Track / Stage 6 / 70 of 400 #1070 of 1964

Problem 1070

National olympiad, first round

Number theory Difficulty 6.1 Prove it

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

untranslated text remains the same as requested. However, if you need any further assistance or a different format, please let me know!

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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 and solution reproduced as published; topic, difficulty and ordering added by this site.