number theory - Show $3x^2 + 2 = y^2$ has no solution in integers.

Show $3x^2 + 2 = y^2$ has no solution in integers.

I've seen from similar problems, the idea is to reduce the equation to a congruence $\mod{3}$ and show that the congruence $y^2 \equiv 2 \pmod{3}$ has no solutions.

Why is one able to reduce the problem in this manner?

Answer

Start from basics,

What does the representation $a \equiv b \pmod c$ mean in the first place?

Answer : It means $(b-a)$ is divisible by $c$, or in a fancy way, it's written as $$c \mid (b-a)$$

For your question, you can clearly see that if $~3x^2 + 2 = y^2$ is true it would imply $~ y^2-2=3x^2$. Which, therefore implies that $y^2-2$ is a multiple of $3$.

Therefore $3 \mid y^2-2 \implies y^2 \equiv 2 \pmod{3}$

So if you could prove, somehow, that this ain't possible, it would prove that the equation has no solution in integers.