Square Subtract Multiply Repeat

It can be seen that $(1.3)(2.4)(3.5)....(n-1)(n+1) = 1.2.3.4.5...(n-1).3.4.5.6....(n+1) = (n-1)!.(n+1)!/2 = (n-1)!^2 .n(n+1)/2$

Now since $(n-1)!^2$ is already a square the remaining part has to become a square for the hole expression to become a square. So, the smallest vallue for which $n(n+!)/2$ is a square is $8$

The format of this expression is $(A^2 - 1)(B^2 - 1)...$ . This can be factored as $(A+1)(A-1)(B+1)(B-1)$ . Factoring the first few numbers we get $(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)$ or $1\times 3 \times 2 \times 4 \times 3 \times 5$ . To get a perfect square, each number must be repeated an even number of times. Here, 3 is repeated twice, 4 is a perfect square, but 5 and two are not repeated. The last number in our factoring (in this case, 5) will not have a number repeating it. Thus the last number should be a perfect square. We try N as 8 and we see the product is a perfect square.

Prior to the expression, we can factorize and simplify to know the integral factors. F = (2^2 - 1)(3^2 - 1)(4^2 - 1)...(N^2 - 1) = (2 - 1)(2 + 1)(3 - 1)(3 + 1)(4 - 1)(4 + 1)...(N - 1)(N + 1) = (1)(3)(2)(4)(3)(5)...(N - 1)(N + 1)

Observing the pattern, one part gives the product of consecutive integers. In order for F be a perfect square, the prime numbers must occur an even number of times. Doing the case-to-case analysis, we obtain N = 8 as the smallest number for F be a perfect square.

The above product can be expressed as Gamma(N)*Gamma(N+2)/2

Now we know that Gamma(N) = (N-1)! and Gamma(N+2) = (N+1)! = N(N+1)(N-1)!

So the above expression simplifies to ((N-1)!)^2*(N(N+1)/2)

All we need to show is that the number N(N+1)/2 which is a triangular number is ALSO a square

The first three triangular numbers which are squares are 0, 1 and 36

Since N >= 2, then it means that 0 and 1 are ruled out.

So equating (N(N+1)/2) = 36 and solving for N, will give us the value of N = 8.