Question 974899: Prove: The sum of the cubes of two consecutive positive numbers is odd.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If you have previously proven a theorem that asserts the sum of an odd integer and an even integer is odd, you can use that here. If you haven't proven such an assertion, then you have two choices at this point: Prove it for yourself or take it on faith.
The  th even integer is  , the th odd integer is  , and the ) th odd integer (the odd integer 1 larger than the th even integer) is  .
^3\ =\ 8n^3) . Since  , ^3) is even.
^3\ =\ 8n^3\ -\ 12n^2\ +\ 6n\ -\ 1) and ^3\ =\ 8n^3\ +\ 12n^2\ +\ 6n\ +\ 1) . Either one has a remainder of 1 when divided by 2, hence ^3) is odd.
If the first of two consecutive integers is even, then the second is odd. If the first of two consecutive integers is odd, then the second is even.
So for any given we either have ^3\ +\ (2n\ +\ 1)^3) or ^3\ +\ (2n)^3) which is even plus odd = odd or odd plus even which is odd by the Theorem mentioned at the outset of this discussion.
Therefore the sum of the cubes of any two consecutive integers is odd. Q.E.D.
John
My calculator said it, I believe it, that settles it
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