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I have this assertion: if $p$ is a prime such that $p\equiv 11 \pmod{56}$, then $p$ splits in $\mathbb{Z}[\sqrt{14}]$ (the discriminant of $\mathbb{Z}[\sqrt{14}]$ is $56$.)

Why? Does $p\equiv 11\pmod{56}$ imply $14$ is a quadratic residue mod $p$?

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Sorry, I have asked wrong my question. I wanted to say: if p=11(mod 56) why 14=a^2 (mod p) – user17346 Oct 9 '11 at 15:35
There is an "edit" button right beneath the tags "number theory", "prime numbers", etc. - you can edit this question to be correct. – Zev Chonoles Oct 9 '11 at 15:39
up vote 4 down vote accepted

If we have a quadratic field $K = \mathbb{Q}(\sqrt{d})$ with $d$ squarefree, then an odd prime $p$ splits if and only if $\left(\frac{d}{p}\right) = 1$.

Claim: If $q$ is an odd prime with $q \equiv 3 \pmod 4$, then $q$ is a quadratic residue mod $p$ if and only if $p \equiv \pm b^2 \pmod {4q}$, where $b$ is an odd integer prime to $q$. The proof is straightforward from the law of quadratic reciprocity.

So, $\left(\frac{14}{p}\right) = \left(\frac{2}{p}\right) \left(\frac{7}{p}\right)$.

$2$ is a quadratic residue mod $p$ if and only if $p \equiv \pm 1 \pmod 8$. Since $p \equiv 11 \pmod {56}$ we see that $p \equiv 3 \pmod 8$. So $\left( \frac{2}{p}\right) = -1$.

On the other hand, we can use the claim above to calculate $\left(\frac{7}{p}\right)$. So we need only check that $p \equiv \pm b^2 \pmod {28}$. Since $p \equiv 11 \pmod {56}$, $p \equiv 11 \pmod {28}$. And simple computation shows that $\pm 11$ are non-quadratic residues mod $28$. Hence $\left(\frac{7}{p}\right) = -1$.

So $\left(\frac{14}{p}\right) = 1$, and so $p$ splits in $\mathcal{O}_K = \mathbb{Z}[\sqrt{14}]$.

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Many thanks Brandom, is very well explained – user17346 Oct 9 '11 at 16:04
The first sentence of this answer does not cover the case $p=2$. – KCd Oct 9 '11 at 17:29
@KCd: Thanks, fixed now. – Brandon Carter Oct 9 '11 at 17:40
I would upvote, but this comment is forever going to be when you had 2,222 reputation... :-) – Asaf Karagila Oct 9 '11 at 22:42

$11$ is not a square modulo $56$. Suppose $11\equiv a^2 \pmod{56}$. Then reducing modulo $8$ we see that $$3\equiv a^2 \pmod{8}$$ which is impossible, since $1$ and $4$ are the only squares modulo $8$.

Here is one way to prove that $p\equiv 11 \pmod{56}$ does not factor. Since the norm is multiplicative, we would need the factors to have norm 11. But then $B^2-14A^2=11$ which implies that $3$ is a square modulo $7$, a contradiction.

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