= Solution
Let $a$ be the equilibrium spacing of this monatomic chain and $u_j$ the displacement from $ja$. The right nearest-neighbour bond exerts $K(u_{j+1}-u_j)$ and the left bond $K(u_{j-1}-u_j)$. The two next-nearest-neighbour bonds similarly connect $j$ to $j\pm2$. Thus
$$
\boxed{m\ddot u_j=K(u_{j+1}+u_{j-1}-2u_j)
+G(u_{j+2}+u_{j-2}-2u_j).}
$$
Every uniform displacement gives zero force, as translation symmetry requires. There is one displacement coordinate per primitive cell. Adding further-neighbour springs does not introduce a second atom or turn this into a diatomic-chain problem.
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