Entry 177

From Entry 176, we gave the fundamental unit $U_d$ $$U_{163}=64080026 + 5019135\sqrt{163}= \left(\frac{\color{blue}{8005} + 627\sqrt{163}}{\sqrt2}\right)^2$$ and from $e^{\pi\sqrt{163}}\approx 640320^3+744$, observed that $$640320=80\,(\color{blue}{8005}-1)$$ This may be just coincidence, but what is not is when $U_{3d}$ for $d = 7,11,19,43,67,163$. This was also observed by H. H. Chan. Given the fundamental units

$$\begin{align}U_{21} &=\left(\tfrac{\sqrt3+\sqrt{7}}2\right)^2\\ U_{33} &=\big(2\sqrt3+\sqrt{11}\big)^2\\ U_{57} &=\big(5\sqrt3+2\sqrt{19}\big)^2\\ U_{129} &=\big(53\sqrt3+14\sqrt{43}\big)^2\\ U_{201} &=\big(293\sqrt3+62\sqrt{67}\big)^2\\ U_{489} &=\big(35573\sqrt3+4826\sqrt{163}\big)^2 \end{align}$$

Define the function

$$F_d = 3\sqrt3\big(\sqrt{U_{3d}}-1/\sqrt{U_{3d}}\big)+6$$

then we get the rather familiar $$\begin{align}F_{7} &= 15\\ F_{11} &= 6(1+\sqrt{33})\\ F_{19} &= 96\\ F_{43} &= 960\\ F_{67} &= 5280\\ F_{163} &= 640320\end{align}$$

which (except for $d=11$) are the cube roots of the j-function (negated). 

P.S. I don't know why $d=11$ does not obey the pattern, but it does yield the integer $42$ if the positive sign of the second square root $\pm 1/\sqrt{U_{3d}}$ is used, though the correct value should be $32$.

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.

Entry 175

For discriminant $d=4m$ with class number $8$ and semiprime $m$: 

If $m =5p$, then distinguish between $p \equiv 1\,\text{mod}\, 8$ and $p \equiv 5\,\text{mod}\, 8$. 

If $m =7p$, then distinguish between $p \equiv 3\,\text{mod}\, 8$ and $p \equiv 7\,\text{mod}\, 8$. 

The semiprime $m =5p$ was in the previous entry. For $m =7p$, there are only four and Ramanujan found the radicals below. 

For the 1st case, $p = 11, 43$, thus $m = 7p = 77,\, 301$  

$$\begin{align}G_{77} &= \big(8+3\sqrt7\big)^{1/8}\left(\frac{\sqrt7+\sqrt{11}}2\right)^{1/8} \left(\sqrt{\frac{2+\sqrt{11}}4}+\sqrt{\frac{6+\sqrt{11}}4}\right)^{1/2}\\ G_{301} &= \big(8+3\sqrt7\big)^{1/8}\left(\frac{57\sqrt7+23\sqrt{43}}2\right)^{1/8} \left(\sqrt{\frac{42+7\sqrt{43}}4}+\sqrt{\frac{46+7\sqrt{43}}4}\right)^{1/2}\end{align}$$

For the 2nd case, $p = 31, 79$, thus $m = 7p = 217,\, 553$ 

$$\begin{align}G_{217} &= \left(\sqrt{x_1-\tfrac12}+\sqrt{x_1+\tfrac12}\right)^{1/2} \left(\sqrt{y_1-\tfrac12}+\sqrt{y_1+\tfrac12}\right)^{1/2}\\ G_{553} &= \left(\sqrt{x_2-\tfrac12}+\sqrt{x_2+\tfrac12}\right)^{1/2} \left(\sqrt{y_2-\tfrac12}+\sqrt{y_2+\tfrac12}\right)^{1/2} \quad\end{align}$$ where $$x_1=\frac{10+4\sqrt{7}}2,\quad y_1=\frac{14+5\sqrt{7}}4$$ $$x_2=\frac{142+16\sqrt{79}}2,\quad y_2=\frac{98+11\sqrt{79}}4$$

But it seems not noticed that a fundamental unit is imbedded in these radicals as

$$x_1+2y_1 = \frac32(8+3\sqrt{7}) = \frac32\left(\frac{3+\sqrt7}{\sqrt2}\right)^2 =\frac32\,U_7$$

$$x_2+2y_2 = \frac32(80+9\sqrt{79}) = \frac32\left(\frac{9+\sqrt{79}}{\sqrt2}\right)^2 = \frac32\,U_{79}$$

