Let $X$ and $Y$ be independent random variates with the same probability distribution, $P(x)$. Assuming that the product $Z=XY$ is a random variate with normal distribution, say $$f_Z(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x^2}$$ what is the form of $P(x)$? Does it exist in closed form?

Let $X$ and $Y$ be independent random variates with the same probability distribution, $P(x)$. Assuming that the product $Z=XY$ is a random variate with normal distribution, say $$f_Z(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x^2}$$ what is the form of $P(x)$? Does it exist in closed form?

EDIT: I just realized that a product distribution is necessarily divergent at zero, which means that the normal distribution cannot be a product distribution (at least when $P(x)$ is interpreted as a normal function; maybe there is a generalized function that solves the problem?). Do you agree?