Math CentralQuandaries & Queries


Dear Sirs,

Please could you help me with an arithmetic problem?

I was told that if x > y (integers); then x would never exactly divide y^n (n integer > 1) if (x,y) = 1; or if x is "square free". Is the latter true and why?

Many thanks,



If (x,y) = 1 then no prime divisor of x appears in y and thus no prime divisor of x appears in yn 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 yn and dropping the (x,y) = 1 condition? I think here you may want the x > y condition. If x was to divide yn and x was squarefree then what would that say about x dividing y? Now if x > y ... .




The Fundamental Theorem of Arithmetic (yes, that's its real name) says this:

Every natural number greater than 1 can be written as a unique product of prime numbers.

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 yn if n is an integer > 1. This is because the prime factors of yn 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 yn 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 yn = 2n 3n. 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 yn can be written as 2i 3j 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 2times3 which is 6. But that means x > y is violated. Thus, if x is square free, then it cannot divide yn.

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,
Stephen La Rocque.

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