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?
Your technique should have worked, but if you don't know which expansions to do first you can get yourself in a tangle of algebra and make silly mistakes that bring the whole thing crashing down.

The way I reasoned was, well, I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. So I'll try multiplying the big one with the small one, and the two middle ones.

$$p(p+1)(p+2)(p+3) + 1 = (p^2 + 3p)(p^2 + 3p + 2) + 1$$

Now those terms are nearly the same. How can we force them together? I'm going to use the basic but sometimes-overlooked fact that $xy = (x+1)y - y$, and likewise $x(y + 1) = xy + x$.

$$\begin{align*}
(p^2 + 3p)(p^2 + 3p + 2) + 1 &amp;= (p^2 + 3p + 1)(p^2 + 3p + 2) - (p^2 + 3p + 2) + 1 \\
 &amp;= (p^2 + 3p + 1)(p^2 + 3p + 1) + (p^2 + 3p + 1) - (p^2 + 3p + 2) + 1 \\
 &amp;= (p^2 + 3p + 1)^2
\end{align*}$$
Tada.
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?