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>>> Katherine Stange: So thank you to the organizers. And also thank you to Microsoft.
Hello to Kristin, wherever she is out there in videoland.
So I'm going to talk about work which is joint work with Joseph Min [phonetic] actually
when he was at Microsoft Research. He spent a semester at the Cambridge branch,
piece, whatever it is. And this was from then.
And so I'm going to start with a motivation which I'm then going to forget for the rest of
the talk, because I think the question that comes out of it sort of takes on its own interest.
So but let's start with this. So suppose we have an integer sequence. And I want to look
at the indices and such that N divides the nth term.
So I'll call it the index divisibility set. And obviously the interest of such a question
depends greatly on what sequence you're looking at. So, for example, if your sequence
is B to the N minus B, these already have a name, pseudo prime to the base B. And I
want to look at the structure of the set by making it a directed graph.
So you can take the elements, the indices which divide their terms as the vertices and
draw an arrow from N to M if N divides M and N is smaller than M, and if it's a direct
connection, so there's nothing in between, which is also in the set.
So this was actually Chris Smyth took a look at this for Lucas sequences. So these are
just linear recurrent sequences. Anything that satisfies the linear recurrence like that
where A and B are whatever coefficients you want.
If you call delta the discriminant A squared minus 4B, then you can describe this set in
terms of all the arrows in the graph. So suppose you have -- suppose you have some N
in the index divisibility set, and then you can classify all the arrows originating at that
vertex. So you have an arrow heading from N to N times P where P is prime, whenever
that prime is a devisor of that particular term L sub N or divides the discriminant delta.
Then you have a few extra ones in there, which are classified down here. Sometimes
you can go from N to 6N or 12N.
So we saw this paper, and we thought of asking the same question for elliptic divisibility
sequences, which are related to elliptic curves. So suppose we have an elliptic curve
and a point on the curve, both defined over rational numbers.
Then the point, the X coordinate of the nth multiple of the point. So I'm told that not
everybody in this room is a number theorist. So I'll sort of toss out the basic facts about
elliptic curves. Elliptic curve, its points form a group, which I'm sure you've all heard,
even if you don't enjoy studying them all the time.
This means the nth point looking at the X coordinate. So because of the form of the
equation, the X and Y coordinates will have denominator something squared and cubed.
We'll call that something DN. We'll just extract those numbers, those integers, take the
positive integer and make that a sequence.
And this is called an elliptic divisibility sequence. I won't talk about them for the rest of
the talk, but I'll just mention that they carry a lot of the information about the elliptic curve.
So you can see a lot of things going on in the elliptic curve just by looking at these
sequences.
Okay. So these were the sequences we were interested in looking at. And we got a
theorem which is not dissimilar to what Chris Smyth got. So you can ignore some of the
technical stuff here.
But let's suppose we have an elliptical divisibility sequence. You do get an arrow from N.
So looking at part 1. I guess I'll lose the pointer. Part 1 here. You get an arrow from N
to N times the prime, when the prime divides the nth term, just like in the other theorem.
>>: Can you go back one slide? One question. So the index M indicates the multiple.
>>> Katherine Stange: Yes.
>>: Then where is the sequence, it's referring also to one fist ->>> Katherine Stange: Yes, exactly. You pick a fixed curve and pick a fixed point, then
you get a sequence of integers. So for every point and curve. Okay. Yes. So we're
picking a particular elliptic curve in a point and looking at an associated sequence.
If you have some vertex in your graph, then you get an arrow from that, from N to N times
P, if P divides that element of the sequence.
And you also get some extra arrows. You get N going to N times D where D is what I've
highlighted in red there, an aliquot number. This is not aliquot in the classical sense. We
decided to call these aliquots for elliptic curves.
I'll explain what that is in a second. But I wanted to highlight the relationship to the
previous theorem. Here we get when you divide the term and then some extra special
stuff. The special stuff for these sequences is aliquot numbers.
And then parts 3 and 4 here are sort of the converse. So if you have an arrow going from
N to N times the prime, then it did usually come from either dividing the term or being
aliquot number, the prime itself.
