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I came across a math question today which asked for the solutions to $$x!=12!y!$$ I was wondering if there were any non-integer solutions to the equation using the extended definition of the factorial as $\Gamma(n) = (n-1)!$

Blue
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Collon
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  • An evident solution (a trivial one) is $(x,y)=(12,1)$ (also with $y=0$). It is not hard, I think, to prove there are not non-trivial solutions. – Piquito Nov 05 '19 at 18:26
  • If I was to try, I would convert the problem to binomial (m,n) form and try from there. The binomial coefficients are forced to be an integer. There is also the pochhammer function (a)n which has integer results and a _lot of proven properties. – rrogers Nov 05 '19 at 18:33
  • HINT.- $11|x!\Rightarrow x\ge12\Rightarrow 13\cdot14\cdot15\cdots x=y!$ – Piquito Nov 05 '19 at 18:38

2 Answers2

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There are a non-trivial solution besides the trivial ones $(x,y)=(12,1),(12,0)$ and this solution is unique. From $x!=12!y!$ we have $x\gt y$ so putting $x=y+h$ we get $$y!(y+1)(y+2)\cdots(y+h)=12!y!\iff (y+1)(y+2)\cdots(y+h)=12!$$ Clearly the $LHS$ has no more than eleven factors (really less!).

When $y+1=12!=2^{10}\cdot3^5\cdot5^2\cdot7\cdot11$ the first degree equation give us always a solution.We have $$\color{red}{(x,y)=(12!,12!-1)}$$

NOTE.- Always, for all $n$ one has $(n!)!=n!(n!-1)!$ wich is easy to prove.

Proving now with the successive equations of degree two, three, etc, Wolfram calculator gives non-integer (real) solutions.Thus the given non-trivial solution is unique.

Piquito
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If you look at this question of mine asking for the solution of $$n!=a^n 10^k$$ you will see a magnificent approximation built by @robjohn for this last equation. $$ n\sim ea\exp\left(\operatorname{W}\left(\frac k{ea}\log(10)-\frac1{2ea}\log(2\pi a)\right)\right)-\frac12$$ where appears Lambert function.

For your cas, let $n=x$, $a=1$, $10^k=12!\, y!=12!\,\Gamma(y+1)$ to get

$$\color{blue}{x=e \exp\left(\operatorname{W}\left(\frac 1 e\log \left(\frac{12! \,\Gamma (y+1)}{\sqrt{2 \pi }}\right)\right)\right)-\frac12}$$ which will work for any value of $y$.

Trying for a few integer values of $y$, here are some results for $x$ $$\left( \begin{array}{ccc} y & \text{approximation} & \text{solution} \\ 0 & 11.9987 & 12.0000 \\ 1 & 11.9987 & 12.0000 \\ 2 & 12.2720 & 12.2732 \\ 3 & 12.7005 & 12.7017 \\ 4 & 13.2336 & 13.2348 \\ 5 & 13.8429 & 13.8439 \\ 6 & 14.5099 & 14.5109 \\ 7 & 15.2221 & 15.2231 \\ 8 & 15.9705 & 15.9714 \\ 9 & 16.7484 & 16.7492 \\ 10 & 17.5505 & 17.5513 \\ 20 & 26.3227 & 26.3232 \\ 30 & 35.7007 & 35.7010 \\ 40 & 45.3096 & 45.3098 \\ 50 & 55.0341 & 55.0342 \\ 60 & 64.8259 & 64.8261 \\ 70 & 74.6611 & 74.6612 \\ 80 & 84.5261 & 84.5262 \\ 90 & 94.4127 & 94.4128 \\ 100 & 104.316 & 104.316 \\ 200 & 203.764 & 203.764 \\ 300 & 303.500 & 303.500 \\ 400 & 403.333 & 403.333 \\ 500 & 503.214 & 503.214 \\ 600 & 603.123 & 603.123 \\ 700 & 703.050 & 703.050 \\ 800 & 802.989 & 802.989 \\ 900 & 902.937 & 902.937 \\ 1000 & 1002.89 & 1002.89 \end{array} \right)$$

Claude Leibovici
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