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Prove that there are no integer solutions to the equation $x^2=3y+2$

How do you prove this $even^2\equiv 0, odd^2\equiv 1$ with $\mod 4$?

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Considering the equation mod $3$.

$$x\equiv0,\pm1 \ \text{mod}\ 3$$

$$x^2\equiv0,1 \ \text{mod}\ 3$$

$$x^2-2\equiv 1,2 \ \text{mod}\ 3$$ Thus $x^2-2$ is never a multiple of $3$ so the equation has no integer solutions.

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Just to give a (slightly) different approach, let's use the fact that $3$ divides one of any three consecutive numbers, i.e., $3\mid (x-1)x(x+1)$ for any $x$.

Now if $x^2=3y+2$, then $(x-1)(x+1)=x^2-1=3y+1$, which implies $3$ divides $(3y+1)x$, which in turn implies $3$ divides $x$. But that implies $3$ divides $x^2=3y+2$, which implies $3$ divides $2$, which is a contradiction.

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