proof with quadratic residue

**Zalren** · July 17th 2011, 07:45 AM

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.

Thanks in advance,

**Bruno J.** · July 17th 2011, 11:51 AM

Well, if $p \equiv 3 \mod p$ , that doesn't leave you much choice for the value of $p$ , does it?

**Zalren** · July 17th 2011, 01:03 PM

Oops, I typed it wrong. I corrected it!

**Zalren** · July 17th 2011, 01:20 PM

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 $(((p-1)/2)/p)=(p/((p-1)/2))=(1/((p-1)/2))=1$

**Bruno J.** · July 17th 2011, 02:22 PM

$p=2\times ((p-1)/2)+1$ .

And $(p-1)/2$ is not even if $p \equiv 3 \mod 8$ ...

**Zalren** · July 17th 2011, 03:17 PM

So can I assume $p>3$ , for example $p = 11, 19, ...$ , and just show the specific case where $p=3$ and show that $(((p-1)/2)/p) = (((3-1)/2)/3) = (1/3) = 1$ ?

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