Track / Stage 6 / 92 of 400 #1092 of 1964

Problem 1092

National Olympiad, first round

Number theory Difficulty 6.1 Prove it

3 Let $(a, b)=1$ , prove: $\left(a^{2}+b^{2}, a b\right)=1$ .

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Official solution

3. Since $(a, b)=1$ , we have $\left(a^{2}, b\right)=1$ , thus $\left(a^{2}+b^{2}, b\right)=1$ . Similarly, $\left(a^{2}+b^{2}\right.$ , $a)=1$ . Therefore, $\left(a^{2}+b^{2}, a b\right)=1$ (using (6) of this unit).

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.