Track / Stage 3 / 93 of 260 #93 of 2000

Problem 93

AMC 10/12, early questions

Algebra Difficulty 3.2 Prove it Junior Macedonian Mathematical Olympiad · North Macedonia

Let $x$ be a real number such that the numbers $x^3$ and $x^2 + x$ are rational. Prove that $x$ is rational.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let $a = x^3$ , $b = x^2 + x$ . Then $a = x^3 = x(x^2 + x) - (x^2 + x) = x b - b = b(x - 1)$ . It is clear that $b \neq -1$ , since, if this wasn't the case, $x^2 + x = -1$ or $(x + \frac{1}{2})^2 - \frac{1}{4} = -1$ . Then $(x + \frac{1}{2})^2 = -\frac{3}{4}$ which is impossible. We get that $x = \frac{a + b}{b + 1}$ which is a rational number since the numbers $a$ and $b$ are rational.

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