So when I say good "reduction here", what I'll talk about in the rest of the talk, I'll be
talking about taking an elliptic curve, defined over Q, and reducing it modulo P and
getting something defined over FP.
Then the question is whether you still get an elliptic curve. If you do, then you call it good
reduction. You could get something singular, and that's called bad reduction.
And the last one here says, if you have an arrow with a composite multiple, then in
certain cases you can say that's an aliquot number. It's not a full converse to this.
So when we were looking at this, we came up with these aliquot numbers naturally from
working on that project. Now I'm going to forget that project completely and define
what -- move from the definition of an aliquot number to something just talking about
elliptic curves that's a question in and of itself.
What do I mean by an aliquot number? So if you have some primes of good reduction,
all distinct primes, and the first index which PI appears as a devisor is PI plus one.
For each link in the entire loop. So including that the first index at which PL appears as a
devisor is P1. So moving right around. Then the product of those things is called an
aliquot number.
And the important piece of the elliptic curves that you see in the sequence is that if your
prime divides the nth term of the sequence, that means that the nth multiple of the point
to which the sequence was associated is zero mod P on the reduced curve. So where
the primes appear in the sequence is telling you the order of the point.
So, in particular, if you just have the order of your curve being prime, then your point has
to have, if it has nontrivial order, it has to have that order.
So you can get aliquot numbers. You can build such things by finding elliptic curves that
when you reduce mod the ith prime have a number of points equal to I plus the first
prime.
And the special case of this where there's just one prime, and it references itself, is where
the number of points when you reduce mod P is actually P. And that's already got a
name. That's an anomalous prime. That's in particular an aliquot number by this
definition here.
So you can forget all of the stuff about sequences and just ask the question for pairs to
start with. So when can you have two primes so that when you reduce modulo one of
them the number of the points on the reduced curve is equal to the other? And so this is
in vague analogy to the classical number theoretic meaning of amicable numbers. But
not in any meaningful analogy.
Okay. So here's a couple examples. So these were just searches. We took a couple of
curves, a small conductor, and looked for such pairs up to 10 to the 7. And in one
example we found one pair. In the other example we found four.
And so this is a little background slide for those who want a little bit more about elliptic
curves. If we have an elliptic curve over a finite field, let's say FP, then we define the
trace of Frobenius to be the difference between the number of points on the reduced
curve and P plus one because the reduced curve should have about P plus one points,
but there's some variation. The variation can be bounded. We call this the Hasse
interval in which that thing called the trace of Frobenius can lie.
And we have a theorem of during that says that everything in that Hasse interval can be
obtained if you pick the right curve for a particular prime. And also we have some
conjectures which govern the distribution within that house interval, as you vary over
different curves.
So we can ask about -- we can ask about the number of amicable pairs up to a given
bound. So this is for a particular curve E. So I'm giving this counting function Q, which is
indexed by our particular elliptic curve E, and the first question is how does this grow with
X. How many amicable pairs do you expect there to be. And the question is sort of a
strange one. You don't really expect amicable pairs to occur for any reason other than
just kind of randomly when you ask the question. Right? But it turns out this kind of has
an interesting answer.
The second question here is maybe we can relate it to a counting function which has
already been studied, which is, I'll call N, which is the number of primes so that the
reduction is prime.
If you're going to have an amicable pair at all, then in particular when you reduce modulo
P you have to get some other prime Q.
So we can count the number of primes for which there's prime reduction up to some
bound and we could maybe compare the two counting functions.
So here's a sort of -- sorry. I put the counting function up at the top just for reference
there.
And there's a conjecture due to Koblitz, which gives a conjectural growth for this with
respect to the bounds. X over log X squared. And the constant there, he actually gave
some conjectural values for that context in different cases and Zywina gave some
corrections to that portion of the conjecture, so I stuck his name up there as well.
And, in particular, the conjecture is that that constant is 0 when there's only finite many
primes, obviously, but that's the only time when it's 0.
