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


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.


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    $\begingroup$ This nomenclature seems to disagree with wikiwand.com/en/Hilbert_number $\endgroup$
    – Igor Rivin
    Commented May 17, 2017 at 22:49
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    $\begingroup$ Also posted to m.se, math.stackexchange.com/questions/2285274/hilbert-numbers without notice to either site. Please don't do that. $\endgroup$ Commented May 17, 2017 at 23:21
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    $\begingroup$ @IgorRivin I called them this name because i like David Hilbert, they are numbers with the conditions stated above $\endgroup$
    – Ahmad
    Commented May 17, 2017 at 23:56
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    $\begingroup$ Intriguing question. It sometimes helps if you explain where the problem came from. Also, what were your own ideas to attack this problem? And do you have any observations which may be useful? As for me, I am faintly reminded of the famous sixth problem of the 1988 IMO. $\endgroup$
    – R.P.
    Commented May 18, 2017 at 10:54
  • $\begingroup$ @René I managed to prove that if $n$ is $-d \mod 4m-1$ such that $d|m$ and $m \geq 1$ then $n$ is sure a Hilbert Number, also every number of the form : $4k+2$,$8k+5$,$12k+8$,$16k+15$ for $k \geq 0$. then $n$ is a Hilbert Number, but the problem with my approach is that i look for patterns in modular arithmetic which i highly doubt that it will provide a complete answer(i may be wrong). $\endgroup$
    – Ahmad
    Commented May 18, 2017 at 21:33

2 Answers 2


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.)

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    $\begingroup$ Very nice analysis. $\endgroup$ Commented May 19, 2017 at 17:05
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    $\begingroup$ I second that. +1 $\endgroup$
    – R.P.
    Commented May 19, 2017 at 17:27

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 $10^7$.


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