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I think the right approach would be to observe that $R^n$ is the leading term in $j(n\tau)$ as above. Then there is a modular polynomial $\Phi_n$ which satisfies $\Phi_n(j(x),j(nx))=0$. For small $n$ you could presumably just solve this for $j(n\tau)$ in terms of $j(\tau)$, especially since modular curves are rational for small n. The conclusion would be that $j(n\tau)$ is, again for small $n$, an integer.