Question

[University Math] Proof

Answer 1

If i understand correctly, it sounds like you are asking if there are two different prime numbers p and q such that for integers m and n, p\^n =q\^m . The answer to this is no and this is basically because of how prime numbers are defined.

 We say that an integer p is prime if for any a and b such p divides a\*b, then p divides at least one of a and b. For example 6 isn't prime because 2\*3=6, so it is divisible by 6, but 6 doesn't divide either 2 or 3.

So suppose there was an equation p\^n=q\^m, then we can write q\^m=q\*q\^m-1 and since q\^m =p\^n , it is divisible by p. However this means p must divide either q or q\^m-1, if p divides q then q is either equal to p or not prime. On the other hand if p divides q\^m-1, we can do the same thing, p must divide either q or q\^m-2 and so forth. So we conclude that p must divide q and then once again either q=p or q is not prime, so we are done.

Answer 2

There is a proof already posted but the result also follows directly from the Fundamental Theorem of Arithmetic:

Every integer greater than 1 can be represented uniquely as a product of Prime Numbers, up to the order of the factors.

If k = p_1^n = p_2^m ( for integer k, primes p_1 and p_2 ) then k is not represented uniquely as a product of prime numbers because there are 2 prime factorizations: p_1^n and p_2^m.

Answer 3

Hmm oh wait I think I see it now...

Well that was a waste of time.

Edit: no wait I don't see it... Ya I don't see the solution so if anyone can prove it go for it.