Square roots and !'s!

Let's first consider the case where $a, b,$ and $c$ are all equal. This implies that $c = 0$ , and thus so are $a$ and $b$ . It is easy to see that $(0, 0, 0)$ is a solution.

Now notice that if $c$ is not the smallest in the sequence, we have that either $c = a + c$ or $c = a + 2c$ , implying in either case that all of the integers are equal, which we have already dealt with, and thus $c$ is the smallest in the sequence.

Next, notice that if $a < b$ we can rewrite this equation as

$\sqrt{ (2c)! } = (3c)!\sqrt{c!} \:\Rightarrow$

$(2c)! = c!(3c)!^2$

The RHS (right-hand side) clearly grows far faster than the LHS, and already at $c = 1$ , the RHS is greater than the LHS.

Finally, consider $a > b$ , allowing us to rewrite the equation as

$\sqrt{ (3c)! } = (2c)! \sqrt{ c! } \:\Rightarrow$

$\frac{(3c)!}{c!} = (2c)!^2$

Now we will prove that the LHS cannot be a square number. Consider the largest prime number $p$ such that $c < p \leq 3c$ . In order for the LHS to be a square number, $c < 2p \leq 3c$ so that the LHS will have at least two factors of $p$ , but by Bertrand's Postulate, there must exist another prime number $q$ such that $p < q < 2p$ , making $q > p$ , which is a contradiction.

Thus, we have but $\boxed{1}$ solution, $(0, 0, 0)$ .