Eulers and Fermats Theorems Help

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Homework Statement



I'm having a hard time solving two problems I've been looking at for a while, they both involve Eulers theorem and Fermat's Little Theorem, here they are:

Let p ≡ 3 (mod 4 ) be prime. Show that x^2 ≡ -1 (mod p) has no solutions. (Hint: Suppose x exists. Raise both sides to the power (p - 1)/2 and use Fermat's theorem).


The other problem is this:

Let n = 84047 * 65497. Find x and y with x^2 ≡ y^2 (mod n) but x ≠(the symbol should be not congruent, couldn't find the symbol) +- (plus or minus) y mod n.


I've been looking at these for 2 hours and I don't know how to solve these. Any ideas? If you show the solution, please explain how if possible, thanks!

Homework Equations



Fermats little theorem: a^p-1 ≡ 1 mod p if p is a prime and p does not divide a.

\phi(pq) = (p -1)(q - 1) when n = pq and p and q are two distinct primes.


The Attempt at a Solution



For the first one, I took both sides to the power of (p - 1)/2 and got...

x^p-1 ≡ 1 mod p

The reason the -1 goes to 1 is because (p - 1)/2 must be even since p is prime, there -1 raised to an even power goes to 1. From this point on I don't know what to do, I tried making the claim that LaTeX graphic is being generated. Reload this page in a moment. x^p-1 was a in Eulers theorem and using his theorem and showing it was wrong, thus making the whole argument above wrong, but it did in fact work anyway...unless I did something wrong.

For the 2nd problem, I don't know where to begin
 
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Are you sure (p-1)/2 is even? We're given that p=3(mod4), i.e. p=4k+3.

For the second problem, maybe try experimenting with a smaller n.
 
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Given that p = 4k + 3, wouldn't p have to be an odd? Anything a multiple of 4 is even, +3 will always result in an odd right? (even plus an odd is always an odd).
 
Oh nevermind, an even divided by an even can be even or odd, its an even * even that is always even. So now, I get the following::

x^p-1 ≡ -1 mod p

which isn't right because fermat stated:

x^p-1 ≡ 1 mod pso x^2 ≡ -1 mod p has no solutionsman I suck.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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