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.
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