Question 1207642
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As tutor Edwin points out in his long division table, the 1 at the very bottom represents remainder 1.


This means that computing {{{((n+1)^2)/(n+2)}}} will give some quotient with remainder 1.


Here are some examples:
n = 5 ---> {{{((n+1)^2)/(n+2) = ((5+1)^2)/(5+2) = 36/7 = matrix(1,3,5,"remainder",1)}}}
and
n = 6 ---> {{{((n+1)^2)/(n+2) = ((6+1)^2)/(6+2) = 49/8 = matrix(1,3,6,"remainder",1)}}}
and
n = 7 ---> {{{((n+1)^2)/(n+2) = ((7+1)^2)/(7+2) = 64/9 = matrix(1,3,7,"remainder",1)}}}


Getting remainder 1 means (n+2) is not a factor of (n+1)^2, and furthermore means {{{((n+1)^2)/(n+2)}}} is never an integer when n is a positive integer.



If we got remainder 0, then it would prove the fraction is an integer.


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Another approach that doesn't use long division


{{{((n+1)^2)/(n+2)}}}

 
= {{{(n^2+2n+1)/(n+2)}}} 


= {{{((n^2+2n)+1)/(n+2)}}} 


= {{{(n^2+2n)/(n+2)+1/(n+2)}}} 


= {{{(n(n+2))/(n+2)+1/(n+2)}}} 


= {{{n+1/(n+2)}}} 


In short,
{{{((n+1)^2)/(n+2) = n+1/(n+2)}}} 


We get a whole number n, but an additional fractional amount 1/(n+2) to prove that {{{n+1/(n+2)}}} is never an integer.
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