You could have, for example, you could have, over your original curve over the rationals,
you could have some torsion and that torsion is going to survive reduction mod P and
divide the order of the reduced curve all the times, in which case this you wouldn't expect
to be growing.
So here's a sort of vague heuristic for how many amicable pairs you should expect. We'll
do it in terms of probability. So the probability that P is part of an amicable pair, well, first
of all, you have to reduce modulo P and get some prime order. Call that Q and then
reduce mod Q and see if you get back P. Let's just assume those two things are sort of
independent.
If that's the case, then we can analyze each of those probabilities separately. And
turning around the conjecture of Koblitz, we get that the probability that when you reduce
mod P you get something prime should be somewhere around log P.
And we also have that the probability when you reduce modulo in some particular prime
Q you get the particular value you had in mind which is P. Should be about 1 over root
Q, which is about the size of root P. P and Q here, if you're going to be an amicable pair,
they have to be about the same size, because they have to be in each other's Hasse
intervals.
>>: Estimates are up to like order ->> Katherine Stange: I'd have to look. I'd have to go back. I don't remember.
>>: [inaudible].
>> Katherine Stange: I think that, yeah, I think that's an unusual notation that we used in
that. It means that there's a constant in both directions, I think.
Yeah, I think the referee said that's not normal notation for this.
>>: I think it's usually like the smile and the frown, that one.
>> Katherine Stange: Yeah, that's a more usual notation.
So putting these two things together, then, we should have the probability that P is part of
an amicable pair should be around 1 over root P log P.
So we turn that back into a growth function. And we get something like that. The number
of amicable pairs should grow something like root X over log X squared. You remember
the conjecture from Koblitz was X over log X squared. So this is smaller which you would
expect.
So the first conjecture then is, well, let's just put this together. There's the counting
function for amicable pairs, assuming you get infinitely many primes so you have prime
reduction, because otherwise you're not going to be able to count more than finite many
pairs. And we expect it to grow like that, where these constants should somehow depend
on the curve.
But who knows how. So I think that if somebody asked you this question about amicable
pairs, you might do this heuristic. Come up with this growth rate and be like, okay, well,
probably can't prove this and probably that's about the story. It kind of happens
randomly, the amount of time you expect it to happen.
I guess we can check against a little bit of data. But we don't have very many pairs to
count. So for this particular curve, and thanks to Andrew Sutherland, who extended our
calculations beyond what we could do and none of this was done with Sage as far as I
know, we can count the number of pairs, and we only get even 117 up to the 10 to the
12. So we can compare our counting function to root X over log X squared. And I'm not
sure that the data is very convincing.
We can compare it to log X there. This should go to like a half or something. Not very
convincing. So we just don't have enough data to even know. But let's look at another
example.
So these are the two that I had on the first time I put up some examples. So up to about
10 to the 7 we get one in one case and four in the second case, but for this curve down
here we get 5,578 in the same range, of which I have not listed them all.
So that curve, you may, if you're familiar with elliptic curves you might notice that's a
curve with complex multiplication. So something special might be happening in the case
of complex multiplication. So I'll just tell you what complex multiplication is.
Usually if you look at the endomorphism with your elliptic curve, it's usually isomorphic to
Z. You have an endomorphism given by multiplication by M for each possible integer M.
But sometimes your endomorphism ring is the order of a class number one in some
quadratic imaginary number field. Sometimes you get curves that have more interesting
endomorphism ring and those are called curves with complex multiplication.
So we can actually prove something. We can say suppose we have a curve with
complex multiplication. And suppose we have the complex multiplications even by this
field joined by minus D, some order in that field. And let's throw out the J equals 0 case,
that particular case.
And let's suppose P and Q are primes of good reduction. So not too small. And suppose
we already know that when we reduce mod P we get a prime order Q. The question is,
does Q reciprocate, does it return the love? Right? When you reduce mod Q do you still
get P? If so, then that would be an inamicable pair. In fact, when you reduce mod Q, two
things can happen. Either you do in fact get P back and you have a pair. In this case we
have a pair or you can get this other quantity.
