Quadratic Reciprocity / Legendre symbol/ Congruence Question

I know that $(\frac{5}{p})=1$ $\leftrightarrow$ $p=1,4mod5$

I want to know $(\frac{5}{p})=-1$ $\leftrightarrow$ ?....

My Attempt

I know that $(\frac{5}{p})=-1$ will be the complementary congruence conditions on prime numbers reduced mod 5. I think I need to know what can $p mod 5$ be for a prime (other than 1 or 4) if p is NOT 1 or 4 mod 5? Not sure how to approach this

  • $\begingroup$ The Legendre symbol is always $\pm 1$, except for its value at zero modulo $5$. $\endgroup$ – user296602 Apr 25 '16 at 2:23

Let $p$ be an odd prime other than $5$. Then $(5/p)=-1$ (meaning that $5$ is not a quadratic residue of $p$) if and only if it is not the case that $(5/p)=1$. Thus for any odd prime $p$, we have $(5/p)=-1$ if and only if $p\equiv \pm 2\pmod{5}$.

Or else we can compute. We have $(5/p)=(p/5)=(r/5)$ where $r$ is the remainder when we divide $p$ by $5$. Now we can check directly that $1$ and $4$ are quadratic residues of $5$, and $2$ and $3$ are not, by squaring $1$, $2$, $3$, and $4$ modulo $5$.

More generally, if $p$ is an odd prime, then exactly half of the numbers $a$ between $1$ and $p-1$ are quadratic residue of $p$, giving Legendre symbol $(a/p)=1$, and the remaining half are quadratic non-residues of $p$, giving Legendre symbol $(a/p)=-1$.


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