I think the right approach would be to observe that Rⁿ is the leading term in j(n\tau) http://latex.mathoverflow.net/png?j%28n%5Ctau%29, for \tau http://latex.mathoverflow.net/png?%5Ctau as above. Then there is a modular polynomial \Phi\sb n http://latex.mathoverflow.net/png?%5CPhi%5Fn which satisfies \Phi\sb n(j(x),j(nx))=0 http://latex.mathoverflow.net/png?%5CPhi%5Fn%28j%28x%29%2Cj%28nx%29%29%3D0. For small n you could presumably just solve this for j(n\tau http://latex.mathoverflow.net/png?%5Ctau) in terms of j(\tau http://latex.mathoverflow.net/png?%5Ctau), especially since modular curves are rational for small n. The conclusion would be that j(n\tau http://latex.mathoverflow.net/png?%5Ctau) is, again for small n, an integer.

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.