It is easy to show that for all odd x that:
(x^n-2) = 7 (mod 8)
but why are the factors of above either congruent 1 or 7 modula 8, i.e
factor(x^n-2) = {1,7} (mod 8)
can anyone give me a hint.
regards
Anton
Love your joke :rofl: - by your theory I should be rich - alas - far from it :frown: .
Yes I know Fermats small theorem
Mod(5^(13-3),13) = Mod(5^10,13) = 5
Mod(5^(26-4),13) = Mod(5^22,13) = 5
And Fermats Little theorom
Mod(5^13,13) = 5 or Mod(5^12,13)=1
Ok now I see -...
If p is a prime or psuedo-prime base b then the following equality holds for all values \mbox{p , b} , but how to prove it?
b^{2p-4}\equiv\mbox{x (mod p)}\Longleftrightarrow b^{p-3}\equiv\mbox{x (mod p)}
we are talking about one and the same value for x in the above two congruencies...
We all know the definition of prime numbers and the first prime number is always 2.
Why is -1 not listed as a prime number? , it qualifies as it passes all tests for a prime number.
A helicopter was flying around above Seattle when an electrical malfunction disabled all the navigation and communications equipment. The pilot saw a tall building, flew toward it, circled, drew a handwritten sign saying "Where am I?" and held it in the helicopter's window.
People in the tall...