Question

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say $m$, of these three integers?

Accepted Answer

A. Let the three consecutive positive integers be $n - 1, n, n + 1$. Sum of powers: $(n - 1)^1 + n^2 + (n + 1)^3$. Sum of integers $= (n - 1) + n + (n + 1) = 3n$. Given: $$(n - 1) + n^2 + (n + 1)^3 = (3n)^2$$ $$n - 1 + n^2 + n^3 + 3n^2 + 3n + 1 = 9n^2$$ $$n^3 - 5n^2 + 4n = 0 \implies n(n - 1)(n - 4) = 0$$ Since $n$ is a positive integer greater than 1, $n = 4$. The three integers are $3, 4, 5$. The minimum is $m = 3$. This satisfies $1 \le m \le 3$.

Powers of Three Consecutive Integers

Explanation

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