SOLUTION: Prove that for every positive integer n, n3 + n is even

Question 672884: Prove that for every positive integer n, n3 + n is even
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
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.
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