2
$\begingroup$

The question asks to find congruence conditions on prime $p$ such that $7$ is the least quadratic nonresidue mod p. Also, find the least such prime.

I solved it for $1,2,3,4,5,6$ mod $p$ and got the following congruences: $$p\equiv \pm1\pmod8, p\equiv \pm1(mod12) , p\equiv \pm1\pmod5$$ for numbers 2,3 and 5. I combined the congruences to make $$p\equiv \pm1 (mod 60)$$ Is this correct? Then by this method will the least such prime be $61$? $59$ is not possible. Thanks.

$\endgroup$
2
  • $\begingroup$ You seem to have forgotten about $7$ $\endgroup$
    – user14972
    Commented Nov 10, 2014 at 11:22
  • $\begingroup$ You're right. Is this correct that $p\equiv -1 (mod 4)$ for 7? $\endgroup$
    – Paradox 101
    Commented Nov 10, 2014 at 11:28

1 Answer 1

Reset to default
1
$\begingroup$

The lowest common denominator of $8$,$12$ and $5$ is 120, so we are looking $\pmod{120}$. There are two independent conditions, that is $\pm1\pmod{24}$ and $\pm1\pmod5$, which give rise to four possible conditions: $p\equiv\pm1,\pm49\pmod{120}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .