Competition problem — Number theory

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Number theory Difficulty 5.7 AIME, harder Prove it Romania

The positive integers $a < b < c$ are such that $a + b + 2c$ is a common multiple of $b$ and $c$ (*).

a) Prove that the greatest common divisor of $a + b$ and $c$ is $c$ .

b) Find the positive integers $k < 1000$ , so that $abc = k^2$ and $a, b, c$ fulfill (*).

Solution

a) Since $c \mid a + b + 2c$ and $c \mid 2c$ , $c \mid a + b$ . Now $a < b < c$ implies $a + b < 2c$ , hence $a + b = c$ . Then $(a + b, c) = c$ .

b) From $b \mid a + b + 2c$ and (a), $b \mid 3a + 3b$ . Since $b \mid 3b$ , $b \mid 3a$ . From $3a < 3b$ follows $3a \in \{b, 2b\}$ .

Case I: $3a = b$ . Then $a = n$ , $b = 3n$ , $c = 4n$ , whence $abc = 12n^3$ . The values of $n$ so that $2^2 \cdot 3 \cdot n^3 = k^2$ , $k < 1000$ are: $n = 3$ , implying $abc = 18^2$ ; $n = 3 \cdot 2^2$ , implying $abc = 144^2$ ; $n = 3^3$ , implying $abc = 486^2$ .

Case II: $3a = 2b$ . Then $a = 2p$ , $b = 3p$ , $c = 5p$ . So $abc = 30p^3$ , yielding $p = 30$ and $abc = 900^2$ .

In conclusion, $k \in \{18, 144, 486, 900\}$ .

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