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I wish to show that if for integers $x,y$

$$ x = y^n$$

if the total number of factors of $x$ is $1 \mod n$. When $n=2$ this reduces to showing that square numbers have an odd number of factors, this is argued by claiming any factor $f$ of a number $X$ has a complement $g$ such that $fg = X$. Therefore if one looks at all the factor pairs of $X$, one will find 2 pairs for every $f,g$ that are unequal. Squares carry a factor $n$ such that $n^2=X$. Thus square numbers will only receive a single pair from this factor, resulting in an odd number of factors.

Now how to easily generalize this to $n=3, ... $

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  • $\begingroup$ Why the downvote? $\endgroup$ – frogeyedpeas Oct 20 '17 at 4:56
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Write the prime factorization of y as $y=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$

Then x has the factorization $p_1^{ne_1}p_2^{ne_2}\cdots p_k^{ne_k}$

Each factor of x is divisible by $p_i$ somewhere between 0 and $ne_i$ times. By trying every possibility of this for each $p_i$, we can enumerate the factors of x. This gives us that the number of factors of x is $(ne_1+1)(ne_2+1)\cdots(ne_k+1)$, which is 1 mod n since it is a product of numbers that is 1 mod n

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