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Romanian Mathematical Olympiad - District Round

Number theory Difficulty 5.7 AIME, harder Prove it Romania

Determine the positive integers $a$ , $b$ , $c$ with the following properties:
(i) $(a^2 + b^2)(c^2 + 2023^2) = (ab + 2023c)^2$ ;
(ii) $(a^2 + 2023^2)(b^2 + c^2) = (2023a + bc)^2$ ;
(iii) the greatest common divisor of $a$ , $b$ , $c$ and $2023$ equals $1$ .

Solution

By subtracting the equality (i) from (ii) we obtain $(a^2 - c^2) \cdot (b^2 - 2023^2) = 0$ . Because $a$ , $b$ , $c$ are positive integers, we deduce that $a = c$ or $b = 2023$ .

If $a = c$ , using (i) we find $(a^2 - 2023b)^2 = 0$ , that is $a^2 = 2023b = 7 \cdot 17^2 \cdot b$ . Consequently $7 \mid b$ and $7 \mid a = c$ , therefore the greatest common divisor of $a$ , $b$ , $c$ and $2023$ is at least $7$ , contradicting assumption (iii).

If $b = 2023$ , from (i) we obtain $(ac - 2023^2)^2 = 0$ , that is $ac = 2023^2 = 7^2 \cdot 17^4$ . From (iii) we deduce that $a$ and $c$ are relatively primes, therefore the solutions are: $(a, b, c) \in \{(1, 2023, 2023^2), (7^2, 2023, 17^4), (17^4, 2023, 7^2), (2023^2, 2023, 1)\}$ .

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