<s>[INST] Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How do I proceed from here to end up with something squared $- 1$?
 [/INST] let the nos be 
n-1, n, n+1, n+2
we need to prove that (n-1)(n)(n+1)(n+2)+1= square no.
rearranging it : (n-1)(n+2)(n)(n+1)+1= square no.
LHS= (n-1)(n+2)(n)(n+1)+1
   =(n^2+n-2)(n^2 +n)+1
   = (k-2)(k) + 1                    (by replacing (n^2 +n) by k)
   =  k^2 -2k +1
   = (k-1)^2 
 hence proved , LHS is a square no.
on an interesting note,
(k-1)^2 = (n^2 +n-1)^2= {(n)(n+1)-1} ^2
therefore, the no. formed is always the square of, product of the two middle terms -1.  
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