This is an idea that I had about 3 months ago. I tried some college professors, they didn't care. I tried to solve, but with no luck. I ask for help to find the closed form of the following product series. I know it would be safer only to put the product, but I just would like to see this solved, and if I put why I think this is important, the chance of someone helping would be higher.

My idea is:

If $\prod_{k=2}^{X-1}{\sin(\pi\frac{X}k)}≠0$, than the number $X$ is prime. Else, $X$ is a non-prime.

Why:

- $\sin(\pi\frac{X}k)$ tests if a number k divides X. If the value returned is $0$, than k divides X. Because for a number to divide another, the result must be an integer, and $\sin(\pi\alpha)$ is equal to $0$ if and only if $\alpha$ is an integer. So, to test if a number, for example, k=3 divides 9, we could try that like $\sin(\pi\frac{9}3)=\sin(3\pi)=0$. Now let's try to see if 2 divides 11 this way: $\sin(\pi\frac{11}2)=\sin(5.5\pi)=-1≠0$
- Using logic, we could extend this to the product above. We only need one sine being $0$ for the product to be $0$. That means that with only one divisor from range $2$ to $X-1$, the number X is a non-prime. And the only way for the product to be $0$ is if one or more of the sines is $0$: You can't multiply numbers $≠0$ and get $0$.

For example, 7 is prime if: $$\sin(\pi\frac{7}2)\times \sin(\pi\frac{7}3)\times \sin(\pi\frac{7}4)\times \sin(\pi\frac{7}5)\times \sin(\pi\frac{7}6)\ne 0$$ and that's the case.

I know I could test only until $\sqrt{7}$ rounded up, but if someone can get the closed form of the product, that wouldn't matter, and we wouldn't have to calculate the square root.

Thank you for reading this.