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There was a question in a past exam paper that asked Find criteria $\mod(20)$ for determining the Legendre symbol $(\tfrac{-5}{p})$ where $p\geq 7$.

I am very confused by what this means as I understand it the Legendre symbol is implicitly working in $\mod p$, how can we learn anything about $\mod 20$

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    $\begingroup$ This is a (routine) question about quadratic reciprocity. Are you familiar with that? $\endgroup$ – lulu Jul 20 at 11:42
  • $\begingroup$ Ah I see, That's actually next on my list to study , at least now I know where to look to solve such problems though thank you :) $\endgroup$ – excalibirr Jul 20 at 11:44
  • $\begingroup$ In case the formulation is what confuses you: The question asks you to find $\left(\frac{-5}p\right)$ when you only know $p\bmod 20$. $\endgroup$ – Hagen von Eitzen Jul 20 at 11:44
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    $\begingroup$ I think you'll find that you can work this problem very quickly once you've got reciprocity down. Good luck! $\endgroup$ – lulu Jul 20 at 11:45
  • $\begingroup$ @lulu I posted an answer could you take a look at it for me please :) ? $\endgroup$ – excalibirr Jul 20 at 13:29
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From Hagen von Eitzen's Comment I believe the following is the correct method :

We're looking at all $p \geq7,$ in $\mod 20$ so then we have to consider $7,11,13,17,19$

say $p\equiv 7$ then we compute $(\tfrac{-5}{7})$

$$(\tfrac{-5}{7})=(\tfrac{2}{7})=1$$ as $7\equiv-1 \mod8$

Now consider $p\equiv11$

$$(\tfrac{-5}{11})=(\tfrac{6}{11})=(\tfrac{2}{11})(\tfrac{3}{11})$$

Note that $11\equiv3 \mod 8$ so we have

$$-(\tfrac{3}{11})=(\tfrac{11}{3})=(\tfrac{2}{3})$$

Where we took the first step here because by the quadratic law of reciprocity $(\tfrac{3}{11})(\tfrac{11}{3})=(-1)^{5\times1}=-1\Rightarrow (\tfrac{3}{11})=-(\tfrac{11}{3})$

$$(\tfrac{2}{3})=-1$$

as $3\equiv 3\mod 4$.

Now we just follow the same method outlined above for $p=13,17,19$and we're done!

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    $\begingroup$ Not sure I follow the game plan here. The point is not to search the primes up to $20$. The answer (they are implying) is that $\left(\frac {-5}p\right)=1$ for $p\equiv a,b,c \pmod {20}$ and $\equiv -1$ for $p\equiv f,g,h\pmod {20}$. Your job is to find $a,b,c,f,g,h$ (note: I don't mean to suggest that there are $3$ each here). Granted, it is not obvious that $20$ is the correct modulus here, but that will drop out of reciprocity. $\endgroup$ – lulu Jul 20 at 13:33
  • $\begingroup$ @lulu I don't isn't the above method finding those a,b,c's , f,g,h's for instance for $(\tfrac{-5}{7})=1$ so then why can't we say that $7=a$, I'm very confused by the formulation of this question... $\endgroup$ – excalibirr Jul 20 at 13:54
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    $\begingroup$ To get you started: You want to know, say, all the primes $p\neq 2,5$ for which $\left( \frac {-5}p\right)=1$. Well $\left( \frac {-5}p\right)=\left( \frac {-1}p\right)\times \left( \frac {5}p\right)$ so we either need both factors to be $1$ or both to be $-1$. Now, $\left( \frac {-1}p\right)=1\iff p\equiv 1 \pmod 4$, so we understand that term. QR tells us that $\left( \frac {5}p\right)=\left( \frac {p}5\right)$ so you just have to analyze that (which is easy!). $\endgroup$ – lulu Jul 20 at 13:59
  • $\begingroup$ I'm sure your text goes through some similar examples in detail...this is an absolutely standard exercise in reciprocity. If nothing else, the wiki article goes through several examples, including this one as it happens. The technique is amazingly powerful...which is why Gauss' proof that it worked was such an impressive achievement. $\endgroup$ – lulu Jul 20 at 14:02
  • $\begingroup$ @lulu would you by any chance have any links to sites or pdfs where such problems are solved, the only text I have is a set of lecture notes and our lecturer didn't like to provide any examples $\endgroup$ – excalibirr Jul 20 at 14:50

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