Take $b_1=1,b_2=p,b_3=np$ for some $n>1$ and prime $p$ Then $a_1,a_p,a_{np}$ is an arithmetic progression, so $a_{np}=2a_p-a_1$ not depends from choice of $n$ So we have for every prime $p$ that $a_{2p}=a_{3p}=...=a_{np}$ Now take $b_1=1, b_2=2p, b_3=4p$ and $b_1=p, b_2=2p, b_3=4p$ to get that $a_1=a_p=a_{np}$ for every $n$ and every prime $p$ And so $a_i$ is constant