Questions on Dudley's paper "Formulas for Primes"

Question

I found the very well-written and interesting paper "Formulas for Primes" by Underwood Dudley (https://www.jstor.org/stable/2690261).

In this paper Dudley gives plenty of examples of totally useless formulas for finding primes and primality tests etc.

A few things are unclear to me and since I am no number theory expert, I thought maybe someone can explain them to me.

Question 1

On the very first page, Dudley cites a formula from 1895 which says that for an integer $n\geq2$ the expression

$$ \frac{e^{2\pi i(n-1)!/n}-1}{e^{-2\pi i/n}-1}$$

is $1$ if $n$ is prime and $0$ if $n$ is not prime.

The problem is that it is true for all numbers except $4$, where the expression evaluates to $1-i$ which is neither $1$ nor $0$. So, my question is: Why is he stating a wrong formula here?

My guess is that this is just a small mistake, but I do not know wether the original formula from 1895 was already wrong (I could not find the original source) and noone saw that which would be confusing since the counter-example $4$ is not so big. Probably, noone cared because usually you want primality tests only for big numbers and everyone knows that $4$ is composite.

Question 2

On the second page, Dudley says that for an integer $n>1$ the expression

$$ f(n)=\sin^2 \pi n + \sin^2\pi\left(\frac{1+(n-1)!}{n}\right)$$

is integer if and only if $n$ is prime. I believe that this is true but I do not understand why we need the first summand which is always zero. So, why did he include the first summand making this formula unnecessarily complicated?

Question 3

My last question is related to the second formula on the second page. Here, Dudley says the function $f$ can also be written as

$$ f(n)= \frac{\sin^2 \pi n}{(\pi n)^2(1-n^2)^2}\cdot \sum_{k=2}^\infty \frac{\pi n}{k \sin\pi n/k}.$$

This makes no sense whatsoever since now the factor $\sin^2 \pi n$ which is always zero is a factor in front making $f(n)$ always zero if $n$ is an integer. My question here is: Am I not seeing anything and the formula is correct or - if the formula is wrong - what was meant here?

I hope someone can help me. Many thanks in advance!

Answer 1

It is easier to respond to questions like these if you limit yourself to one question per post :)

Re 1, the $n = 4$ case was omitted perhaps because Dudley was quoting Dickson, who also omitted it. As you note, it is fairly common to omit small counterexamples when they are the only counterexamples (it is also possible that Dudley and/or Dickson did not notice the small counterexample). The original source, H. Laurent in Comptes Rendus vol. 126 (1898), pp. 809-810, does include the restriction $n > 4$. See http://gallica.bnf.fr/ark:/12148/bpt6k3082d/f809.item (a direct link to the relevant page).

Re question 2, Dudley is specializing a slightly more complicated statement in Dickson (which is itself a summary of a paper that Dickson is only loosely describing), and maybe Dudley is just not realizing that it does indeed simplify further. You can see the original in page 428 of Dickson, which is also out of (at least US) copyright and available freely at https://archive.org/details/historyoftheoryo01dick. While Dudley's oversight does make the formula more complicated than it could be, one broader point of Dudley's paper seems to be that it is easy to produce "formulas for prime numbers" if you don't specify many restrictions on what a "formula" can be or otherwise care about what your "formula" might look like (the two examples discussed in this answer, for example, seem to be putting fairly accessible facts about divisors of factorials in very complicated dress as "formulas"). In this context, the overly esoteric appearance of the formula may (unintentionally?) support Dudley's point.

Answer 2

The other answer made everything clear. I will just add why the formula in your second question holds (I will go over the second summand, since the first is always $0$, as you stated): $$ f(n)=\sin^2\pi\left(\frac{1+(n-1)!}{n}\right)$$ This needs to be an integer. It can be only $0$ or $1$, because the only integer values of $\text{sin}(\theta)$ for $\theta\in\mathbb{R}$ are $0$ and $\pm1$. What this implies, is that $$g(n)=\frac{1+(n-1)!}{n}$$ is an integer or a half-integer. It can be a half-integer, if the denominator is even, but it's possible only for $n=2$ - the only even prime. And $g(2)=1\in \mathbb{N}$, so we want to prove that for all prime $n$-s $g(n)\in\mathbb{N}$, which would imply $f(n)=0$.

By Wilson's theorem we have that for all primes $n$: $(n-1)!\equiv-1(\text{mod }n)$. An equivalent statement of it is: There exists an integer $x$, such that $(n-1)!=nx-1$. After rearrangement you get that $$x=\frac{1+(n-1)!}{n}=g(n)\in\mathbb{N}.$$ The reverse statement is prooved in the same way, just with the steps in reversed order.