As written on Wikipedia a problem of Brocard is a problem of finding solutions in natural numbers of an equation $$n!+1=m^2$$

There are three known solutions: (4,5), (5,11) and (7,71).

Erdos conjectured that there are no other solutions and I also believe that there are no other solutions.

I thought about an approach that characterizes m by noting that m^2 must be of the form $m^2=m_1...m_k\underbrace{0... 0 ...0}_{l\text{ times}}1$.

That is, as n grows bigger and bigger, an n! will have more and more zeroes at its end and n!+1 will have 1 as the last digit before which there will be some number of zeroes.

Thus, a question begs for characterization of m´s for which m^2 has a lot of zeroes before the last digit, which is one.

I am thinking whether it is true that only natural m`s whose squares end in $\underbrace{0... 0 ...0}_{l\text{ times}}1$ are these ones: 101,1001,10001,100001,... and, more generally, these ones:

101,10b01,100b001,..

If this is really true then we have a solution of a problem.

So, what do we can tell about m if we know that, in decimal notation, m^2 has a form $m^2=m_1...m_k\underbrace{0... 0 ...0}_{l\text{ times}}1$?

Can we characterize those m`s?