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I’m currently reading Cox’s Primes of the Form $x^2+ny^2$, and when giving congruences regarding forms of $x^2+ny^2$, they are given $\pmod{4}$. E. g., when stating the primes that satisfy $\left(\dfrac{-7}p\right)=1$, it is stated $p \equiv 1,9,11,15,23,25 \pmod{28}$. But this is equivalent to $p \equiv 1,2,4 \pmod 7$, which is a much more natural answer as this is what quadratic reciprocity immediately reveals.

I assume this was done for a reason, as when the solutions to $p=x^2+ny^2$ are discussed, they are always $\pmod{4n}$. E. g., if $p=x^2+5y^2$, $p \equiv 1,9 \pmod{20}$. On page 13, it says that:

The reciprocity step can be restated as the following question: is there a congruence $p \equiv a,b,\cdots \pmod{4n}$ which implies $\left(\dfrac{-n}p\right)=1$ when $p$ is prime?

But I don’t understand why it is necessary to use $4n$ rather than $n$. I’m sorry if this is a silly question but I feel that it’s essential to understand before continuing. Thanks for any help in advance.

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The mod $4$ relation comes from the minus sign: from standard properties, $\left(\frac{-7}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{7}{p}\right)$ and it is well known that $\left(\frac{-1}{p}\right)=1 \Leftrightarrow p \equiv 1 \mod{4}$. One then applies the main theorem that allows you to connect $\left(\frac{7}{p}\right)$ with $\left(\frac{p}{7}\right)$ but you should note that this again depends on $p \mod{4}$.

In particular $\left(\frac{7}{p}\right)=\left(\frac{p}{7}\right)$ if and only if $p \equiv 1 \mod{4}$. However, this is the same condition we had before so actually $\left(\frac{-7}{p}\right)=\left(\frac{p}{7}\right)$ but we didn't know without checking first.

However I should remark at this point that if $n \equiv 1 \mod{4}$ is prime then $\left(\frac{-n}{p}\right) \neq \left(\frac{p}{n}\right)$ and this mod 4 relation is important.

A more general explanation is that this comes from factorising $p=x^2+ny^2$ as $(x+i\sqrt{n}y)(x-i\sqrt{n}y)$. To guarantee this splitting, you want both $i=\sqrt{-1}$ and $\sqrt{n}$ to be elements mod $p$ and $i$ is an element mod $p$ exactly when $p \equiv 1 \mod{4}$. This is sufficient but by no means necessary; in the case $n=7$ above, we have $\sqrt{-n}$ is an element mod $p$ in those cases, but you'll find that $\sqrt{n}$ and $i$ are not.

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  • $\begingroup$ Is there any reason why the fractions are in parentheses? $\endgroup$
    – PinkyWay
    Apr 21, 2020 at 10:07
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    $\begingroup$ They're not fractions: this is notation for legendre symbols. See here en.wikipedia.org/wiki/Quadratic_reciprocity $\endgroup$
    – Matt B
    Apr 21, 2020 at 10:08
  • $\begingroup$ Thank you for responding! $\endgroup$
    – PinkyWay
    Apr 21, 2020 at 10:09
  • $\begingroup$ 8 does not need to be on the list since $p$ is prime $\endgroup$
    – joachxm
    Apr 21, 2020 at 11:49
  • $\begingroup$ @häxq thanks I thought I was missing something; I just lifted the residue classes without thinking about coprimality. I've edited accordingly. $\endgroup$
    – Matt B
    Apr 21, 2020 at 14:43

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