The term in the equation is known as the factorial of . Basically, for any positive integer , factorial is defined as the multiplicative product of all of the integers from 1 to . 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 is the equation satisfied for some positive integer ?
There are only three known solutions. Ordered pairs (,) 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 other than for = 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 ” in The Ramanujan Journal showing computational verification that no other solutions exist for up to . 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.
I have proved that there is no solution except known 4, 5 and 7 for Brocard question.
I find a couple (20;1559776269) m=1559776269,n=20
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.
Yeah… even Python code says that (4,5),(5,11),(7,71) are the only possible pairs (even if we can’t believe it).
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).