1
$\begingroup$

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

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

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 $(a/p)=-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.