# The Problem Of Brocard

Down through the river of time, rapids of mathematical imagination have emerged curving the path of linear thought.   Often times, problems arising from these rocky waters branch off into smaller streams and drift indefinitely avoiding the trappings of resolution.  One such problem was introduced by the French mathematician Henri Brocard in 1876 and later, in a separate paper, in 1885.  Brocard inquired about a set of possible positive integers $n$ such that the equation $n! + 1 = m^{2}$ is satisfied.

The $n!$ term in the equation is known as the factorial of $n$.  Basically, for any positive integer $n$, $n$ factorial is defined as the multiplicative product of all of the integers from 1 to $n$.   For example, 4 factorial is simply 4 x 3 x 2 x 1 = 24.  With this definition it is to see the obscurity surrounding the problem of finding a set of solutions that satisfy the equation.  The basic question still remains: For what positive integers $n$ is the equation $n! + 1 = m^{2}$  satisfied for some positive integer $m$?

There are only three known solutions.  Ordered pairs ( $n$, $m$) that satisfy the equation are known as Brown Numbers.  The only three known pairs of Brown Numbers are (4, 5), (5, 11), and (7, 71).  More formally, Brocard’s Problem asks the question: Does $n$ and $m$ exist such that $n! + 1 = m^{2}$ other than for $n$ = 4, 5, and 7?  In other words, Do more pairs of Brown Numbers exist other than (4,5), (5,11), and (7,71)?

The prolific mathematician Paul Erdős conjectured that no other pairs of Brown Numbers exist.  In 2000, mathematicians Bruce C. Berndt and William F. Galway published a paper, “The Brocard-Ramanujan diophantine equation $n! + 1 = m^{2}$” in The Ramanujan Journal  showing computational verification that no other solutions exist for $n$ up to $10^{9}$.  Furthermore, it was proved by mathematicians A. Dabrowski (1996) and Florian Luca (2002), that the equation only has finitely many solutions assuming another major unsolved problem in Number Theory, the abc conjecture, is true.

So, does there exist any more pairs of Brown Numbers?  With the computational verification of Berndt and Galway one is tempted to say no.  However, past mathematical inquiry has shown that computational verification does not always mean proof.   For some conjectures, counterexamples have been shown to creep up just beyond extensive computational rendering.  In any case, Brocard’s Problem is still gushing in the rapids of mathematical imagination. Avery Carr is a senior analyst and past senior editor for the American Mathematical Society Grad Blog. He and his wife, Alison, live in Olive Branch, MS.
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### 5 Responses to The Problem Of Brocard

1. Nurzhan says:

I have proved that there is no solution except known 4, 5 and 7 for Brocard question.

2. karim says:

I find a couple (20;1559776269) m=1559776269,n=20

• Noah says:

If the results of my Python code is to be believed, n = 20 does not give a couple. We get (20!+1)^(1/2) approximately equals 1559776268.6284978.

• Spectre says:

Yeah… even Python code says that (4,5),(5,11),(7,71) are the only possible pairs (even if we can’t believe it).

3. Spectre says:

I think there do exist more Brown numbers. Those factorials that can be expressed as the product of the successor and the predecessor of a particular m should work.

This is from a trivial rewrite I tried

Since n! + 1 = m^2 , n! = m^2 – 1 = (m + 1)(m – 1)

Taking either of the terms in the RHS as x, you’ll get n! = x(x + 2) or n! = x(x – 2).