As [written](https://en.wikipedia.org/wiki/Brocard%27s_problem) on Wikipedia a problem of Brocard is to find solutions of 

$$n!+1=m^2$$

in natural numbers.

There are three known solutions: $(4,5)$, $(5,11)$ and $(7,71)$.  I believe Erdös' conjecture 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 $1$.

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: $101$, $1001$, $10001$, $100001$, $\ldots$ and, more generally, these ones:

$101$, $10b01$, $100b001$, $\ldots$

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?