Entry 174

Continuing from the previous entry, for $d=4m$ with class number $8$, the semiprime $m =5p$ with $p \equiv 5\,\text{mod}\, 8$ is also well-behaved. And it involves the golden ratio. There are only four, namely $p = 13, 29, 53, 101$, thus $m = 5p = 65, 145, 265, 505$. Ramanujan found the radicals below and the $G$-function have a common form

$$G_{5p} = \phi^{k}\,U_{p}^{1/4}\,x_{p}^{1/2}$$

with powers of the golden ratio $\phi$, fundamental unit $U_n$, and $x_{p}^2$ a root of a unit quartic

$$\begin{align}\frac{G_{65}}{\phi} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4} \left(\sqrt{\frac{1+\sqrt{65}}8}+\sqrt{\frac{9+\sqrt{65}}8}\right)^{1/2}\\ \frac{G_{145}}{\phi^3} &= \left(\frac{5+\sqrt{29}}2\right)^{1/4} \left(\sqrt{\frac{9+\sqrt{145}}8}+\sqrt{\frac{17+\sqrt{145}}8}\right)^{1/2}\\ \frac{G_{265}}{\phi^3} &= \left(\frac{7+\sqrt{53}}2\right)^{1/4} \left(\sqrt{\frac{81+5\sqrt{265}}8}+\sqrt{\frac{89+5\sqrt{265}}8}\right)^{1/2}\\ \frac{G_{505}}{\phi^7} &= \big(10+\sqrt{101}\big)^{1/4} \left(\sqrt{\frac{105+5\sqrt{505}}8}+\sqrt{\frac{113+5\sqrt{505}}8}\right)^{1/2}\end{align}$$

The case $p \equiv 1\,\text{mod}\, 8$ or $p = 41, 89$, thus $m = 5p = 205, 445$ behaves slightly differently though

$$\begin{align}\frac{G_{205}}{\phi} &= \left(\frac{43+3\sqrt{205}}2\right)^{1/8} \left(\sqrt{\frac{-1+\sqrt{41}}8}+\sqrt{\frac{7+\sqrt{41}}8}\right)\\  \frac{G_{445}}{\phi^{3/2}} &= \,\left(\frac{21+\sqrt{445}}2\right)^{1/4}\, \left(\sqrt{\frac{5+\sqrt{89}}8}+\sqrt{\frac{13+\sqrt{89}}8}\right)\end{align}$$

How Ramanujan found these is a mystery.

Entry 173

There are many in the set $d=4m$ with class number $8$. When $m$ is a prime or a semiprime (a product of two primes) like $m =3p$, then it may be well-behaved. For this set, there are only three, namely $p = 23, 47, 71$, thus $m = 3p = 69, 141, 213$. Ramanujan found the radicals below and the $G$-function have a common form

$$G_{3p} = U_{p}^{1/24}\,U_{3p}^{1/16}\,x_{p}^{1/2}$$

with fundamental unit $U_n$ and where $x_{p}^2$ is a root of a unit quartic

$$\begin{align}G_{69} &= \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^{1/12} \left(\frac{3\sqrt3+\sqrt{23}}2\right)^{1/8} \left(\sqrt{\frac{2+3\sqrt3}4}+\sqrt{\frac{6+3\sqrt3}4}\right)^{1/2}\\ G_{141} &= \left(\frac{7+\sqrt{47}}{\sqrt2}\right)^{1/12}\; \big(4\sqrt3+\sqrt{47}\big)^{1/8}\; \left(\sqrt{\frac{14+9\sqrt3}4}+\sqrt{\frac{18+9\sqrt3}4}\right)^{1/2}\\ G_{213} &= \left(\frac{59+7\sqrt{71}}{\sqrt2}\right)^{1/12} \left(\frac{5\sqrt3+\sqrt{71}}2\right)^{1/8} \left(\sqrt{\frac{19+12\sqrt3}2}+\sqrt{\frac{21+12\sqrt3}2}\right)^{1/2} \end{align}$$ How Ramanujan found these is unknown as it is uncertain if he was aware of class field theory. Note also that without the factor $3$, then $d=23,47,71$ are the smallest $d$ with class number $h(-d) = 3,5,7$, respectively, while $h(-3d) = 8$.

P.S. Checking $h(-3d) = 16$, one finds the only primes are $d=167,191,239,383,311$ which are the smallest $d$ with class number $h(-d) = 11,13,15,17,19$, respectively. Makes me wonder if their $G_{3d}$ would be analogous.