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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