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?
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Hint $3$ divides one of $y-1,y,y+1$.
If $3|y$ then $3|y^2-3x^2=2$.
If $3 \nmid y$ then $$3|y^2-1=3x^2+1$$
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The main power of modular arithmetics relies in disproving facts. Your problem is kind of trivial, so let us study a similar but less trivial one - there are no integer solutions of $$ a^5+b^5+c^5 = 224. $$ Proof: any fifth power is either $-1,0$ or $1\pmod{11}$ and $224\equiv 4\pmod{11}$.
Finding a prime (or a set of primes) allowing to perform such disproofs might be tricky in some cases, but experience is a good teacher. As a rule of thumb, observing the involved exponents and the involved coefficients gives many leads. In $3x^2+2=y^2$ to study what happens $\!\!\pmod{3}$ comes natural.
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It is same as proving that no perfect square leaves remainder 2 when divided by 3. Let n be an integer, by EDL we know that $n = 3q, 3q+1$ or $3q+2$ $\implies n^2=3(3q^2), 3(3q^2+2q) + 1$ or $3(3q^2+4q+1)+1$. So $n^2$ leaves only 0 or 1 as remainder when divided by 3. Hence, there is no perfect square which leaves 2 remainder when divided by 3.
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.