# Proof of Test of primality from E.Dickson's book "History of theory of numbers vol.1"

E.Dickson mentioned the following result by E.Zondadari in his book "History of theory of numbers vol.1"(Chap XVIII):

Consider the following 'function':

$$\frac{\sin²(πx)}{(πx)²(1-x²)^2} \prod_{n=2}^\infty \frac{(πx)} {n\sin(πx/n)}$$

It's zero if |x|= prime and else otherwise .

How to prove the above identity ?

• Is $x$ real or complex? Commented Jan 31, 2020 at 19:52
• @saulspatz real Commented Jan 31, 2020 at 19:54
• I don't understand. If $x=4$ say, then the factor outside the product is $0$ so if the product converges, the value must be $0$. What am I missing? Commented Jan 31, 2020 at 20:00
• @saulspatz the doubt is mutual ; that's why ,earlier ,I added the image of the page of the book in the first place . But got downvotes saying don't add images! Commented Jan 31, 2020 at 20:03
• if n divides x we get a division by 0 under my interprwtation. but it still fails for primes then unless it's an empty product.
– user645636
Commented Jan 31, 2020 at 22:30

A copy of Zondadari's paper can be found here.

The key to understanding the stated result is an earlier one that the infinite product $$P(x)=\prod_{n=1}^\infty \frac{\sin\frac{\pi x}{n}}{\frac{\pi x}{n}}$$ converges absolutely and uniformly to an analytic function whose zeros are precisely the non-zero integers, with multiplicities equal to the number of their positive divisors. By $$\ \frac{\sin\frac{\pi x}{n}}{\frac{\pi x}{n}}\$$ Zondadari obviously here means the analytic function $$\ \sum_\limits{i=0}^\infty \frac{(-1)^i}{(2i+1)!}\left(\frac{\pi x}{n}\right)^{2i}\$$, which assumes the value $$1$$ at $$\ x=0\$$, and adopts the convention that an infinite product with only a finite number of zeroes and a tail that converges to a non-zero quantity is convergent.

Enumerating the zeros of this product, we have $$\ \pm 1\$$, each of multiplicity $$1$$, primes (positive or negative), each of multiplicity $$2$$, and and composites, each of multiplicity $$\ d\$$, where $$\ d\$$ is the number of its positive divisors (equal to $$\ \left(n_1+1\right) \left(n_2+1\right)\dots \left(n_r+1\right)\$$, where $$\ p_1^{n_1}p_2^{n_2}\dots p_r^{n_r}\$$ is its prime decomposition).

Now the function $$f(x) =\frac{\sin^3\pi x}{(\pi x)^3\left(1-x^2\right)^2}$$ has zeros of multiplicty $$3$$ at every integer, except $$0$$, where it has value $$1$$, and $$\ \pm1\$$, where it has zeros of multiplicity $$1$$. Therefore, the function \begin{align} g(x)&= \frac{\sin^3\pi x}{(\pi x)^3\left(1-x^2\right)^2}\cdot\frac{1}{P(x)}\\ &= \frac{\sin^2\pi x}{(\pi x)^2\left(1-x^2\right)^2} \prod_{n=2}^\infty \frac{\pi x}{n\sin\frac{\pi x}{n}} \end{align} has zeros of multiplictity $$1$$ for prime $$\ x\$$, a pole of multiplicity $$\ d-3\$$ at any composite $$\ x\$$ with $$\ d\$$ positive divisors, and has a finite non-zero value at $$\ x=0\$$ and $$\ x=\pm1\$$.

• thank you for the answer . Also , I'm asking if the product could be twisted in such a way that it gives zero for primes and finite values for composite ? Commented Feb 2, 2020 at 8:55
• The function $\ \frac{g(x)}{g(x)+x}\$ will vanish for prime $\ x\$ and have value $1$ for composite $\ x\$. Commented Feb 2, 2020 at 10:20
• Is this a "good" test ?( Efficiency) Commented Feb 2, 2020 at 11:16
• @lonzaleggiera: I see, sorry, I hadn't reloaded; I've deleted the comment. Commented Feb 2, 2020 at 11:40
• I can't claim any expertise in tests for primality, but it looks to me that to distinguish whether an integer $\ x\ne\pm1, 0\$ is a zero or a pole of $\ g(x)\$ you would need to evaluate $\ \sin\frac{\pi x}{n}\$ at least for $\ n= 2, 3,\dots,\left\lfloor\sqrt{|x|}\right\rfloor\$, which would be even less efficient than using trial division as a test for primality. Commented Feb 2, 2020 at 23:13