A Perfect Square? No Way!

We have to find the values of $n$ for which $2*n!*(n+2)!$ is a perfect square which is clearly equivalent to proving $2(n+1)(n+2)$ to be a perfect square.

So we have for some integer $k$ , $2(n+1)(n+2) = k^{2}$ $2n^{2}+6n+4 = k^{2}$ $2n^{2}+6n+4- k^{2} = 0$

So the discriminant of this equation should be a perfect square to yield integer solutions(that is the sufficient condition in this case and it can be proved by checking) for $n$ ,hence,

$36 - 4*2*(4-k^{2}) = p^{2}$ $4+8k^{2} = p^{2}$ $p^{2} - 8k^{2} = 4$ Now if pis odd,the RHS can never be even,so $p$ must be even,say $p =2p_1$ , $4p_1^{2} - 8k^{2} = 4$ $p_1^{2} - 2k^{2} = 1$ $p_1^{2} = 2k^{2} + 1$ Now k must be even,otherwise $p_1^{2}$ will become of the form $4k+3$ which cannot be a perfect square.So let $k = 2k_1$ $p_1^{2} = 8k_1^{2} + 1$ $p_1^{2} - 8k_1^{2} = 1$ Now clearly have a Pell's equation ,whose one trivial solution is $p_1 = 3$ and $k_1 = 1$ . So the other solutions can be obtained from the recurrence $x_{k+1} = x_1x_k+ny_1y_k$ and $y_{k+1} = x_1y_k+y_1x_k$ .[put $p$ in place of $x$ and $k$ in place of $y$ .

And each of values correspond to a positive value of integer $n$ .

Now since we are given $n<10^{6}$ ,the largest possible permissible value in the recurrence relation will be $p_8 = 665857$ and $k_8 = 235416$ and this corresponds to $n$ having a value of $332927$ .But $p_1 = 3$ and $k_1=1$ makes $n=0$ and hence it should not be counted.

So the total number of values of $n$ possible is $\boxed{7}$ .

This equals $2 (n!^{2})(n+1)(n+2)$ , so it is a square when $2(n+1)(n+2)$ is. Notice that this value is 4 times the $(n+2)$ th triangular number, so the problem reduces to counting the triangular numbers from the 3rd (6) up to the 1,000,001st (some 12-digit number) that are perfect squares.

I remembered a recurrence for square triangular numbers (after 0 and 1); the next after a and b is $34b - a + 2$ . So this gives

36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056 and then we get to a 13-digit number.

So the answer is 7. I hope there's a more direct way to solve this one.

I used a table and was able to find a few values for n that would make n a perfect square. After I found 7, 48 and 287, I realized that from (n+1) and (n+2), one had to be a perfect square and one had to be a perfect square divided by 2. I looked at the (n+1), or (n+2), (whichever one was the perfect square) that I found and searched for a sequence that matched the square roots of (n+1) or (n+2).

One sequence did. Numerators of continued fraction convergent to sqrt(2). Look at Look athttps://oeis.org/search?q=3%2C+7%2C+17&language=english&go=Search.

In fact, sqrt(n+1) or sqrt(n+2) always equaled one of these numbers. As a result, I was able to use this sequence I found and was able to determine n by reversing the process.

I found 7, 48, 287, 1680, 9800, 57120 and 332927.