A positive integer $n$ is called a *Hilbert number* if $\exists a,b,d \in \mathbb{N}$ such that $ 4ab-a-b = d n$ and $d|a b$.


I ran an algorithm checking divisors for all $0\lt a,b\le500$, and the only numbers $n\le500$ for which I did not find a solution are


$\{1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,288,289,324,336,361,400,441,484\}$ 


Aside from $288,336$ they are all square numbers. It might be that a wider check would exclude these numbers, as well as some (or all) of the squares, from the list.


My Question is : **Does every prime number also a Hilbert number ?**


Basic hunch tills me that its true that all **Prime Numbers** are **Hilbert Numbers**.


Any idea how small it is, would help my a lot.


Thanks.



Hilbert Numbers
Your equation $$4ab-a-b=dn$$ gives
$$(1)\quad \frac{4}{n} =\frac{1}{ab/d}+ \frac{1}{an}+\frac{1}{bn}.$$


This is (almost) the Erdos-Straus equation.
See here for the well known [Erdos-Straus conjecture](https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture), which asks
if the equation
$$\frac{4}{n} =\frac{1}{x}+ \frac{1}{y}+\frac{1}{z}$$
has positive integer solutions for all $n\geq 2$, especially the primes.
The case $n$ is a prime consists of two subcases, namely
$n$ divides one or two of the denominators on the right hand side.


There is a standard parametrization, (going back to Rosati, 1954),
but also reproduced in Mordell's book Diophantine equations,
that the case of $n$ prime, with two denominators divisible
by $n$ on the right hand side
can be written as
$$(\*) \quad \frac{4}{n} =\frac{1}{abd}+ \frac{1}{acdn}+\frac{1}{bcdn}.$$


Now with
$$e=acd, f=bcd, g=c^2d, \frac{ef}{g}=\frac{abc^2d^2}{c^2d}=abd$$ you come
exactly to
$$ \frac{4}{n} =\frac{1}{ef/g}+ \frac{1}{en}+\frac{1}{fn},$$
which is (with an exchange of letters) exactly equation (1).


It is known (among people who did computations on the Erdos-Straus
conjecture) that this special equation $(\*)$ does not have solutions
for the squares, and for $288, 336$ and $4545$. But proving that there are no more
such numbers would imply the well known Erdos-Straus conjecture.
(I did actually not check if your version is equivalent to equation $(\*)$
or only a special case of it.)


(As you ask if these numbers can be represented with larger
$a,b$: no they can't. For every fixed $n$ there are explicit search bounds
such as $ab/d \leq n, an \leq c\_1 n^2, bn \leq c\_2 n^4$
(with some explicit constants $c\_1,c\_2$.) The case of squares has been handled by several authors in various forms, e.g. Schinzel, Yamamoto, Mordell, Elsholtz-Tao.)


(I do wonder if you were trying to solve the Erdos-Straus equation
and just phrased your question as some variant of it.)


