Solutions for x!/y!=(y+1)!

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.

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.)

Consider the primes that occur in the range $ \frac{x}{2} < p_i < x $.

If $x! = (y!)^2 (y+1)$, then each prime $p_i$ must occur exactly once, which means that $y < p_i $, and thus $y=p_i + 1$. Hence, if we have 2 primes in the range, then there is no possible value of $y$ that satisfies the equation.

We use a stronger form of Bertrand's Postulate, which state that if $n \geq 12$, then there is a prime between $n $ and $\frac{4}{3}n$. In particular, this gives us 2 primes between $n$ and $\left(\frac{4}{3} \right)^2n < 2n$.

Hence, we need only check for solutions up to $x= 25$. I leave you to check that the only solutions are $(x,y) = (10,6), (2,1), (1,0)$

Note, the generalized version of Bertrand's postulate states that, for any constant $k>1$, there exists an integer $N$ such that for all $n>N$, there is a prime between $n$ and $kn$.

I just happen to know that for $k= \frac{4}{3}$, $N=12$. This gives 2 primes between $n$ and $2n$, which is often useful (like in this case).

I do not know what Wolfram alpha does, but

$$x=2,$$ $$y=1,$$

satisfies $\frac{x!}{y!}=(y+1)!$.