The no. of ways of putting $N = p_{1}^{\alpha_{1}}.p_{2}^{\alpha_{2}}.p_{3}^{\alpha_{3}}......p_{k}^{\alpha_{k}}$ as a product of $2$ Natural no. is

Question

The no. of ways of putting $N = p_{1}^{\alpha_{1}}.p_{2}^{\alpha_{2}}.p_{3}^{\alpha_{3}}......p_{k}^{\alpha_{k}}$ as a product of $2$ Natural no. is

$\displaystyle \frac{1}{2}(\alpha_{1}+1).(\alpha_{2}+1).....(\alpha_{k}+1)\;\;,$ If $N$ is not a perfect Square.

$\displaystyle \frac{1}{2}\left\{(\alpha_{1}+1).(\alpha_{2}+1).....(\alpha_{k}+1)+1\right\}\;\;,$ If $N$ is a perfect Square.

Can anyone explain me how can i prove it.

Thanks

Answer 1

The idea is, take all possible factors of N, of which there are $( \alpha_1 + 1)(\alpha_2 + 1) \ldots (\alpha_k + 1)$. For each factor $x$, $N/x$ is also a factor, so we're double-counting pairs of factors $(x, N/x)$ which multiply to $N$, hence divide by $2$. But, if $N$ is a perfect square $N = x * x$, then we're not double-counting the pair $(x, x)$, so re-compensate for that pair by adding $\frac{1}{2}$.