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As written 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?

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  • $\begingroup$ More generally, a prime power that divides the factorial of n divides exactly one of m+1, m-1, except that 2 divides both and so one of the twos goes in one factor and the rest in the other. So one of the factors is (for large n) a multiple of 10, so m ends in 1 or 9. However, it is not clear that appropriate prime powers will be so allocated among m+1 and m-1 for m larger than 70. Gerhard "This Way Is Not It" Paseman, 2018.09.02. $\endgroup$
    – Gerhard Paseman
    Commented Sep 3, 2018 at 0:33
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    $\begingroup$ $249^2=62001$, $1249^2=1560001$, $18751^2=351600001$, .... $\endgroup$
    – Gerry Myerson
    Commented Sep 3, 2018 at 2:42

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$m^2 \equiv 1 \mod 10^k$ iff $m \equiv \pm 1 \mod 2^{k-1}$ and $m \equiv \pm 1 \mod 5^k$ (not necessarily the same $\pm$).

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