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I need to find out if there is a solution to the congruence relation $x^2 \equiv 160 \mod 51$ by using the quadratic reciprocity law. Given that $51$ is not prime, I do not know how to use the results about the quadratic reciprocity law. I will appreciate your help. Thanks.

Bill Dubuque
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user996159
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    $a\equiv b\bmod{51}$ iff $a\equiv b\bmod3$ and $a\equiv b\bmod{17}.$ – Anne Bauval Feb 27 '23 at 22:51
  • @AnneBauval Thanks. I have a remarque: $a = b \mod 51 \iff a = b \mod 3$ or $a = b \mod 17$ since both $3$ and $17$ are primes. Given $pq|s,$ where $p,q$ are primes, then $p|s$ or $q|s$. Do you agree ? I am asking this since if one of the congruences fails to hold and the other holds, I am wondering if the initial congruence will have a solution ? – user996159 Mar 30 '23 at 23:00
  • When $p,q$ are coprime, $pq\mid s$ iff $p\mid s$ and $q\mid s.$ Otherly said: the l.c.m. of two coprime integers is their product. – Anne Bauval Mar 30 '23 at 23:04

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For $x^2-160$ to be divisible by $51$, we need it being divisible by $17$, so $x^2\equiv 160\mod{17}$ is necessary. However, using reciprocity, the quadratic symbol shows $$\left(\frac{160}{17}\right)=\left(\frac{7}{17}\right)=\left(\frac{17}{7}\right)=\left(\frac{3}{7}\right)=-\left(\frac{7}{3}\right)=-\left(\frac{1}{3}\right)=-1.$$ Therefore, we have no solution for $x$.

Ayaka
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