Are there non-integer solutions to $x!=12!y!$ that make use of the Gamma function?

Question

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 $\Gamma(n) = (n-1)!$

Answer 1

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

Answer 2

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.