Entry 176

Ramanujan found the exact value of the Ramanujan \(G\)-function $$G_{69} = \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^{1/12} \left(\frac{3\sqrt3+\sqrt{23}}2\right)^{1/8} \left(\sqrt{\frac{2+3\sqrt3}4}+\sqrt{\frac{6+3\sqrt3}4}\right)^{1/2}$$Note the fundamental units \(U_n\) $$\begin{align}U_{23} &= 24+5\sqrt{23} = \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^2\\ U_{69} &= \frac{25+3\sqrt{69}}2 = \left(\frac{3\sqrt3+\sqrt{23}}2\right)^2 \end{align}$$ and how he uses the squared version. As a second example

$$G_{77} = \big(8+3\sqrt7\big)^{1/8}\left(\frac{\sqrt7+\sqrt{11}}2\right)^{1/8} \left(\sqrt{\frac{2+\sqrt{11}}4}+\sqrt{\frac{6+\sqrt{11}}4}\right)^{1/2}\quad$$

and fundamental units $$\begin{align}U_7 &= 8+3\sqrt7 =  \left(\frac{3+\sqrt{7}}{\sqrt2}\right)^2\\ U_{77} &= \frac{9+\sqrt{77}}2 = \left(\frac{\sqrt7+\sqrt{11}}2\right)^2 \end{align}$$

Ramanujan mostly uses the squared version of the \(U_n\) to get "simpler" expressions with smaller integers. For prime \(p \equiv 3\,\text{mod}\,4\), one can always do since 

$$x^2-py^2=-2\\ x^2-py^2=+2$$ are solvable by \(p \equiv 3\,\text{mod}\,8\,\) and \(p \equiv 7\,\text{mod}\,8\), respectively. Checking  \(U_{67}\) and \(U_{163}\), yields the reductions $$U_{67}= 48842 + 5967\sqrt{67} = \left(\frac{\color{blue}{221} + 27\sqrt{67}}{\sqrt2}\right)^2\\ U_{163}=64080026 + 5019135\sqrt{163}= \left(\frac{\color{blue}{8005} + 627\sqrt{163}}{\sqrt2}\right)^2$$And from \(e^{\pi\sqrt{67}}\approx 5280^3+744\) and \(e^{\pi\sqrt{163}}\approx 640320^3+744\), we find the relations $$5280=24\,(\color{blue}{221}-1)\\ 640320=80\,(\color{blue}{8005}-1)$$ though it may be just coincidence.