>>10245151Any number can be uniquely factored into powers of primes: 2^a*3^b*5^c*7^d*.... (this is known as the fundamental theorem of arithmetic).
Let e(x,p) denote the exponent of prime p in the factorisation of x. If e(y,p)<=e(x,p) for all p then y divides x. If e(y,p)>=e(x,p) for all p and e(y,p)>e(x,p) for at least one p then y is a multiple of x (the latter requirement is only necessary if you don't consider x itself to be a multiple of x). If e(y,p)<e(x,p) for any p then y isn't a multiple of x. Also, e(x^n,p)=n*e(x,p).
So any number y for which:
e(y,p)<=n*e(x,p) for all p (1)
e(y,p)>e(x,p) for some p (2)
e(y,p)<e(x,p) for some p (3)
satisfies your requirement.
(1) ensures that y divides x^n, (2) ensures that it is not a factor of x, (3) ensures that it is not a multiple of x.
It's possible to satisfy all 3 constraints for any x whose factorisation includes at least two primes. If x is a prime power p^k, then the only numbers which divide x^n=p^(k*n) have the form p^m where m<=k*n, and these are either x (if m=k), factors of x (if m<k) or multiples of x (if m>k).
If x has at least two prime factors p,q (x=p^a*q^b), then y=q^(b+1) divides x^n, isn't a factor of x, and isn't a multiple of x.
