The Problem Of Brocard

The first page of Brocard's  original 1897 Paper.  Photo Credit [ Wikimedia Commons]

The first page of Brocard’s original 1897 Paper. Photo Credit [ Wikimedia Commons]

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.

About Avery Carr

I am a grad student studying mathematics at Emporia State University. I obtained a B.S. in Mathematics from the University of Memphis in 2010 where I served as President of the math club my senior year. My research interests include Graph Theory, Number Theory, Recreational Mathematics, Mathematical Physics, and Operator Theory.
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13 Responses to The Problem Of Brocard

  1. chrisnamastephys11 says:

    Great post, Avery! Thank you for sharing–I learned about a new concept: Brown numbers, and I learned about a new French mathematician: Henri Brocard!

    One thing, though, you write in your second paragraph that 4 factorial is 120; it should be 24. 5! = 120.

  2. Avery Carr says:

    Thank you for your comment and correction. It is corrected now. I originally used 5 factorial and overlooked changing the right hand side when I switched to 4 factorial. Good catch and thanks again.

    • chrisnamastephys11 says:

      You’re welcome, and I thought that you may have started out with 5 factorial. Keep up the great and interesting posts! Look forward to your posts, always, Avery!

      -Christian Luca

  3. Avery Carr says:

    Thank you Christian for your kind comments.

  4. Ray says:

    Hello guys,
    I might have proved it that there is no other pairs to the equation; but I don’t know where I can post my solution for others to check. I am not a professional mathematician so I don’t like to pay my own money to publish it on a mathematical journal unless it is free to me ^-^. I solve math problems just for fun. Is there any free website that I can post its solution (proof) to the problem so other people to check it?

    • Avery Carr says:


      One thing I would do is write up an electronic version of your proof and then search an appropriate journal ( like the Journal of Number Theory) and submit it to them electronically. They will rigorously review it and you shouldn’t be out of any of your own money if they accept it for publication. Also, a good site,, is a place you can post your results before official publication for peer review. I hope this helps and if anyone else has advice please post it. Thank you.

      • Ray says:

        Thank you for your advices. They really help me!

        • Avery Carr says:

          You are welcome. Please keep us posted on your progress.

          • Ray says:

            Sure. Are you a mathematician?

          • Ray says:

            Avery, You have posted your background over there. Sorry, I didn’t see that. That’s why you know so much.

          • Avery Carr says:

            Thank you. I am grad student studying mathematics and enjoy learning more everyday. Math is one of those fields that the more you learn, the deeper and more generalized it gets. There is always something to learn that challenges the limits of your imagination and knowledge. And there are interesting problems to explore that stretch there abstract tentacles across time and manifest themselves as something useful in many applied fields.

          • Ray says:

            Since you are a math student with good training in math. I can post my proof here first so you can help to check if there is some bugs in it because the solution is too simple to believe.

          • Avery Carr says:


            It would probably serve you better to submit to arXive or a formal publication, but you are welcome to post it here. But realize that this a very public forum and it will not be the least bit private. Thank you.

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