Factorial-Quadratic Diophantine Equation

Erdos proved that $\binom{n}{k}$ , for $3 \le k \le \tfrac12n$ is only ever the power of an integer in the case $\binom{50}{3} = 140^2$ . Thus $\binom{2n}{n}$ is never a perfect square for $n \ge 3$ , and it is easy to check that $\binom{2}{1}=2$ and $\binom{4}{2} = 6$ are not perfect squares. Thus there are $\boxed{0}$ solutions.

By direct calculation, we can verify that the equation does not have any solution for $n=1$ and $n=2.$ Thus, we can assume that $n\geq 3$ and are going to prove that the equation does not have any solution in this case either. Using Bertrand's Postulate , we can draw the conclusion that for any integer $n\geq 3,$ there is a prime number $p$ such that $n<p<2n.$ Since $n\geq 3,$ then $p^2 > n^2>2n.$ Now, we can use the property that the exponent of the prime number $p$ in the prime factorization of any number of the form $r!$ is given by the sum $\sum_{k=1}^{\infty}\lfloor \frac{r}{p^k}\rfloor.$ where $\lfloor \cdot \rfloor$ represents the floor function (for an explanation of this formula, use the following link ). Then the exponent of $p$ in the prime factorization of the number $\frac{(2n)!}{(n!)^2}$ is $\sum_{k=1}^{\infty}(\lfloor \frac{2n}{p^k}\rfloor-2\lfloor \frac{n}{p^k}\rfloor).$ Using the inequalities mentioned above for $p$ and $n$ , we obtain that the exponent must be 1. Hence the number $\frac{(2n)!}{(n!)^2}$ cannot be a perfect square. Therefore, the number of integer solutions of the given equation is $\boxed{0}.$

Note : It is easy to see that $\frac{(2n)!}{(n!)^2}$ is a positive integer number for any positive integer $n,$ because it gives the total number of combinations of $2n$ elements taken $n$ at the time.