Question 672884
Case 1: n is an even integer


Let n be an even integer.


So n = 2k for some integer k.


So if n = 2k, then n^3 = (2k)^3 = 8k^3


and


n^3 + n


becomes


8k^3 + 2k


which partially factors to


2(4k^3 + k)


which is in the form 


2q


where q = 4k^3 + k (which can be proven that it is also an integer). 


Since 2q is even for any integer q, this proves that if n is an even integer, then n^3+n is even.


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Case 2: n is an odd integer


Let n be an odd integer.


So n = 2k+1 for some integer k.


So if n = 2k+1, then n^3 = (2k+1)^3 = 8k^3 + 12k^2 + 6k + 1


So


n^3 + n


becomes


(8k^3 + 12k^2 + 6k + 1) + (2k + 1)


8k^3 + 12k^2 + 6k + 1 + 2k + 1


8k^3 + 12k^2 + 8k + 2


which partially factors to


2(4k^3+6k^2+4k+1)


which is in the form 


2q


where q = 4k^3+6k^2+4k+1 (which can be proven that it is also an integer). 


Since 2q is even for any integer q, this proves that if n is an odd integer, then n^3+n is even.


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We've exhausted all possibilities and scenarios because any integer is either even or odd (cannot be something else or both).


So these two cases prove that n^3 + n is an even integer for every integer n.