Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Prove it Romania

a) Prove that there are infinitely many natural numbers $n$ such that $2 \cdot n$ is a perfect square and $3 \cdot n$ is a perfect cube.

b) Prove that there is no natural number $m$ such that $2 + m$ is a perfect square and $3 \cdot m$ is a perfect cube.

Solution

a) Consider the numbers $n = 72 \cdot a^6$ , with natural $a$ . Then $2 \cdot n = (12 \cdot a^3)^2$ and $3 \cdot n = (6 \cdot a^2)^3$ , which shows that every such $n$ is 'good'.

b) If $3 \cdot m$ is a perfect cube, then $3 \cdot m$ is a multiple of $27$ . In this case $m$ is a multiple of $9$ , so $m + 2 = M_3 + 2$ . Since the perfect squares are $M_3$ or $M_3 + 1$ , $m + 2$ cannot be a perfect square.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.