Exercise of Quadratic Reciprocity

Question

This is an exercise in Burton :

Prove that $$(5/p) =1\ iff\ p\equiv 1,\ 9,\ 11,\ or\ 19\ (20) $$

Note that $5=4+1$ so that $(5/p)=(p/5)$. In further $$ (p/5)^2=(5/p)(p/5) = (-1)^{1\cdot \frac{p-1}{2}} $$

So we conclude that $$p=4m+1$$ In other words, $$ p=20k+1,\ 5,\ 9,\ 13,\ or,\ 17$$

$$(p/5)=1 \ iff\ p=20k+1,\ 5,\ or,\ 9$$

Am I miss something ?

Answer 1

We have $(p/5)=1$ if and only if $p\equiv\pm 1 \pmod{5}$. But $p$ has to be odd, so we want $p\equiv \pm 1 \pmod{10}$.

There is no particular virtue to translating that to congruences modulo $20$. But if one cares to, one gets $x\equiv 1$ or $11$ or $9$ or $19$ (modulo $20$).