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.
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.
On
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.
$584 = 2^{3} * 73$
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.