I am attempting to construct a field containing 625 elements and should be in the form Zn[x] mod f(x).
Factoring 625 leads to 5^4. So I'm guessing my field will be GF(5^4). So in order for me to construct a field with all elements in it, I need f(x) to be some irreducible polynomial mod 5...
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.
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 p
so x^2 ≡ -1 mod p has no solutions
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).
How can (p - 1) / 2 be odd if p is prime? All prime numbers are odd, so minus one from an odd is an even, divided by 2 is another even.
Or, are you saying the whole expression (p ≡ 3 mod 4) is a prime, and not p by itself? In that case, I don't understand...
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...
I'm not sure I understand what you are saying. I did raise the both sides to the (p-1)/2 and I got this:
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...
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 -...