Prove that the product of four consecutive positive integers plus one is a perfect square I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.

If $n = 1 + m$, where $m$ is the product of four consecutive positive
  integers, prove that $n$ is a perfect square.

Now since $m = p(p+1)(p+2)(p+3)$;

$p = 0, n = 1$ - Perfect Square
$p = 1, n = 25$ - Perfect Square
$p = 2, n = 121$ - Perfect Square

Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?

Recursion on the square root, k(n)
n = lead integer
k(n) = square root [n(n+1)(n+2)(n+3) + 1]

k(1)^2 =   25
k(2)^2 =  121
k(3)^2 =  361
k(4)^2 =  841
k(5)^2 = 1681

k(1) =  5
k(2) = 11 =  5+ 6 =   5 + 2*(2+1)
k(3) = 19 = 11+ 8 =  11 + 2*(3+1) = 5 + 2 * ((2+1) + (3+1))
k(4) = 29 = 19+10 =  19 + 2*(4+1) = 5 + 2 * ((2+1) + (3+1) + (4+1))
k(5) = 41 = 29+12 =  29 + 2*(5+1) = 5 + 2 * ((2+1) + (3+1) + (4+1) + (5+1))

k(n) = 5 + 2 * (  (n+1)(n+2)/2    -   (1+2)  ) 
 = (n+1)(n+2) - 1


induction step
k(n+1) = k(n)             + 2(n+2)
   = (n+1)(n+2) - 1   + 2(n+2)

   = (n+2)(n+3) - 1


k(n)^2 - 1  = ((n+1)(n+2))- 1)^2  -  1
        = ((n+1)(n+2))^2 - 2(n+1)(n+2)

        = ((n+1)(n+2)) * ((n+1)(n+2) - 2)

        = ((n+1)(n+2)) * (n^2 + 3n)

        = n(n+1)(n+2)(n+3)


!!!

Thinking a little more about k(n).

A rough approximation of the root of ( n(n+1)(n+2)(n+3) + 1 )   would be n^2.

We might improve it a bit by averaging the products of the outer and inner pair of factors,

k(n)   = (n(n+3) + (n+1)(n+2))  /  2
   = (n+1)(n+2) -1


Dead on.  Surprising, no?

