According to NASA the sidereal orbital period, or ‘year’ for short, of Venus is 224.701 days, while that of the Earth is 365.256 days. We want the sidereal rather than tropical years here. It would be nice to have more decimal places, but this will do.

So, 8 Earth years is 2922.048 days, while 13 Venus years is 2921.113 days.

More importantly for us, 8 Earth years is 8 × 365.256 / 224.701 = 13.00416 Venus years. So that matches what I said. I vaguely remember doing this calculation myself before. Maybe it’s even on the blog article! I haven’t reread it.

So, every 8 years the points of the Pentagram of Venus ‘slip forward’ by .00416 × 360 degrees. That’s about 1.5 degrees. So, for the pentagram to rotate 360 degrees takes about 360 / 1.5 = 240 of these 8-year periods, meaning about 240 × 8 = 1920 years.

Or I could avoid using the number 360 twice by saying: each 8 years the pentagram makes .00416 of a full turn, so to make a full turn it takes 8 / .00416 years, or about 1923 years. Good! It’s nice to do a calculation twice and get the same answer. (I don’t really have good enough numbers to start with, to settle the last decimal place here.)

I’ll fix my article! Thanks, I guess nobody had checked this out.

]]>If it’s true, one could write a nice Whiggish history of Diophantine equations organized around the theme of ‘group laws’. I certainly would have understood number theory a bit better if someone had told me that tale.

If you can dig up anything on group structures on degenerate cubics, I’d be very interested. I’ll try to learn a bit about the history of the stereographic projection trick.

One can’t help wonder what’s in the lost works of Diophantus. (Right now I’m having a huge amount of fun reading *The Archimedes Codex* by Reviel Netz and William Noel, so my interest in the lost history of Greek mathematics is re-energized.)

As for (1): you may have to help me a bit because I am not on intimate terms with the terminology.

Neither am I, which is why I put ‘real form’ in quotes. I know about real forms of complex Lie or associative algebras or (a bit more to the point) complex algebraic varieties, but clearly that idea generalizes a lot…

If I understand the question, it’s whether the complexified versions of the norm 1 groups are isomorphic, and the answer has to be ‘yes’.

Great, that’s what I figured. I suppose to be more conservative in this Diophantine context one might tensor with an imaginary quadratic extension of rather than . For example, if you look for solutions of in you’ll find Pythagorean triples. But if you adjoin and then look for solutions of this equation, you’ll also find rational solutions of .

]]>Additionally, what is nice is that the group of units in , identified with certain integer pairs , is a *discrete* subgroup of the two-hyperbola group (each hyperbola has two branches, so the two-hyperbola group is isomorphic to . So the group of units is finitely generated. And if we focus on just the group of integer pairs sitting on the branch of where , then this branch is isomorphic to , so we are dealing with a discrete subgroup of . But we know that discrete subgroups of are either trivial or infinite cyclic! So basically we know that the solutions to Pell’s equation are (modulo the torsion elements ) generated by a single element, namely the integer pair that is closest to the identity without actually being the identity. It’s easy to convince oneself that that’s or (the inverse). And the same is true for the general Pell’s equation : although you have to convince yourself somehow that non-identity solutions exist, once you have one you know that you are dealing with an infinite cyclic group and you can generate all the solutions from a single solution. (Besides fancy theorems like Dirichlet’s unit theorem, the only way I know that guarantees existence of a nontrivial generator for the Pell group is by analyzing continued fraction expansions and especially their periodicity.) Oh, let me not forget to say that actually is the *square* of which sits on the other hyperbola , so actually it’s which generates the group of units in , modulo its torsion subgroup.

Okay, onto your questions now!

As for (1): you may have to help me a bit because I am not on intimate terms with the terminology. I think what we are dealing with is the group of units in the algebra , dealing with cases like and . (Or elements of norm 1, depending on what we want to look at.) The uniform group law can be written as

If I understand the question, it’s whether the complexified versions of the norm 1 groups are isomorphic, and the answer has to be ‘yes’ since for every nonzero , the algebra is isomorphic to , and the norm 1 group corresponds to . I think. (But let me know if that’s not what you’re asking.)

As for (2), I think you are alluding to the stereographic projection trick. In the case of a circle, we take a point like on the circle and draw the straight line between that and another point and watch where that line intersects a line like . Of course by point-slope it hits it at a point where is rational if is. But conversely, if is rational, then the line connecting to intersects the circle in two points. Obviously one of those points is . Then we argue that the other point *must* have rational coordinates — essentially because if a rational quadratic polynomial has one rational root, then the other root must be rational as well, by e.g. the quadratic formula. So stereographic projection sets up a bijective correspondence between rational solutions to the equation and rational numbers , and by this method we can generate all Pythagorean triples where . (I wonder how old this method is? It’s not exactly Euclid’s since he didn’t have cartesian coordinates or algebraic representations AFAIK, but it’s not ridiculous to think maybe it had occurred to a 17th century mathematician…)

Now this stereographic projection argument is perfectly general, and so it applies to any conic with rational coordinates, such as where is a squarefree integer. So there is a bijective correspondence between rational points on this conic and rational points on the projective line , say. After a little algebra, we see that this parametrization takes a point on the line to . Which is exactly how it works for Pythagorean triples, taking .

