# What's the explanation for these (infinitely many?) Ramanujan-type identities?

Define the function,

$$F(\beta) := \sqrt{\beta+x_1}+\sqrt{\beta+x_2}+\sqrt{\beta+x_3}\tag1$$

where,

$$x_1 =2\cos\big(\tfrac{2\pi }{7}\big),\;x_2 =2\cos\big(\tfrac{4\pi }{7}\big),\; x_3 = 2\cos\big(\tfrac{8\pi }{7}\big)$$

or the $x_i$ are the roots of the cubic $x^3+x^2-2x-1=0$. We have two nice identities,

$$F(0) = \sqrt{5-3\sqrt7}$$

$$F(1) = \sqrt{-4+3\sqrt7}$$

The first one is by Ramanujan. (See The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society, p. 9, by Bruce Berndt, et al.)

These two (and other pairs) can be explained by the negative and positive cases of $\pm \sqrt{d}$ in this answer. However, davidoff303 found a plethora of others,

$$F\Big(\frac{74}{43}\Big) = 2\,\sqrt\frac{7^2}{43}$$

$$F\Big(\frac{5105}{11349}\Big) = 3\,\sqrt\frac{7}{11349}$$

$$F\Big(\frac{-2306997866696}{1047656140569}\Big) = -10980\,\sqrt\frac{7^2}{1047656140569}$$

$$F\Big(\frac{9658771264742899051}{5361029308457632889}\Big)= 13\cdot 127\cdot 1381\sqrt{\frac{7}{5361029308457632889}}$$

and so on. Note that, for general rational $\beta$, then $\big(F(\beta)\big)^3$ is a root of a $9$th deg equation.

Q1: Are there infinitely many rational $\beta$ such that $\big(F(\beta)\big)^3$ is also rational, like the ones found by davidoff303? Is there a formula for $\beta$?

$\color{blue}{Update:}$

I formed the $9$th deg equation satisfied by $\big(F(\beta)\big)^3$. Acting on a hunch, it turns out that the $\beta$ used by davidoff303 satisfies the simple relation,

$$\beta^3-\beta^2-2\beta+1 =w^3\tag2$$

This can be transformed into an elliptic curve, and it has infinitely many rational points. And the $LHS$ of $(2)$ in fact is a factor of the discriminant of the $9$th deg eqn.

Q2: Is it true that if rational $\beta$ satisfies $(2)$, then $\big(F(\beta)\big)^3$ is also rational?

• This has definitely something to do with Kummer sums (en.wikipedia.org/wiki/Kummer_sum) and the fact that the Galois group of the minimal polynomial of $\cos\left(\frac{2\pi}{n}\right)$ is a cyclic group with $\frac{\varphi(n)}{2}$ elements. – Jack D'Aurizio Jun 27 '16 at 18:45
• @davidoff303: How did you find the $\beta$? It's a vague feeling, but does the growth of $\beta$ involve an elliptic curve? – Tito Piezas III Jun 28 '16 at 1:51
• @davidoff303: I figured it out. There is an elliptic curve involved. However, I still don't know why it works. – Tito Piezas III Jun 28 '16 at 3:47
• In other words, (2) means $(\beta+x_1)(\beta+x_2)(\beta+x_3)=w^3$. – ccorn Jun 28 '16 at 8:17
• correct me if I'm wrong but $\beta=0$ satisfies $(2)$, but $(F(0))^3$ is not rational ? (though, it is a root of a degree $3$ polynomial) – mercio Jun 28 '16 at 9:43

Since there are $3$ ways to choose each of the cube roots, in general, $F(\beta)$ is a root of a degree $27$ polynomial whose coefficients are expressions in $\beta,(x_1+x_2+x_3),(x_1x_2,x_2x_3,x_3x_1),(x_1x_2x_3)$.

Since multiplying each root by $\zeta_3$ doesn't change the value of $F(\beta)^3$, this polynomial is in fact a polynomial of degree $9$ in $F(\beta)^3$.

Assume now that we have $(\beta+x_1)(\beta+x_2)(\beta+x_3) = w^3$.

Then the degree $9$ polynomial factors after extending our coefficients field with $w$, into a cubic and a sextic : gather in the cubic the three choices of roots giving $(\beta+x_1)^{1/3}(\beta+x_2)^{1/3}(\beta+x_3)^{1/3} = w$

Then, $F(\beta)^3 = (3\beta+x_1+x_2+x_3) + 6w + 3\sum_{i \neq j} (\beta+x_i)^\frac 23(\beta+x_j)^\frac 13 \\ = (3\beta+x_1+x_2+x_3) + 6w + 3w\sum_{i \neq j} (\beta+x_i)^\frac 13(\beta+x_j)^{-\frac 13}$

And so $(F(\beta)^3$ is rational if and only if the sum is. At this point, a choice of $F(\beta)$ among the $3$ possibilities is equivalent to a choice of only one term if the sum. Or equivalently, you go from one value to the others by multiplying $(\beta+x_2)^\frac 13$ by $\zeta_3$ and $(\beta+x_3)^\frac 13$ by $\zeta_3^2$

Split the sum into two parts $z_1 + z_2$ (one for each coset of $A_3$ in $S_3$). When applying this automorphism, one is multiplied by $\zeta_3$ and the other is multiplied by $\zeta_3^2$, so their product $z_1z_2$ has to be invariant, and indeed, $w^2z_1z_2 = 3\beta^2-2\beta-2+(3\beta-1)w+3w^2$.

