document.write( "Question 974899: Prove: The sum of the cubes of two consecutive positive numbers is odd. \n" ); document.write( "

Algebra.Com's Answer #596717 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "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

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. Since

is even.\r
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and

. Either one has a remainder of 1 when divided by 2, hence

is odd.\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "So for any given

we either have

which is even plus odd = odd or odd plus even which is odd by the Theorem mentioned at the outset of this discussion.\r
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\n" ); document.write( "\n" ); document.write( "Therefore the sum of the cubes of any two consecutive integers is odd. Q.E.D. \r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it\r
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