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

I believe I need to show that $\displaystyle (((p-1)/2)/p) = 1$. 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 $\displaystyle p \equiv 3 \mod p$, that doesn't leave you much choice for the value of $\displaystyle p$, 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: $\displaystyle (((p-1)/2)/p)(p/((p-1)/2)) = (-1)^{((((p-1)/2)-1)/2)((p-1)/2)} = 1$ since $\displaystyle (p-1)/2$ is an even exponent. That tells me that $\displaystyle (((p-1)/2)/p)=(p/((p-1)/2))$

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

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