Solution
= Solution
By <modular arithmetic>, $a_1-b_1=nu$ and $a_2-b_2=nv$ for some <integers> $u,v$. Adding gives
$$
(a_1+a_2)-(b_1+b_2)=n(u+v).
$$
Since $u+v$ is an <integer>, this proves
$$
\boxed{a_1+a_2\equiv b_1+b_2\pmod n.}
$$
= Solution
By <modular arithmetic>, $a_1-b_1=nu$ and $a_2-b_2=nv$ for some <integers> $u,v$. Adding gives
$$
(a_1+a_2)-(b_1+b_2)=n(u+v).
$$
Since $u+v$ is an <integer>, this proves
$$
\boxed{a_1+a_2\equiv b_1+b_2\pmod n.}
$$