Carreidas Algebra Challenge

[Carreidas]

Easy:
1)Take "p" a prime number and "a" any number. Divide a by p and take it's remainder, which we will call "r". Now, take a^p (that is, a elevated to the pth-power) and divide it by p, prove that it's remainder will also be r.
More succintly (and in more mathematical terms), a and a^p have the same remainder modulus p.
2) Prove that the proposition is false if p is NOT a prime number (for example 4).

Hard: prove that any even number greater than 2 can be written as the sum of two prime numbers. For example, 8 can be written as 5 + 3. 12 can be written as 7 + 5.

[Lul Thyme]

2) Prove that the proposition is false if p is NOT a prime number (for example 4).

Carmichael numbers say hi!

(I suppose you could be meaning that the proposition is false for SOME composite p, but it's not very clear.)

[Carreidas]

(I suppose you could be meaning that the proposition is false for SOME composite p, but it's not very clear.)

Good question. Could it be that if p is composite then it is always false? Or it only happens with some numbers? I add it to my already incredibly-successfull challenge!

[Lul Thyme]

http://en.wikipedia.org/wiki/Carmichael_number

http://en.wikipedia.org/wiki/Fermat_pseudoprime

There exists (in fact infinitely many) composite numbers p such that a^{p-1} is congruent to 1 mod p for every coprime a.

[Carreidas]

Excellent! Thanks for the info, never heard of Carmichael numbers before. Will look more into it, maybe base an exam question out of them.