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$\newcommand{\floor}[1]{\lfloor #1 \rfloor}$ I have the equation $m!=100x^2+20x$ where $x$ and $m$ are real non-negative integers. I wish to disprove for when $m\geq20$ how can I do this? I had an idea that is based on ending digits. For example m! can be divided by $10^{\floor{\frac{m}{5}}-1}$ and still end in a $0$. I was wondering if one could put an argument for that the equation doesn't exist. Using iteration. For example if I substitute $x$ with $10x_1+5$(other correct substitution is $10x_1+10$ one gets $m!/200=50x_1^2+51x_1+13$ then one can proof that $x_2$ ends in a $7$ so $\therefore$ I can substitute $ x_1$ for $10x_2+7.$ Can one prove the following equation doesn't equal possibly using a lower bound on the approximation of the factorial function.

My method $\sqrt{2\pi}m^{m+0.5}e^{-m}$is the lower bound for $m!$

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  • $\begingroup$ For big $m$, approximations for $m!$ do not do well to the lower digits. $\endgroup$ – Hagen von Eitzen Apr 22 '18 at 17:44
  • $\begingroup$ I think that $m!+1$ cannot be square $\endgroup$ – Hagen von Eitzen Apr 22 '18 at 17:48
  • $\begingroup$ @hagenvoneitzen I understand. Can I give you a simpler yet hard question. If one has the equation (for thisexample it doesn't have to be intergers) $100x^2+20x=m!$ and $100y^2+20y=k*m! $ Imagine $x=10y$. Obviously $k>1$. Can you give be a lower bound for $k$ using $m$&$k $. $\endgroup$ – PintOfMilk Apr 22 '18 at 17:54
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    $\begingroup$ @HagenvonEitzen en.wikipedia.org/wiki/Brocard%27s_problem it is still an open problem. But that was a nice idea to use the discriminant. $\endgroup$ – zwim Apr 22 '18 at 17:56
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    $\begingroup$ We have $m!+1 = (10x+1)^2 = y^2$, which is Brocard's problem and still open. $\endgroup$ – Serikbolsyn Duisembay Apr 22 '18 at 17:56
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Just eyeballing this with the first m! divisible by 20 (5!) results in 20(5x + x - 6) = 0. (5x^2 + x - 6) = (5x + 6)(x - 1). So x = 1 when m = 5 is a solution.

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