If the original Galois group between $x_1,x_2,x_3$ had been $S_3$, things would have been more complicated, but here, since the Galois group is cyclic, $x_1/x_2+ x_2/x_3 + x_3/x_1$ is rational, and this implies that $z_1^3$ and $z_2^3$ are rational :

$\{(wz_i)^3\} = 9\beta^3-9\beta^2-11\beta+3w(3\beta^2-2\beta-2)+3w^2(3\beta-1) + \{2;9\}$

If one of them is $0$, then the other is $\pm 7$, and this corresponds to the $\beta = 0,1$ cases.

If not, then $F(\beta)^3$ is rational if and only if one of those two rational numbers is a cube (since their product is a nonzero cube, this is equivalent to both being a cube).

The map $(\beta,w) \mapsto \\ (X = 3\beta^2-2\beta-2+(3\beta-1)w+3w^2 ; \\ Y = 9\beta^3-9\beta^2-11\beta+3w(3\beta^2-2\beta-2)+3w^2(3\beta-1))$

is a rational isomorphism between the elliptic curves $E_1 : w^3 = (\beta+x_1)(\beta+x_2)(\beta+x_3)$ and $E_2 : (Y+2)(Y+9) = X^3$ (it can't be ramified, and the only preimage of the point at infinity from $E_2$ is the point at infinity with $\beta/w = 1$)

Since $Y+2$ has a triple pole at the point at infinity of $E_2$ and a triple zero at $(X=0,Y=-2)$, $Y+2$ is not a cube (there is no degree $1$ map from an elliptic curve into $\Bbb P^1$), but the extension obtained by adding a cube root of $Y+2$ is a $3$ fold unramified covering $f : E_3 \to E_2 \cong E_1$

Then the rational points on $E_1$ whose $F(\beta)^3$ is rational are (aside from the two exceptions with $X=0$) the images of the rational points on $E_3$.

Choose a rational point on $E_3$ (such as any example where $F(\beta)$ is a cube, or the rational point above the point at infinity of $E_2$) so we can have a group law on $E_3$ and $E_2$.

If $\hat f : E_2 \to E_3$ is the dual map to $f$ then $f \circ \hat f = $, so $f(E_3(\Bbb Q))$ is a subgroup of $E_2(\Bbb Q)$ that has to contain $(E_2(\Bbb Q))$

If you name $u = (Y+2)^\frac 13,v = (Y+9)^\frac 13$, then $E_3$ is given by $v^3=u^3+7$ and $f : E_3 \to E_2$ is given by $(X = uv, Y = u^3-2)$. After composing this with the isomorphism $E_2 \to E_1$ (that I didn't compute), you get a nice rational algebraic parametrization by $E_3$ of the points you want, $(u,v) \mapsto (\beta,w,F(\beta)^3)$

From here on, things might be different with other choices of $(x_1,x_2,x_3)$ or other base fields.

According to http://www.lmfdb.org/EllipticCurve/Q/441/d/2

$E_2(\Bbb Q)$ 's torsion subgroup has order $3$ and is generated by $Q_1 = (X=0, Y=-2) = (\beta = 1, w = -1)$. Since $Y+9$ is not a cube, it doesn't have a rational preimage in $E_3$ and $F(1)^3$ is not rational.

The other point of order $3$ is $2Q_1 = Q_2 = (X=0, Y=-9) = (\beta = 0, w = 1)$ and is in the same case.

$E_2(\Bbb Q)$ has rank $1$ and its free part is generated by $R = (X=2,Y= -1) = (\beta = \frac {74}{43}, w = - \frac {29}{43})$. Since $Y+2$ and $Y+9$ are nonzero cubes, it has a rational preimage in $E_3$, and $F(\frac {74}{43})^3$ is rational.

This is enough to determine that the subgroup of "good" points is generated by $(X=2,Y= -1)$. For example $2R = (X = -\frac{20}9, Y = -\frac{118}{27}) = (\beta = \frac {5105}{11349}, w = - \frac {2521}{11349})$, and so on.

Since $E_3$ also has a group law, there are addition formulas that combine triplets of rationals $(\beta,w,F(\beta)^3)$, so if you compute those formulas, you get an easy way to generate the points you want.

As you noticed, $\tau : (\beta,w) \mapsto (-\frac 1{\beta-1},\frac w{\beta-1})$ is the translation by $Q_2$ (a point of order $3$), and thus for any rational point $P$ on $E_1$, exactly one of $\{P,\tau(P),\tau^2(P)\}$ has a rational preimage in $E_3$ (has $F(\beta)^3$ rational).

