2

When I first discovered this problem, I rewrote $n!+1=m^2$ like this: $$n!=m^2-1$$ which can be rewritten like this: $$n!=(m+1)(m-1)$$

From this, I've gathered the following:

  • Because n! must contain a factor of 2 to satisfy the equation (smallest possible $n$ is $4$), $(m+1)(m-1)$ must be even; $m$ must be odd to satisfy this.
  • Only one of either $(m+1)$ or $(m-1)$ can be divided by 2 to a power of more than one; the other one can be divided by 2, but to no greater power.
  • Because prime factors greater than 2 can only occur once between 3 consecutive numbers, and because statement 2 implies that $(m+1)$ and $(m-1)$ have a GCF of 2, $\frac{m+1}{2}$ and $\frac{m-1}{2}$ must be coprime.

These statements hold true for all currently known solutions: $$4!=5^2-1=4*6=6*4=(2*3)*(2^2)$$ $$5!=11^2-1=10*12=(2*5)*(2^2*3)$$ $$7!=71^2-1=70*72=(2*5*7)*(2^3*3^2)$$

So, is this information at all useful to solving/helping get around Brocard's problem?

weatherman115
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    They are fine observations, but a generally useful principle is that if short and simple observations are all that are necessary to solve something, it wouldn't have been open for 152 years. –  Jan 23 '18 at 19:49
  • Just one thing, (m-1)/2 and (m+1)/2 are coprime because one of them is the other plus 1, that's all. – Arnaud Mortier Jan 23 '18 at 20:02

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