# Proof that $5$ is a quadratic residue $(\mod p)$ with $p$ odd prime iif $p \equiv \pm 1 \mod 10$

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 $$p \equiv \pm 3 \mod 10$$.

Dim:

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 $$p=-3(\mod10)$$

$$\Box$$

I appreciate any kind of critics and corrections.

Thank you

• There are more cases. A prime can be $2$ or $4 \pmod{5}.$. – B. Goddard Nov 30 '18 at 12:42
• 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? – Alessar Nov 30 '18 at 13:08
• Notice that $7 \equiv 2 \pmod{5}$ and $19\equiv 4 \pmod{5}$. – B. Goddard Nov 30 '18 at 13:31
• Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version – Alessar Dec 1 '18 at 11:28

See my other post use Gauss lemma to find $$(\frac{n}{p})$$:
$$(\frac{5}{p}) = 1 \iff p \equiv \pm 1 \mod 5$$