If you're familiar with this sort of thing, you'll probably recognize this formula. What this is
saying actually is the nontrivial quadratic twist over the finite field has P points. So you
always get E and E tilde and one of the two has P points.
So I'll quickly give you the proof. It actually fits on a slide. First of all, we're interested in,
going back to the statement here, we're interested in getting prime reduction.
We know when you reduce mod P, you get Q. So we're going to throw out curves with
two torsion. Because those will survive the reduction mod P and we won't get prime
order for the curve. So that leaves that D is congruent to 3 mod 4. There's only finitely
many different cases. So I forget how many this leaves. But anyway, when we look at P,
it has to split. This is because the inert case corresponds to supersingular reduction, and
in supersingular reduction the number of points would be in particular even.
So P splits. And if we look at the number of points, so pick -- so P splits back P in its
conjugate. If we look at the number of points over FP, then this is just the usual formula.
The number of points in the curve is P plus 1 minus the trace of Frobenius where the
Grossencharacter gives you the Frobenius as an element of the order number field that
you're identifying with your endomorphism ring.
So this expression up here in number three there, that's the norm of 1 minus psi of P.
So any way, in particular, we know this reduction is Q. This reduction modulo P is Q.
That says Q is the norm of something in this order. That means that Q itself splits. And
then we can look at the norm of psi of Q, which is also Q. So that means those are two
different things which have the same norm. Now, up to picking which of the Q or Q bar
we want here, that means that those two things are the same up to a unit.
And in all the cases that we're in, because we threw out the J equals 0 case, it turns out
that the units available to us are just plus or minus 1.
So then you can just calculate the trace of Frobenius for Q, and in each of the two cases
you get two different values and they correspond to these two cases. It comes from the
complex multiplication structure which is this rich structure we have in the case of CM
curves.
So let's look at the data. So this is -- these are the different possible orders for the
endomorphism ring. D was Q adjoined root minus D. And F there is the index of the
order. That's the bound and those are the number of pairs. This is just the counting
function up here.
And then let's actually look at the ratio of the amicable pair counting function over the
counting function for prime reduction.
So what the theorem said was if you already have prime reduction then there's two
possible cases. And one of those cases corresponds to you being an amicable pair. So
you'd expect this to go to a half, except that the pair counting function has two primes for
each pair. You actually expect it to go to a quarter. And here it does look, I'd say, at
least encouraging.
Okay. So that looks good. It looks like basically what this table is saying is that in the
statement of theorem, going back for a second, it looks like these two cases really do sort
of happen equally often, which would be your first guess, I suppose.
So here's the conjecture we get to put together, we have the not complex multiplication
case, which is what we had before. And if it has complex multiplication, then we expect it
to go to a quarter of the counting function for prime reduction.
Okay. So you could ask about aliquot cycles. You could ask about more than just pairs.
So suppose you have a whole sequence of primes such that when you reduce modulo
one of them you get the next one, all the way around and coming back to the beginning.
So here's a couple of examples. This one was found just by searching. And this one
shows signs of being constructed. Take a bunch of primes that fit inside each other's
Hasse intervals so they at least could plausibly be a cycle and then you can construct
such a thing. Actually it's not too hard to do something like this. First, make sure they all
satisfy their sizes are good enough so that the Hasse interval condition is verified.
And then for each prime you can find by doing some elliptic curve that has the correct
reduction over FPI. So you get one curve for each of them, but of course then you can
use Chinese remainder theorem to get some curve that reduces to all of those, that's why
you get these giant coefficients. That's how we made that example.
Now, in the CM case, it's actually interesting that you don't get any -- in the CM case, we
had lots of pairs but you don't get any longer cycles. That's again from that same CM
theorem we had before. So supposing you were trying to make a long cycle. Long
meaning three or more primes. Well, we know from that theorem that there's two choices
in the reduction at each stage. So when you put those together, you get a recurrence
relation telling you what the next prime could be based on the previous two.
So if you put together the recurrence relation, you can get a general term. You can find
out in fact that each term that P1 is P2 plus some positive amount. L is the length of the
cycle. If L is at least 3, this is some positive number.
