Question

For which p is 5 a quadratic residue by evaluating the Legendre symbol (5/p) using Gauss's Theorem.

Answer 1

Ok, so by Gauss' Theorem
\[ (5/p) = (p/5) . (-1)^{\frac{5-1}{2}}(-1)^{\frac{p-1}{2}} = (p/5) \]
(provided p is not 2)

Then if 5 is a quadratic residue of p then by the above result p is a quadratic residue of 5 where by definition
\[ (p/5) \equiv p^2 \equiv 1 \mod 5 \]
For example p = 11 (but not 3,5 or 7; nor 13, 17, 23) and p = 29 ...

So now I think you need to characterize all of the prime numbers for which this last relation is true. The examples above should point you in the right direction.

Answer 2

james can u help me adter this???

Answer 3

Yes, in fact ....

Answer 4

Thanks, but it's not quite what I'm talking about. I think, by Gauss's Theorem, you can evaluate the Legendre symbol by (-1)^s. In this case, take the number of integers such that \[10k+r/2 < 5a < 20k + r\]
Do you know what I'm talking about? (I know I just typed it, so it sounds like I know what I'm talking about, but I'm really confused)

Answer 5

hm, this is not a Gauss' Theorem for this result with which I am familiar. But in any case, even if you don't have the result I'm calling Gauss' Theorem, you can derive the result in this case using (a/p) = a^((p-1)/2) mod p