• Jul 17th 2011, 07:45 AM
Zalren
Prove that if $p \equiv 3 (\mod{8})$ and $(p-1)/2$ is prime then $(p-1)/2$ is a quadratic residue $(\mod{p})$.

I believe I need to show that $(((p-1)/2)/p) = 1$. Will someone show me where to start? I have been thinking about this problem all weekend.

• Jul 17th 2011, 11:51 AM
Bruno J.
Well, if $p \equiv 3 \mod p$, that doesn't leave you much choice for the value of $p$, does it?
• Jul 17th 2011, 01:03 PM
Zalren
Oops, I typed it wrong. I corrected it!
• Jul 17th 2011, 01:20 PM
Zalren
What if I started like this:

Using Quadratic Reciprocity Theorem: $(((p-1)/2)/p)(p/((p-1)/2)) = (-1)^{((((p-1)/2)-1)/2)((p-1)/2)} = 1$ since $(p-1)/2$ is an even exponent. That tells me that $(((p-1)/2)/p)=(p/((p-1)/2))$

$p \equiv 1 (\mod{(p-1)/2})$, I believe this is a true statement, how do I back it up?

So $(((p-1)/2)/p)=(p/((p-1)/2))=(1/((p-1)/2))=1$
• Jul 17th 2011, 02:22 PM
Bruno J.
$p=2\times ((p-1)/2)+1$.
And $(p-1)/2$ is not even if $p \equiv 3 \mod 8$...
So can I assume $p>3$, for example $p = 11, 19, ...$, and just show the specific case where $p=3$ and show that $(((p-1)/2)/p) = (((3-1)/2)/3) = (1/3) = 1$?