And it's P1 is P2 plus some positive amount. P1 is bigger than P2 and similarly P2 is
bigger than P3. P3 is bigger than P4. You can see where I'm going. It's like the extra
staircase. You can't come around and be bigger all the way around your cycle. Come
back to where you started.
You can't make bigger cycles in the CM case. All right. So the last -- the case that I left
out of the CM theorem was the case of the J equals 0 curve.
So this curve has a little more interesting number theory. So I'm going to look a little
more closely at what happens. In fact, it turns out that there's a really interesting purely
elementary number theory question that comes out of this, which I'm still looking for an
answer to.
So here we're working with Q ajoins square root minus 3 which has a sixth root of unity.
That's why the previous proof there required that the units were just plus or minus 1. So
here we have more units, the unit group has six. In fact, the ring of integers can be made
by just joining that [inaudible] there, and this has a nice property that the unit group was
six elements. If you reduce modulo three, they each go one each into the invertible
classes.
So that's an isomorphism. We can also talk about the sextic residue symbol when we're
working down here, defined down here. It's quadratic residue quadratic cubic symbol. In
particular, for any element you're getting some sextic root unity as you see. So this is the
number field we're working with for the J equals 0 case, for the endomorphism ring.
And so we get a similar theorem to the CM theorem that we had for the other CM curves.
Now, I've stuck some of the proof into the statement here because we can give explicitly
what the unit is. So in the CM proof, I just said there's some units either plus or minus 1
in the middle of that proof. And here we can actually describe what that unit is. Going
back over the setup here, we have some elliptic curve. For J equals 0 we can write Y
squared equals XQ plus K. And suppose that P and Q are primes of good reduction. P
is not too small. And when you reduce modulo P you get Q points.
Then as in the proof before, P has to split and you can take Q to be the ideal generated
by 1 minus the Frobenius. So then Q also splits. Then we have these P and Q to talk
about, related by a unit. Just as in the previous proof.
Now that unit is actually given in terms of the sextic residue symbols of K, just from the K
as in the equation for the curve.
So if this unit comes out to be one, that's the case where we get an amicable pair. So if
we already had when you reduce mod P we get Q, we also get, when you reduce Q, you
get P, when this thing comes out to one.
So there's six -- the moral of the story, there's six possible options for this unit. And one
of them is the case of an amicable pair. So you would expect that a good first guess, just
like in the other CM case, is that these six units occur evenly often. So you would get the
relationship between the payer counting function and prime reduction counting function
would be 1/12th, counting for the fact there's two primes in a pair rather than one-sixth.
Okay. So just a couple of extra remarks about this. Just as in the original CM case,
those six different units correspond to twists. So that's six different possible values also
just like in the statement of the other CM theorem that you could get for reduction mod Q.
So you can state it that way if you'd like. The reason we can actually give this explicitly is
from the theorem of CM we know Frobenius as a sextic residue.
So here's the data. We've got different values of K across the top. We expect the ratio of
Q over N to be going to about 1/12th. And it looks suspicious. For K equals 2 we seem
to be getting a quarter again like before. For K equals 5, it looks reasonable. And the
other one it's not clear what they're doing. So something more interesting is happening
with this unit in this case.
So then the question is just to take a look at this unit and see if we can analyze what we
would expect it to be doing. If it's not going to be distributed evenly as -- distributed
evenly for different P, than for a fixed K, then what do you expect it to be doing?
So let's look a little more closely. These are the six possible values. Here's a giant
amount of data. So this corresponds to the unit being one. And this corresponds to the
unit being minus one. And these are the other four possibilities for that unit.
And down here we have the different values of K and then these are just the frequencies
of falling into those six different possible cases.
So I think looking at the data here, well, what I noticed, and then Joe wasn't so
convinced, was that it seemed to me that 1 and 2 were always a little bit more than the
other cases in each of them. But it certainly seems as if these two columns are about the
same as each other. But then they differ somehow from the other four cases.
Look at the data. On some of them it's really clear. For example, for K equals 2 you
don't even get those other cases and those two cases seem to divide about evenly. And
for this one, K equals 3, for example, it also looks fairly clear. The other one not so sure.