• Thanks! A small question though. I've realized that $\beta^3-\beta^2-2\beta+1 = w^3$ is a necessary but not sufficient condition such that $\big(F(\beta)\big)^3$ is rational. For example, $\beta=9/5$. Is there an explicit rational Diophantine equation that $\beta = 74/34$ satisfies, but $\beta = 9/5$ does not? – Tito Piezas III Jun 28 '16 at 17:21
• after looking at lmfdb.org/EllipticCurve/Q/441/d/2 the generator they give corresponds to $\beta = 74/43$, it gives $X=2,Y=-1$ which has a rational preimage because $Y+2$ and $Y+9$ are cubes. However the torsion points $(X=0,Y=-2)$ and $(X=0,Y=-9)$ don't have rational preimages because $7$ and $-7$ are not cubes, so the "good" subgroup has index $3$ in $E_2(\Bbb Q)$. Using the multiplication formulas with the torsion subgroup should give you, from a $(\beta,w)$, formulas for the $3$ candidate $\beta$, only one of which is good – mercio Jun 28 '16 at 17:33
• instead of a diphantine equation, I think it's possible that there is a simple modular condition on $\beta$ that tells you by which torsion element you have to compose to get to a good rational point. – mercio Jun 28 '16 at 17:40
• By the way, I noticed that if $\beta_1 = \frac{p}{p+q}$ satisfies $\beta^3-\beta^2-2\beta+1 = \text{cube}$, then so does $\beta_2 = \frac{p+q}{q}$. For example, $\beta_1 = \frac{31}{74}$ and $\beta_2 =\frac{74}{43}$ though usually (excluding the case $\beta = 1/2,\,2$) only one of them yields a cubic with a rational root. – Tito Piezas III Jun 28 '16 at 17:58
• @davidoff303: I found the general formula. It works for any cubic $x^3+ax^2+bx+c=0$. For example, for $d=13$ we have $$\sqrt{p+x_1}+\sqrt{p+x_2}+\sqrt{p+x_3} = \sqrt{15^3 q}$$ where $p = \frac{-12901}{20367}$ and $q = \frac{-13}{20367}$ and the $x_i$ are the roots of $x^3+x^2-4x+1=0$ and can be expressed as sums of $\cos\big(\tfrac{\pi\,k}{13}\big)$. – Tito Piezas III Jul 4 '16 at 17:35

This is a long comment/addendum to mercio’s answer. Given rational solutions to, $$\beta^3-\beta^2-2\beta+1 = w^3\tag1$$ As pointed out by mercio, if we define, $$X = 3\beta^2-2\beta-2+(3\beta-1)w+3w^2 \\Y = 9\beta^3-9\beta^2-11\beta+3w(3\beta^2-2\beta-2)+3w^2(3\beta-1)$$ then this obeys the elegant relation, $$(Y+2)(Y+9) = X^3$$ However, if we require $\beta$ such that $$Y+2 = u^3\tag2$$ $$Y+9 = v^3\tag3$$ holds separately, then we can define $\big(F(\beta)\big)^3$ as, $$\big(F(\beta)\big)^3 = \frac{3(t+19)(u+v)+(3uv+7)(27u^4v-63u^3+105uv-196)}{t+19}\tag4$$ where, $$t=27(u^3+3)(u^3+4)$$ Thus, if $\beta$ satisfies the rational Diophantine conditions $(2),(3)$ in addition to $(1)$, then $\big(F(\beta)\big)^3$ is rational.

Example. Let $\beta = \tfrac{74}{43}$, we get $u=1,\;v=2,\;t=540,$ and using formula $(4)$, $$\big(F(\tfrac{74}{43})\big)^3 = \frac{2^3\cdot7^2}{43}$$ the same as in the post above. All six examples in the post obey $(1)$, but only the last four obey all $(1), (2),(3)$. Presumably there are infinitely many rational $\beta$ that obey all three conditions, but I do not know how to prove it so.

• @mercio: By the way, how did you find $X,Y$? For example, if we use the cubic involved with $\cos(\pi/19)$, and solve, $$t^3-t^2-6t+7 = w^3$$ this has infinitely many solutions. Does this also have a $X,Y$ such that $$(Y+a)(Y+b) = X^3$$ and $-a+b=19$? – Tito Piezas III Jun 29 '16 at 18:23
• @mercio: I figured out the general case. :) So I'm making a question for $d=19$. – Tito Piezas III Jun 29 '16 at 21:25
• (I didn't get pinged) I don't think anything should change much in the first two paragraphs with $d=19$. I found $Y$ by looking carefully at $F(\beta)^3$. The sum was the source of the potential non-rationality, and looking at how the Galois group acted on its terms naturally leads to $z_1$ and $z_2$ and figuring out that they live in the same cubic radical extension so they both need to be cubes. – mercio Jun 29 '16 at 22:12
• finding the formulas for $(wz_i)^3$ and $(w^2z_1z_2)$ was just a matter of computation and replacing the symmetric expressions of the $x_i$ by the coefficients of their cubic. – mercio Jun 29 '16 at 22:13
• @mercio: The question is done. Can you kindly look at this post. – Tito Piezas III Jun 29 '16 at 22:44