How do I solve this congruence?

Question

I have some difficulties solving the following congruential equation.

$3n^2 + 2 ≡ 0\pmod 5,\ \forall\ n \in Z$

If I subtract both members by $-2$, I end up getting $3n^2 = -2\pmod 5$ and I can't continue from there. Can you help me?

Answer 1

Modulo $5$, we have

$$3n^2 \equiv -2$$ $$\Leftrightarrow 6n^2 \equiv -4$$ $$\Leftrightarrow n^2 \equiv 1$$

$$\Leftrightarrow n \equiv 1 \vee n \equiv -1.$$

The solutions are therefore all those $n \in \mathbb{Z}$ such that $n = 5k+1$ or $n = 5k-1$ for some $k \in \mathbb{Z}$.

Answer 2

hint: Consider $n = 5k + r, 0 \leq r \leq 4$

Answer 3

Next step is to divide by $3$. Note that $3 \times 2 = 5+1$.

Answer 4

$$ 3n^2+2\equiv 3n^2-3\equiv 3(n-1)(n+1)\equiv 0 \mod{5} $$ since $\gcd(3,5)=1$ and since $5$ is a prime number it follows that $$ n\equiv\pm1 \mod{5} $$

Answer 5

Note: $3n^2\equiv-2\pmod 5\implies3n^2\equiv3\pmod5\implies n^2\equiv1 \pmod5\implies n\equiv \pm1\pmod 5$