1. The problem statement, all variables and given/known data Show that if a, b, n, m are Natural Numbers such that a and b are relatively prime, then a^n and b^n are relatively prime. 2. Relevant equations Relatively prime means 1 = am + bn where a and b are relatively prime. gcd(a,b) = 1 We have a couple corollaries that may be beneficial: 1. Suppose a, n are positive integers. Then a and a^n have the same prime factors. 2. Let a, n be positive integers. a^(1/n) is either an integer or it's irrational. 3. The attempt at a solution By definition of relatively prime, 1 = ax + by where x,y exist as integers. By the fundamental theorem o arithmetic: a = p1*p2*...*pm where this is the unique prime factorization. b = p1*p2*...*pd m,d are natural numbers. By the corollarly, a^n = p1^n*p2^n*...*pm^n b^n = p1^n*p2^n*...*pd^n So we start with 1 = ax + by 1 = (p1*p2*...*pm)*x + (p1*p2*...*pd)y ****All I have left to do is find the way to raise this to the power of n. If I can raise it such that 1^n = (ax)^n + (by)^n the proof is done, easily, with what I have set up. Question: Is this step needed? Can I just use the corollary to say that "by the corollary" since a and a^n have the same prime factorization and b and b^n have the same prime factorization, then we see a^n and b^n must be relatively prime?