I'm supposed to use a prime factorization somewhere, and that the fundamental theorem of arithmetic is to be applied as well.
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It is false without further hypotheses, since $(-4)(-9) = 6^2\,$ but $\,-4,\,-9\,$ are not squares. But it is true for $\,X,Y > 0.\,$ Suppose they are coprime and their product in an $n$-th power. Then
Hint $\ XY = (x p^{i})(y p^j) = p^{kn} z^n,\ p\nmid x,y,z\ \Rightarrow\, i+j = kn\,$ by uniqueness of prime factorizations. $X,Y$ coprime $\,\Rightarrow\,i=0\,$ or $\,j=0\,$ so $\,j=kn\,$ or $\,i=kn,\,$ so $\,p\,$ occurs to power a multiple of $\,n\,$ in either case. By induction this holds true for all primes, hence both $X,Y$ are $n$-th powers.
Remark $\ $ The argument uses both the existence and uniqueness of prime factorizations in many places (not all mentioned above). If the exercise requires you to carefully identify these invocations then you need to examine the argument very closely to ensure that it is complete and correct.
Bill Dubuque
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Each prime factor of xy divides either x or y, but not both (because x and y are coprime)
If p divides x, then $p^{kz}$ divides x because p does not divide y and $p^{kz}$ divides xy.
k is the highest positive integer such that $p^{kz}$ divides xy
The analogue statement holds for the primes dividing y.
So x and y are a product of prime powers $p^{kz}$
If all prime factors of xy are in x, then y is 1, also a z-th power.
Analogue for the case that all prime factors are in y, then x is 1.
Peter
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Let $p$ be a prime factor of $x$. Then $p$ is a prime factor of $xy$, and since $xy$ is a $z$th power, the exponent $e_p$ of $p$ in the decomposition of $xy$ is a multiple of $z$. Then the exponent of $p$ in the decomposition of $x$ is also $e_p$, since otherwise it would mean that $p$ divides $y$, contrary to assumption. Thus the exponents of all the prime factors of $x$ are multiples of $z$, and $x$ is a $z$th power. The same holds for $y$.
fkraiem
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