



 
Andrew, If (x,y) = 1 then no prime divisor of x appears in y and thus no prime divisor of x appears in y^{n} so x cannot divide it  the x > y condition seems superfluous. I'm not clear on what your second question is asking; are you saying that if x is squarefree then it doesn't divide y^{n} and dropping the (x,y) = 1 condition? I think here you may want the x > y condition. If x was to divide y^{n} and x was squarefree then what would that say about x dividing y? Now if x > y ... . Penny
Andrew, The Fundamental Theorem of Arithmetic (yes, that's its real name) says this:
So it is true that if x > y > 1 and the lowest common factor of x and y is 1, then x does not divide y and therefore y^{n } if n is an integer > 1. This is because the prime factors of y^{n } are copies of the prime factors of y, none of which is shared by the prime factors of x. [Your question has a twist to it however. You say x and y are integers, not natural numbers (whole numbers > 0). So does x = 1 and y = 0 work? Certainly x divides y, since 1 divides any integer evenly. Thus, x cannot equal one, so on this technicality, the statement is false.] So can x divide y^{n } if x > y when x, y, n are natural numbers > 1 and x is square free? Let's say y = 6. Any value of n will give a prime factorization of y^{n } = 2^{n }3^{n}. What divides this? Only other numbers divisible by only 2s and/or 3s, due to the fundamental theorem of arithmetic. So that means that any number x that divides y^{n} can be written as 2^{i }3^{j} where i and j are whole numbers. However, if i or j is larger than 1, then the number x is not square free. To keep it square free, i and j must less than 2. But that just means the only numbers possible are 1, 2, 3, or 23 which is 6. But that means x > y is violated. Thus, if x is square free, then it cannot divide y^{n}. So the latter part of the statement is true as a consequence of the Fundamental Theorem of Arithmetic as well as the first. Hope this helps,  


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