Question 1059605
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Prove that if 1 is added to the product of any four 
consecutive integers, the sum is a perfect square.  
Thank you in advance.
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<pre>
The problem asks us to prove that n*(n+1)*(n+2)*(n+3) + 1 is a square of an integer.


Let x = {{{(n+1) + 1/2}}} be the central point for the original four integers n, n+1, n+2 and n+3.  Then

  n*(n+1)*(n+2)*(n+3) + 1 = (x+0.5)*(x-0.5)*(x+1.5)*(x-1.5) + 1 = 

= {{{(x^2-0.25)*(x^2-2.25) + 1}}} = {{{x^4 - 2.5x^2 + (2.25/4) + 1}}} = 

= {{{x^4 - 2.5x^2 + 6.25/4}}} = {{{(x^2-2.5/2)^2}}}.


Next,  {{{x^2 - 2.5/2}}} = (by the definition of "x") = {{{(n+1)^2 + (n+1) + (1/2)^2 - 5/4}}} = {{{(n+1)^2 + (n+1) - 1}}}  is the integer number.


Thus we proved that  n*(n+1)*(n+2)*(n+3) + 1  is the square of the integer number  {{{(n+1)^2 + (n+1) - 1}}} = {{{(n+1)^2 + n}}}.
</pre>

Proved and solved.