As written on Wikipedia a problem of Brocard is a problem of findingto find solutions of
$$n!+1=m^2$$
in natural numbers of an equation $$n!+1=m^2$$.
There are three known solutions: (4,5)$(4,5)$, (5,11)$(5,11)$ and (7,71)$(7,71)$.
Erdos conjectured that there are no other solutions and I also I believe Erdös' conjecture that there are no other solutions.
I thought about an approach that characterizes m$m$ by noting that m^2$m^2$ must be of the form $m^2=m_1...m_k\underbrace{0... 0 ...0}_{l\text{ times}}1$.
That That is, as n$n$ grows bigger and bigger, an n!$n!$ will have more and more zeroes at its end and n!+1$n!+1$ will have 1$1$ as the last digit before which there will be some number of zeroes.
Thus, a question begs for characterization of m´s$m$´s for which m^2$m^2$ has a lot of zeroes before the last digit, which is one$1$.
I am thinking whether it is true that only natural m`s$m$`s whose squares end in $\underbrace{0... 0 ...0}_{l\text{ times}}1$ are these ones: 101$101$,1001 $1001$,10001 $10001$,100001 $100001$,... $\ldots$ and, more generally, these ones:
101$101$,10b01 $10b01$,100b001 $100b001$,.. $\ldots$
If this is really true then we have a solution of a problem.
So, what do we can tell about m$m$ if we know that, in decimal notation, m^2$m^2$ has a form $m^2=m_1...m_k\underbrace{0... 0 ...0}_{l\text{ times}}1$?
Can we characterize those m`$m$`s?
