
Quadratic residue 5
Hey, so the problem is to find all primes for which 5 is a quadratic residue.
If one can tell me the answer I can probably prove it (using Gauss Lemma or Legendre symbol or so), but so far the primes seem to be randomly distributed.
I have that 5 is a quadratic residue for p = 2, 3, 5 (trivial / exceptional cases)
and for p = 7, 23, 29, 41, 43, 47, 61, 67...
and 5 is a quadratic nonresidue for p = 11, 13, 17, 19, 31, 37, 59...
Now while writing this I noticed that its true for even 10th powers and false for odd 10th powers. Is that the solution?

$\displaystyle \left(\frac{5}{p}\right) = \left(\frac{1}{p}\right)\left(\frac{5}{p}\right) = (1)^{\frac{p1}{2}}(1)^{\left\lfloor \frac{p+2}{5}\right\rfloor} $.
http://en.wikipedia.org/wiki/Legendr...egendre_symbol

How do you get
$\displaystyle \left(\frac{1}{p}\right) =(1)^{\frac{p1}{2}} ? $
EDit: I found it on wiki, thanks a lot!