In each case we can transport the group structure on the rational conic to a group structure on . But we get in this way nonisomorphic group structures on , essentially because the algebraic number fields have different arithmetic structures. (For example, the torsion subgroups may differ.)

Because of this, looking at (3) I’m not sure I draw that much of a connection between the group law trick and the stereographic projection trick. One can see commonalities in method when one passes to sufficiently great extensions like or , but down here we are dealing with arithmetic which is more subtle. Note also that I was looking at elements of norm 1 in the algebraic *integers* , as opposed to elements of norm 1 in the algebraic number field .

But, but, …

As for (4): this looks interesting! I think I’m more inclined to view the stereographic projection as referring to a group structure on the union conic + line, which forms a singular cubic. In the case we are more used to thinking about, nonsingular cubics as (unpointed) elliptic curves, the group law is prescribed by iff are collinear. Similarly, with stereographic projection, we are considering collineations of three points (two on the conic, one on the line). And I remember being told (this was in a brief email exchange with Noam Elkies and some others whose names I’d have to dig up) that actually group laws on eilliptic curves *do* pass over to group laws on even such degenerate cubics like the union of a conic and a line, or the union of three lines, although there can be a bit of weirdness going on in these degenerate siutations. Now I’m not sure right now what could be milked out of that observation, but it looks like it might be fun following up on that. (Maybe I should try to find those emails, but that could take some doing…)

What do I not understand here? If the entire pentagram makes a full turn of 360 degrees in 160 years, the slippage must be 18 degrees for every 8 years. While we are at that, how many degrees are Venus and Earth off the mark made by the first conjunction during the sixth conjunction?

Bear with me, I am not an astronomer or mathematician but an interested layman ;-)

Cheers, Herbert

]]>1) Is your group structure on the hyperbola just another ‘real form’ of the group structure on the circle? If so, what’s the complex form of this algebraic group?

2) There’s a famous old way to get new Pythagorean triples from old: you take your triple , get a point on the circle , and draw a straight line from to this point. The line intersects the circle at another point with rational coordinates, and this is a new Pythagorean triple. The resulting algorithm goes back to Euclid. Is there a hyperbolic analogue of *this* trick?

(The only hard part of this trick is seeing that the line intersects the circle at a point with rational coordinates, but if you write an equation for the coordinates of this point you’ll get a quadratic equation with two solutions, one being the one you want and the other coming from the point Since that other solution of the quadratic is obviously rational, so is the one you want!)

3) How is the ‘straight line’ trick related to your ‘group law’ trick?

4) There are similar famous tricks for getting new solutions of *cubic* Diophantine equations from old ones. These involve drawing a bunch of lines, but ultimately they rely on the fact that any elliptic curve is an algebraic group. Do these tricks reduce to either the ‘straight line’ trick or the ‘group law’ trick in some limit where a cubic degenerates to a quadratic?

The question then is how we get such integer points way way out on the hyperbola? This is the second beautiful idea that John alluded to: exploit a group structure on the hyperbola. If I can make an analogy: the graph of is also a hyperbola, with asymptotes given by the – and -axes. It is also a group, where two points and on the hyperbola may be “multiplied” in an evident way to yield a third point on the hyperbola, . Then starting with a point like , repeated powers bring us closer and closer to an asymptote (the -axis for positive , and the -axis for negative ).

Similarly, we may start with the hyperbola . Factoring it as to liken it more to , we can multiply two points on the hyperbola. Here goes: we have

This suggests that the product of two points and on the hyperbola should be

And indeed this third point lies on the hyperbola, because

Note that if are integers, then so is and . Thus, if we start with an integer pair lying on the hyperbola, then we can take “powers” according to this product law. So the square is

The cube is

Already is not a bad approximation to !

]]>Why is ?

We’re trying to calculate the square root of two. If the square root of two were a rational we could find and with

or in other words

But the square root of two is *not* rational! So we can’t get But we can come close: we can find integers and with

An easy example is

(You caught a typo in my post: I wrote , which is nonsense. I’ll fix it now. Thanks!)

The wonderful trick is then to take one solution of

and use it to get another, bigger solution—which gives a better approximation to the square root of two!

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