I was watching a video recently, and I saw how 10*9*8*7 was equal to 7*6*5*4*3*2*1, or to make it clearer, 10!/6!=7!. I was wondering if there were any other solutions, so I checked the web, to find nothing. I also checked Wolfram alpha, but it gave me just two extra solutions for x=10 and point.
So, what kind of solutions are there? Are there infinite solutions for any arbitrary x? Are there infinite integer solutions for x and y?
Anything would help, I have no idea of how to find these kinds of solutions…
EDIT: Thomas Andrews told me that when talking about negative integers I should use the Gamma function. But to make it simple, can you simply extend the question to negative or complex numbers? Thanks.
Best Answer
We are solving $x! = y! (y+1)! = (y+1)(y!)^2$ over the positive integers.
We can prove the following. Given $x$, and $p$ the largest prime less than or equal to $x$, then $y=p-1$. In fact, if $y\geq p$, then $p^2 | y!(y+1)!$, but $p^2$ does not divide $x!$. If $y < p-1$, then $p$ does not divide $y! (y+1)!$, but $p | x!$.
Note the above is the case with $x=10$ and $y=6$.
My hope is that using some fact about prime numbers, such as the prime density theorem, we can prove that for $x \geq M$ for some $M$, there are no solutions. That is, there are a finite number of solutions.
EDIT: Using the generalized Bertrand's postulate, for large enough numbers, there is a prime $p$ with $x \geq p\geq 3/4 x$. Then if a solution exists, $$x! = \Gamma(x+1) = \Gamma(p+1)\Gamma(p+2) \geq \Gamma(p+1)^2 \geq \Gamma(3x/4+1)^2.$$ Substituting Stirling's formula gives $$(x/e)^x \sqrt{2 \pi x} \geq [(3x/4e)^{3x/4} \sqrt{2 \pi (3x/4)}]^2$$ which simplifies to $$\sqrt{2 \pi x} \geq \frac{3 \pi}{2} x \left(\left(3/4e\right)^3 x \right)^{x/2}.$$
However, as $x \to \infty$, the RHS goes to infinity faster than the LHS. Hence, the inequality is violated and we conclude that there are a finite number of solutions. (All of the above can be made rigorous, knowing that Stirling's formula is an asymptotic result, and dealing with limits.)