- #1

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## Homework Statement

Prove that if p is prime and p [itex]\equiv[/itex] 1 (mod 4), then [itex]x^{2}[/itex] [itex]\equiv[/itex] -1 (mod p) has a solution (x).

## Homework Equations

We already have proved (p-1)! [itex]\equiv[/itex] -1 (mod p)

Hint: Use the properties of [itex]Z_{p}[/itex] - a field that partitions the integers into p equivalence classes denoted [itex]\overline{a}[/itex] where [itex]\overline{a}[/itex] = {integers z such that z [itex]\equiv[/itex] a (mod p) } and try x = ([itex]\overline{1}[/itex] )([itex]\overline{2}[/itex] )([itex]\overline{3}[/itex] )...([itex]\overline{(p-1)/2}[/itex] )

## The Attempt at a Solution

The idea I have, judging from the hint and what we already know about (p-1)!, is that I need to show that

x

^{2}= ( ([itex]\overline{1}[/itex] )([itex]\overline{2}[/itex] )([itex]\overline{3}[/itex] )...([itex]\overline{(p-1)/2}[/itex] ) )

^{2}

is equivalent to (p-1)! mod p

I have no idea how to do this though. I figure if I knew why (p-1)/2 was chosen, I could figure it out.

Others things I know:

p = 4k + 1 for some integer k

so p-1 is even

so (p-1)/2 is an integer