I have to compute the Legendre symbol $(\frac{5}{p})$.

If $p=2$, it is immediate to see that we have $5 \equiv 1 (\mod{2})$. let $p$ an odd prime. By quadratic reciprocity we have $(\frac{5}{p})=(-1)^{p-1}(\frac{p}{5})$, so $(-1)^{p-1}=1$ and we have $(\frac{5}{p})=(\frac{p}{5})$. Thus, we have to analyze the cases $p=1 (\mod{5})$, $p=-1 (\mod{5})$ and $p=3 (\mod{5})$.

But why the result in the book is $$ 1 \, \, \, if \, \, \, p=\pm 1 (\mod{10}) $$ and $$ -1 \, \, \, if \, \, \, p=\pm 3 (\mod{10}) $$ How can we reach the number $10$?

  • $\begingroup$ I am not sure how you deduce the first step, I think this is only valid if $p$ is an odd prime too? But anyway $5 \equiv 1 \mod 2$ which is a square. Other than that, why do you only analyze $p \equiv 1,-1,3 \mod 5$? What about $p\equiv 2$? $\endgroup$ – flawr Sep 10 '15 at 9:10

I don't know why the book (what book?) gives the result modulo $10$ but note that if $p$ is an odd prime $$ p\equiv\pm1\bmod5\Longleftrightarrow p\equiv\pm1\bmod10 $$ and $$ p\equiv\pm3\bmod5\Longleftrightarrow p\equiv\pm3\bmod10 $$

  • $\begingroup$ My problem was exactly this one... I found the solutions mod 5, but i don't know why in the book "Elementary number theory, Group Theory and Ramanujan graphs", the authors use mod 10 $\endgroup$ – TheWanderer Sep 10 '15 at 10:30

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