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I have to determine the values of $p \in \{0, \dots, 23 \}$ such that $\left( \frac {-6} p \right) = 1$.

I have that: $$\left( \frac {-6} p \right) = \left( \frac 2 p \right) \left( \frac {-3} p \right)$$

and I know that $\left( \frac 2 p \right) = 1$ if $p \equiv 1,7 \pmod 8$ and $\left( \frac 2 p \right) = -1$ if $p \equiv 3,5 \pmod 8$, and also that $\left( \frac {-3} p \right) = 1$ if $p\equiv 1 \pmod 3$.

Using the Chinese Remainder Theorem, I find that $p\equiv 1,?,7,?\pmod{24}$.

Thanks in advance!

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    $\begingroup$ $10$ is not coprime to $24$ and $\frac{-6}{10}$ is not an integer, so how could $10$ possible be a solution? $\endgroup$ – Noble Mushtak Jun 11 '16 at 16:12
  • $\begingroup$ The $6$ and $10$ should be $5$ and $11$. $\endgroup$ – André Nicolas Jun 11 '16 at 16:20
  • $\begingroup$ @NobleMushtak by chinese remainder you have $2.3$ and $2.5$ $\endgroup$ – Maman Jun 11 '16 at 16:23
  • $\begingroup$ @AlexM. I edited this question because I thought it was hard to read and because it wasn't using $\pmod{24}$ notation, but if you click the "edited [time] ago" link, you can see the original version. I have no idea what the Jacobi symbol is, so my edits may have messed up Maman's notation. I'm really sorry for any ambiguity I caused! $\endgroup$ – Noble Mushtak Jun 11 '16 at 16:25
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    $\begingroup$ @NobleMushtak: Those things that had parantheses around them (that you have deleted) are not fractions, they are Jacobi symbols. $\endgroup$ – Alex M. Jun 11 '16 at 16:32
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We have $(-6/p)=1$ if (i) $(2/p)=1$ and $(3/p)=1$ or (ii) $(2/p)=-1$ and $(3/p)=-1$.

You seem to have taken care of (i). We get the solutions $p\equiv 1\pmod{24}$ and $p\equiv 7\pmod{24}$.

For (ii) we want a) $p\equiv 3\pmod{8}$ and $p\equiv 2\pmod{3}$ or b) $p\equiv 5\pmod{8}$ and $p\equiv 2\pmod{3}$.

a) has the solution $p\equiv 11\pmod{24}$ and b) has the solution $p\equiv 5\pmod{24}$.

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  • $\begingroup$ the second case is also the chinese remaider theorem ? $\endgroup$ – Maman Jun 11 '16 at 16:47
  • $\begingroup$ Yes, in principle for example we use the CRT to find the solution of the system $x\equiv 3\pmod{8}$, $x\equiv 2\pmod{3}$. However, the numbers are very small, so we go through the numbers less than $24$ which are congruent to $3$ mod $8$. These are $3,11,19$. Now we pick the one aong these which is congruent to $2$ modulo $3$. This is $11$. $\endgroup$ – André Nicolas Jun 11 '16 at 16:53
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I saw this the other day as well. Call this a side note, the primes $1,7 \pmod {24}$ are all represented by $x^2 + 6 y^2.$ It is easy to see that if $p = x^2 + 6 y^2$ in integers, then $(-6|p) = 1.$ $$ 1, 7, 31, 73, 79, 97, 103, 127, 151, 193, 199, $$

the primes $5,11 \pmod {24}$ are all represented by $2x^2 + 3 y^2.$ $$ 2, 3, 5, 11, 29, 53, 59, 83, 101, 107, 131, 149, 173, 179, 197, $$ Of course, $2x^2 + 3 y^2$ also represents $2$ and $3,$ but those divide $6.$

Meanwhile, IF $(-6|p)= 1,$ it is easy enough to construct a (positive) binary quadratic form $f(x,y) = p x^2 + Bxy + C y^2$ of discriminant $-24,$ and this form reduces to one of the two indicated forms above; that process tells us how to represent $p.$ For part of that, prove if n - natural number divide number $34x^2-42xy+13y^2$ then n is sum of two square number

Why not: if $(-6|p)= 1,$ then $(-24|p)= 1,$ we have some integer solution to $\beta^2 \equiv -24 \pmod p.$ If the original $\beta$ we found was odd, replace it by $B= p - \beta$ so that $B$ is even, and $B^2 \equiv -24 \pmod {4p}.$ That is, $B^2 = -24 + 4pC.$ Well, $B^2 - 4pC = -24,$ and the positive binary quadratic form $\langle p, B, C \rangle$ has discriminant $-24.$ The notation $\langle p, B, C \rangle$ means the form $f(x,y) = p x^2 + Bxy + C y^2$

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