Proving Euler's Theorem for Prime Divisors of the Form 4q+1

[dancergirlie]

Homework Statement

If the divisor P is a prime of the form 4q+1 then the number -1 or P-1 is certainly a residue.

Homework Equations

The Attempt at a Solution

First the book told me to prove that 1 and -1 are the only two remainders that are their own reciprocals modP.

Next, I'm not really sure what to do though.

I know that there has to be P-1/2 quadratic residues meaning 4q+1-1/2= 2q residues, which is an even number.

I suppose next I would assume that -1 is NOT a quad residue modP so that i could prove by contradiction.

Listing the quad residues modp as x1, x2...xp-1/2

That would mean that the product of -1 and x1... xp-1/2 would yield the quadratic non-residues modP.

However, I really don't know where to go from here, the book says to match the residues with the reciprocals, but I really don't know what to do..

any help would be great!

 

[Kreizhn]

It may just be because I haven't done any number theory in a while, but I don't understand your question. You are saying that P is a prime of the form P = 4q + 1 for some q \in \mathbb N.

You say that you want to show that -1 or P-1 must be a residue. With respect to what modulo? Maybe this is obvious, but I personally don't understand what you're trying to say.

Edit: Do you mean a quadratic residue?

 

[dancergirlie]

I am trying to show that -1 or P-1 is a quadratic residue modP

sorry I am using exact language from Euler's work- which is a bit outdated.

 

[Kreizhn]

Okay, that makes sense.

Sorry for being ignorant, but what "Euler's Theorem" are you trying to prove (there are just so many!). Is it Euler's criterion that states

\left( \frac ap \right) \equiv a^{\frac{p-1}2} \mod p
where
\left( \frac ap \right) = \begin{cases} 1 &amp; a \text{ is a quad. residue of } p \\<br /> -1 &amp; a \text{ is a non-quad. residue of } p \\<br /> 0 &amp; \text{ if } p\mid a \end{cases}
is the Legendre symbol?

 

[Kreizhn]

Also, do you mean to ask "show that -1 (equivalently p-1) is a..." or that there are two cases "-1" and "p-1." Because both are congruent mod p, so I assume you mean the first sentence.

I know I'm asking a lot of questions but I'm trying (part-time) to figure out how to help you.

 

[dancergirlie]

Yes, I am aware that -1 and P-1 are the same in modP.

Thanks for trying to help, but I figured it out on my own!