Can I say the integer solution of this equation is unique?

Question

I want to solve the equation $\frac{y!}{(x-2)!(y-x)!}=\frac{1340!}{659!\times679!}$ where $x,y \in \mathbb{Z}$.
I think it is natural to say $(x,y)=(681,1340)$ and $(x,y)=(661,1340)$ are desired solutions.
But can I assure that there are no other solutions?
I tried MATLAB and Wolframalpha, but they stop working due to the huge number $\frac{1340!}{659!\times679!}$.
Would you help me?

Answer 1

COMMENT.-Note that this is a type of problem formed as follows: if $a + b + 2 = c$ then the equation $$\frac{y!}{(x-2)!(y-x)!}=\frac{c!}{a!b!}$$ has the obvious two solutions $ x = a + 2 $ and $ x = b + 2 $ with $ y = c $ in both cases.

Take $(a,b,c)=(659,679,1340)$ and note that we have for the fixed value $c=1340$ all the possibilities $\dfrac{1340!}{(a+2)!(1338-a)!}$ corresponding to $x=a+2$ with many admissible values of $a$.

Because of all these possible values are binomial coefficients in developing $(x+1)^{1340}$ and these are two equal (for the even $1340$ there is one non repeated which is $\binom{1340}{670}$) it follows that the two given solutions, $x=661$ and $x=681$ are the only ones $\color{red}{\text{ assuming that } y=1340}$.

I don't know if proving $y$ is necessarily equal to $1340$ is easy or difficult (I feel it is not hard and in any case one must have $1327\le y\le1361$ where $1327$ and $1361$ are consecutive primes).