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Yes. This proposition can be proved by the following reasoning: By definition, if an integer is uneven, it can be written in the form 1 + 2x, in which x represents an integer. If this binomial number is cubed, the result will be the sum of 13 plus several other terms that are all integral multiples or powers of 2. Since 13 is 1 itself, and any power or multiple of 2x is even, the cube can be written as another number 1 + 2y,

where y is an integer. However, this satisfies the definition of an uneven integer, so that the cube of an uneven integer can not be even.

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