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So I recently looked at a video on YouTube by Numberphile which is accessible down below:

Brown Numbers $-$ Numberphile

The video was about so-called numbers called Brown Numbers. These types of numbers are a pair of integers $(m, n)$ such that $$n! + 1 = m^2$$ When you have a term looking a little like this $\longrightarrow n!$ it is not read as shouting the term before the exclamation mark (and the term before the exclamation mark isn't called the coefficient either). It is read as, in this case, $n$ factorial and is equal to the product of every number from $1$ to $n$ including $n$ itself. $$n! = 1\times 2\times 3\times \cdots \times n = \prod_{k = 1}^{n}k$$ Understanding what $n!$ means, we can now find a pair of Brown Numbers to substitute for $m$ and $n$ in the equation, however in the video, it was revealed that only three pairs of Brown Numbers have been found, and it has been conjectured that there only exists three pairs of Brown Numbers. $$\begin{align} (5, 4) \qquad 4! + 1 &= 1\times 2\times 3\times 4 + 1 = 24 + 1 = 25 = 5^2 \\ (11, 5) \qquad 5! + 1 &= 1\times 2\times 3\times 4\times 5 + 1 = 120 + 1 = 121 = 11^2 \\ (71, 7) \qquad 7! + 1 &= 1\times 2\times 3\times 4\times 5\times 6\times 7 + 1 = 5040 + 1 = 5041 = 71^2 \end{align}$$ It was also revealed in the video that this conjecture has not been proven, so I accepted the challenge of proving it, but came across a problem.

My Attempt:

If $$n! + 1 = m^2$$ then $$n! = m^2 - 1 = m^2 - 1^2 = (m + 1)(m - 1)$$ Assume that $m = 2r$ for some integer $r$. $$n! = (2r + 1)(2r - 1) = (2r)^2 - 1^2 = 4r^2 - 1$$ $$\implies n! + 1 = 4r^2$$ We arrive at a contradiction. Since $n! = 1\times 2\times \cdots \times n$ then if $n > 1$, $n$ is even. Any even number plus $1$ is an odd number, and an even number cannot divide an odd number, thus $4$ does not divide an odd number, so $m$ cannot be even.

Now let $m = 2r \pm 1$. $$n! = (2r \pm 1 + 1)(2r \pm 1 - 1) = (2r \pm 2)(2r) = 2(r \pm 1)(2r) = 4r(r \pm 1)$$ We now arrive at another contradiction. $4$ now divides $m$, but $4$ is even and $m$ is odd! (This attempt can serve as a Lemma to proving that for any integer $x$, if $x$ is even/odd, then $x^y$ is even/odd.)

Did I do something wrong? I must have. Besides, every Brown Number so far is odd and not even.

Mr Pie
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2 Answers2

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$m = 2r +1$ is enough to generate all odd numbers, no need for the $\pm$.

$$n! = m^2 - 1 = (m+1)(m-1) = (2r + 2)(2r) = 4r(r+1)$$

The problem with your argument is that $4r(r+1) = m^2 - 1$, but you're trying to argue that $m^2$ is divisible by 4, I believe because you've forgotten the $-1$ and continued with $4r(r+1) = m^2$.

orlp
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  • Oh yes true!!! Oh my BLOB I am such a noob. So so sorry if I ever was a waste of time $$\color{green}{\checkmark}$$ – Mr Pie Sep 25 '17 at 12:42
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Note that $4!=4\times (2\times 3)$; $5!=4\times (5\times 6)$ and $7!=4\times (35\times 36)$ all exist - but considering these particular examples might help you to see where your argument is going wrong or getting confused. You might also note that the general odd number can be written $2r+1$, so you don't need the complication of $\pm$.

Mark Bennet
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