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Disprove the statement that every positive integer is the sum of at most two squares and a cube of nonnegative integers.

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Final answer:

To show that not every positive integer is the sum of two squares and a cube, we note that the number 7 cannot be expressed in this form, disproving the statement.

Step-by-step explanation:

To disprove the statement that every positive integer is the sum of at most two squares and a cube of nonnegative integers, we need to find a counterexample. Such a counterexample would be a positive integer that cannot be expressed as the sum of two squares and a cube of nonnegative integers. To illustrate this concept with a concrete example, let's consider the positive integer 7. This number cannot be expressed as the sum of two squares (0, 1, 4, 9, 16, ...) and a cube (0, 1, 8, 27, ...), because the only nonnegative integers whose squares are less than or equal to 7 are 0, 1, and 2 (yielding the squares 0, 1, and 4, respectively), and the only cube less than or equal to 7 is 1. No combination of these squares and the cube sum up to 7 (since 4+1+1=6 and 4+1+0=5), therefore, the original assertion is disproved.

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