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To find when $6$ is a square I divided $6$ in to $2$ and $3$ by The Legendre symbol.

$$\frac{6}{p} = \frac{2}{p} \times \frac{3}{p}$$

I know that we can multiply the cases of when $2$ and $$ are squares because then it is (1)(1) by Legendre.

$2$ is a square when $p \equiv 1 \pmod{8}$

$3$ is a square when $p \equiv 1 \pmod{12}$

so $6$ is a square when $p \equiv 1 \pmod {24}$ ($96$, but the $lcm(8,12)= 24$)

but how do I find the cases when $2$ and $3$ are both NOT squares? because then $6$ would be a square then as well, $(-1)(-1) = 1$ by Legendre.

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You just have to examine each congruence class systematically.

Note that $2$ is a square mod $p$ when $p = 1$ or $p = 7\pmod{8}$, and is not a square when $p = 3$ or $p = 5 \pmod{8}$.

Similarly, using reciprocity we have that $3$ is a square mod $p$ when $p = 1$ or $p = 11 \pmod{12}$ and is not a square when $p = 5$ or $p = 7\pmod{12}$.

This information should be enough to determine what happens for each congruence class mod $24$.

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