Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:

Prove that $5$ is a quadratic residue $(\mod p)$ with $p$ odd prime iif $p \equiv \pm 1 \mod 10$ ; prove also that $5$ is NOT a quadratic residue $(\mod p)$ iif $p \equiv \pm 3 \mod 10$.


To check if $5$ is a quadratic residue $(\mod p)$ I write the equivalent Legendre symbol with the condition:

$(5/p) = 1$

So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}\over 4}=(p/5)(-1)^{(p-1)}$

$\bullet$ The exponent $(p-1)$ must be $(\mod2)$

$\bullet$ $(p/5)$ means to find $p$ : $p(\mod5)$ $\rightarrow$ the choices are $1,3(\mod5)$ because $p$ is prime

The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(\mod5\times 2)=(\mod10)$

Case $1(\mod10)$:

Here $(1/5)=1$ and for the exponent $p=1(\mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(\mod10)$ but also for $p=-1(\mod10)$

Case $3(\mod10)$:

Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(\mod2)=1(\mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(\mod10)$ but also for $p=-3(\mod10)$


I appreciate any kind of critics and corrections.

Thank you

  • 1
    $\begingroup$ There are more cases. A prime can be $2$ or $4 \pmod{5}.$. $\endgroup$ – B. Goddard Nov 30 '18 at 12:42
  • $\begingroup$ I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori? $\endgroup$ – Alessar Nov 30 '18 at 13:08
  • 1
    $\begingroup$ Notice that $7 \equiv 2 \pmod{5}$ and $19\equiv 4 \pmod{5}$. $\endgroup$ – B. Goddard Nov 30 '18 at 13:31
  • $\begingroup$ Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version $\endgroup$ – Alessar Dec 1 '18 at 11:28

See my other post use Gauss lemma to find $(\frac{n}{p})$:

$(\frac{5}{p}) = 1 \iff p \equiv \pm 1 \mod 5$


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