# Compute the sum over all integers n for which (n^3 + 8) / (n^2 - 4) is an integer?

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Compute the sum over all integers n for which (n^3 + 8) / (n^2 - 4) is an integer?
posted Nov 3, 2018

–1 vote

14

n^3+8=(n+2)*(n^2-2n+4) -----> (1)
n^2-4=(n+2)*(n-2) -----> (2)
(1)/(2)= (n^2-2n+4)/(n-2)= n+ 4/(n-2) -----> (3)
For (3) to be an integer 4/(n-2) must be integer, then n-2 = {-4,-2,-1,1,2,4} => n={-2,0,1,3,4,6}
n=-2 is excluded as n^2-4 can't be zero due to divisibility rule
Sum of all n will be 0+1+3+4+6=14

Check for n=1 again! And for n = -2, 0 & 6
I considered only positives... but yes it says all... thanks
Answer will be 14 since n can't be -2as for n=-2 ,n^2-4will be=0
So sum will be 0+1+3+4+6=14
Agreed, thanks
We have divided (n^3+8) with (n^2 - 4) eliminating a common factor (n + 2). Therefore n = -2 is still valid. As long as you get a valid answer for [n+ 4/(n-2)] all values of   n yield valid answers. For n = -2 we have
[n+ 4/(n-2)] = [ (-2) + (4/(-2 - 2)) ]
= (-2) + (-1) = -3 which is valid.

(n^3 + 2^3) = (n + 2)×(n^2 - 2n + 4)
(n^2 - 2^2) = (n + 2)(n - 2)
(n^3 + 2^3)/(n^2 - 2^2) = (n^2 - 2n + 4)/(n - 2) = n + (4/(n - 2))
This means for n = -2, 0, 1, 3, 4 & 6 we have (n^3 + 2^3)/(n^2 - 2^2) = -3, -2, -3, 7, 6 & 7 as for n > 6 & n < (-2), 4/(n - 2) will always yield a fraction making
(n^2 - 2n + 4)/(n - 2) = n + (4/(n - 2)) always a non integer beyond n > 6 & n < (-2).
Therefore the required sum = -2 + 0 + 1 + 3 + 4 + 6 = 12.

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