Perfectly factorial squares

Factor out $(n!)^2$ ; then $2(n+1)(n+2) = 2n^2 + 6n + 4$ is a perfect square $x^2$ . Multiply both sides by $2$ and complete the square to get $(2n+3)^2 - 2x^2 = 1$ . The positive integer solutions to the Pell equation $a^2 - 2b^2 = 1$ are well-known; one way to enumerate them is to start with $(a_0,b_0) = (3,2)$ and apply the formulas $a_{k+1} = 3a_k + 4b_k, b_{k+1} = 2a_k + 3b_k$ . Note that $a$ is always odd and $b$ always even. (Note that $a_0 = 3$ corresponds to $n = 0$ .)

We get $(a_1,b_1) = (17,12), (a_2,b_2) = (99,70), \ldots, (a_7,b_7) = (665857,470832),$ $(a_8,b_8) = (3880899,2744210)$ . This gives solutions $n = 7, 48, 287, 1680, 9799, 57120, 332927, 1940448, \ldots$ . The last one is larger than $10^6$ , so there are exactly $\fbox{7}$ such positive integers.

$\quad2n!\cdot (n+2)! = 2n! \cdot n! \cdot (n+1)(n+2) = 2n!^2 \cdot (n+1)(n+2)$

Since in the prime factorization of $n!^2$ all exponents are even, we can ignore this factor: it is already a square.

$\quad2(n+1)(n+2) = 4\binom{n+2}{2}$

We can discard $4 = 2^2$ as well, for the same reason.

We are then left with, $\binom{n+2}{2} = k^2$ for some integer $k$ .

The $\textrm{n}^{\textrm{th}}$ triangular number is given by $\binom{n+1}{2}$ , and so we must now count the amount of non-negative values of $n < 10^6 - 1$ such that $T_n$ is also a square number.

Unfortunately, I do not know how to continue further by hand other than by chance having the square triangular numbers memorized, and so I used A001108 to count. There are $\boxed{7}$ such square triangular numbers.

2(n!)(n+2)!.
2 $n!^{2}$ (n+1)(n+2).
thus 2(n+1)(n+2) must be square.
2(n+1)(n+2)/4 is square. thus (n+1)(n+2)/2 is square.
the n+1th triangular number is square.
referring to OEIS A001108, 7 values of n satisfy the given conditions.

JAVA CODE :::

import java.io.*;

import java.math.*;

public class PerfectlyFactorialSquares

{

public static void main(String[] args)throws IOException

{

    long a=0,i=0;

    double d=0.0;

    for(i=1;i<1000000;i++)

    {

        d=Math.sqrt(2*(i+1)*(i+2));

        if(Math.ceil(d)==Math.floor(d))

            a++;

    }

    System.out.println(a);

}

}

OUTPUT :: 7

Dividing by $n!^2$ we find that $2(n+1)(n+2)=m^2$ .

Since m is even, set $m=2k$ , now $(n+1)(n+2)=2k^2$ .

Setting $p=n+1$ , we get $k^2= \frac{p(p+1)}{2}$ . So we are actually looking for square triangular numbers.

Since $2<=p <=10^6$ , we have to look for them in the range $3 \le \frac{p(p+1)}{2} \le 500000500000$ .

Wikipedia gives us: "There are infinitely many square triangular numbers; the first few are: 0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025 (sequence A001110 in the OEIS)".

So there are $\boxed{7}$ such numbers in the required range: 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056.