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$x^2+3y^2=p \; \;$ for $p$ prime greater than $3$ has a solution if and only if $p\equiv 1\pmod 3$

I am supposed to use the fact that the class number of $\mathbb Q(\sqrt-3)$ is 1.

I already got the first direction.

Would appreciate it if anyone could point me to the right answer (hints) for the 2nd direction

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Since $p$ is stipulated to be a prime number in $\mathbb Z$ greater than 3, we know that $p \not \equiv 0 \pmod 3$. Therefore $p \equiv 1$ or $2 \pmod 3$. Clearly $3y^2 \equiv 0 \pmod 3$, so we can ignore $y$ for the time being.

Then we need $x$ to be coprime to 3. If $x \equiv 1 \pmod 3$, then $x^2 \equiv 1 \pmod 3$ also. But if $x \equiv 2 \pmod 3$, then $x^2 \equiv 1 \pmod 3$ anyway.

For example, $x^2 + 3y^2 = 5$ has no solutions in integers. But $x^2 + 3y^2 = 7$ does, e.g., $x = -2$, $y = 1$.

Is this the direction you were referring to?

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