)
$\displaystyle(q-1)m^{q-1}\equiv (-1)\cdot1\pmod q\equiv-1\iff (q-1)m^{q-1}+1\equiv0\pmod q$
Here $q=3$
