# Find the natural numbers $n$ in which $n^2$ divides $584$? [duplicate]

I'm trying to find the natural numbers $n$ in which $n^2$ divides $584$ ?

i tried all the ways i know but i get stuck.

• Hint: $584=2^3\cdot 73$. May 8, 2014 at 19:51
• @RandomUser: Also the same OP as the aforementioned question. May 8, 2014 at 19:52
• I agree that it isn't a duplicate, but a good amount of what the highest rated answer states can be applied to this question. In particular, "a number is a square iff all of its prime exponents are even". May 8, 2014 at 20:01

$584 = 2^{3} * 73$
The only way for a square to divide 584 is for the square to appear in the factorization on the right. Thus the only possibility is $2^{2} = 4$, so the only $n$ satisfying the problem is $n = 2$.
Edit: I forgot $1$, which of course also works.
Let $N=\displaystyle\prod_{k=1}^m\big(p_k\big)^{a_k}~$ be the prime factor decomposition of N. Then the number squares that divide
N is $\displaystyle\prod_{k=1}^m\bigg(\bigg\lfloor\frac{a_k}2\bigg\rfloor+1\bigg)$. Same for cubes, only with a $3$ instead of $2$ in the denominator, etc. When
the denominator is $1$, we have the number of factors or divisors of N.