Quadratic residues, mod 5, non-residues mod p 1) If $p\equiv 1\pmod 5$, how can I prove/show that 5 is a quadratic residue mod p?
2) If $p\equiv 2\pmod 5$, how can is prove/show that 5 is a nonresidue(quadratic) mod p?
 A: By quadratic reciprocity, $p$ is a residue modulo 5 precisely when 5 is a residue modulo $p$. Something is a quadratic residue modulo 5 precisely when it is a square modulo 5. So compute all the squares modulo $5, (1^2, 2^2, 3^2, 4^2)$ to see whether or not $p$ is a residue.
A: You do not really need the full power of quadratic reciprocity.
1) If $p\equiv 1\pmod{5}$, by the Cauchy theorem there is $\xi\in\mathbb{F}_p^*$ 
with order $5$, so $\xi$ is a root of $p(x)=x^4+x^3+x^2+x+1$, or $q(x)=\frac{p(x)}{x^2}=x^2+x+1+x^{-1}+x^{-2}$. By defining $\eta=\xi+\xi^{-1}$, you have that $\eta$ is a root of the quadratic polynomial $h(x)=x^2+x-1$, but $4\cdot h(x) = (2x+1)^2-5$, so $h(\eta)=0$ implies that $5$ is a quadratic residue$\pmod{p}$.
2) Conversely, if $\eta\in\mathbb{F}_p^*$ satisfies $\eta^2=5$, then $\xi=\frac{\eta-1}{2}$ is a root of the polynomial $h(x)=x^2+x-1$, that completely splits over $\mathbb{F}_p$. This gives that $p(x)=x^4+x^3+x^2+x+1$ completely splits over $\mathbb{F}_p$ or $\mathbb{F}_{p^2}$, since:
$$ x^4+x^3+x^2+x+1 = (x^2+1-\xi x)(x^2+1+x/\xi).$$
The last condition gives the existence of an element with order $5$ in $\mathbb{F}_p^*$ or in $\mathbb{F}_{p^2}^*$, but if $p\equiv 2\pmod{5}$, 
$ 5\nmid(p-1)$ and $5\nmid(p^2-1)$, so we have a contradiction.
A: 1)  $(5/p) = (p/5)$ since $p$ is $1(mod 5)$ then $(p/5) = (1/5) = 1$. So 5 is a quadratic residue mod p.
2) again $(5/p) = (p/5)$ since $p$ is $2(mod5)$ then $(p/5) = (2/5) = -1$ since 5 is 5(mod8). So 5 is not a quadratic residue
