Hilbert Numbers

Question

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.

Answer 1

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, 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.)

Answer 2

Dubickas and Novikas (2012) define the set $E(4)$ as $$E(4) = \{ n\in\mathbb{Z}_+\ :\ n = 4M - d'\},$$ where $(a'b')\mid M$ and $d'\mid(a'+b')$ for some positive integers $a',b'$.

It can be easily seen that $dn=4ab-a-b$ with $d\mid ab$ implies that $d\mid \gcd(a,b)^2$. Writing $d=xy^2$, where $x$ is squarefree, we have $xy\mid a$ and $xy\mid b$. Defining $a'=\frac{a}{xy}$, $b'=\frac{b}{xy}$, $M=\frac{ab}{xy^2}$, and $d'=\frac{a+b}{xy^2}$, we conclude that $n\in E(4)$.

Vice versa, given $M,d',a',b'$ and $n=4M-d'$ satisfying the conditions $(a'b')\mid M$ and $d'\mid(a'+b')$, we can set $(a,b)=(a',b')\cdot \frac{M}{a'b'}\frac{a'+b'}{d'}$ and $d=\frac{M}{a'b'}\left(\frac{a'+b'}{d'}\right)^2$ to get $dn=4ab-a-b$.

So, the equation $dn=4ab-a-b$ under the condition $d\mid ab$ defines nothing else but the set $E(4)$. As an immediate consequence of the Dubickas and Novikas (2012) result, we have that for all $n\leq 2\cdot 10^9$, the equation $dn=4ab-a-b$ is soluble unless $n$ is a square or $n\in\{288,336,4545\}$.

---

Now, some practical considerations.

First, for a given $n$ and $d$, it is relatively easy to solve $4ab-a-b=dn$ by rewriting it as $$(4a-1)(4b-1)=4dn+1$$ and factoring the number $4dn+1$.

To solve $4ab-a-b=dn$ when $d$ is unknown, we let $d=xy^2$ as above and so $a=xy\hat a$, $b=xy\hat b$. Then the equation $4ab-a-b=dn$ becomes $$4xy\hat a\hat b - \hat a - \hat b = yn.$$ This equation implies that $x\leq \frac{n+2}{4}$ and $y\leq \frac{n+1}{4x}+1$. So, there are only $O(n\log n)$ suitable values of $d$. E.g., it took me several hours to double-check that $4ab-a-b=dn$ is soluble for all prime $n$ below $10^7$.