Certainly seems to depend on K a lot.
Okay. All right. I don't know how I'm doing for time. Oh, there's a clock at the back.
Okay. So there's just the pieces of the info for our number field up at the top there.
And we have cubic reciprocity, which I'm going to use in a moment. So I thought I'd state
it nicely. This is my favorite way of stating quadratic reciprocity, which is to think of the
primes, instead of being just the positive primes, think of the primes as minus 3-5, minus
7, minus 11 and so. If you put the signs on the primes, so they end up being 1 mod 4,
then you can state quadratic reciprocity particularly nicely. This is how I stated cubic
reciprocity. I've chosen the representative for prime ideal so that it's 1 or 2 mod 3. So we
have this unit that we'd like to analyze. Okay. So this is what it looked like from the
statement of the theorem. So from cubic reciprocity we can flip numerator and
denominator of course to check all the details and whatnot. So there's a little bit missing.
But we can turn it into this. And so here I've got sextic residue. But I'm just going to
ignore the quadratic part and just put a plus or minus in front. I'm really interested in the
cubic part. The reason for that is when we looked at that table of data, looked like plus or
minus 1 would be hitting pretty evenly. So it's really the cubic part which is the mystery.
So changing this around, we get this symbol here. Okay. So we're interested in this
symbol where M is our Frobenius for different primes P. So a first guess might be that
the Frobenius ranges evenly through whatever elements it would be allowed to be as P
varies. So you could take all of the M for which M times 1 minus M is invertible. And
then you could look at how many of those actually give this cubic residue coming out to 1,
and you could ask whether that's actually a third or that's some other ratio.
And when this comes out to 1, that corresponds to the case where this thing becomes
plus or minus 1, and we're in those first two columns. We seem to be in the first two
columns more often. Is that simply because there are more elements where, more
elements M in the residue here, the residue ring, that M times 1 minus M is actually from,
has cubic residue 1.
So then the conjecture would be that we expect this counting function ratio to go to that
ratio where we have a 4 counting for the fact that now we're allowing plus or minus 1 and
we're still counting two primes in a pair.
So don't worry about the 4. Really we're interested in this ratio of MK star over MK.
Where MK star is the one where this is cubic residue 1 and MK is just all the things where
that numerator is invertible.
So this becomes a purely number theory question. Now we've sort of forgotten about the
elliptic curve completely. We're just asking for different Ks how can you count this.
So here's an example of how we can count it when K is congruent to 2 mod 3. If we want
to look at when M times 1 minus M is a cube, cubic residue 1, then -- one way to do it
would be to look at it might be to at the associated curve, Y times 1 minus Y is X cubed,
how many points are there on that.
That just happens to be a J equals 0 elliptic curve. I don't think this is for a reason. But it
just happens to be. So we have then that this curve E related to our problem for K -when K is a prime congruent to 2 mod 3, this is actually supersingular and counting the
points is really easy. So if we want to count the points over K, that's actually FK squared
is the residue we're interested in. So has K plus 1 squared points. It's real easy to count
the points over supersingular curve. If this wasn't supersingular, if we had 1 mod 3, then
we would have some trace of Frobenius in here instead of a fixed number in terms of K.
Okay. So then we just want to count the points on that that correspond to solutions. So
we have to throw away the points infinity 0 and 01 because they don't really correspond
to solutions for what we're looking for. We want the numerator to be invertible we have to
throw away Y as 0 or 1.
So we take away those three points. Now we have K squared, K plus one all squared
minus 3 and of course we're triple counting because X can be three different cube roots
for each Y. We just want to count the Ys.
So we divide by 3. That's the number of solutions to the problem then. Okay. So
number of residues. Not equal to 0, 1, having that be a cube. And so that lets us count
those things. Okay. So the question -- so this particular case fits on a slide. But all the
other cases, as we went, just got worse and worse and horribler and horribler. So, of
course, if, as I said, you can get traces of Frobenius in there that you have to be counting
somehow. You have the strange thing that happens which is sometimes your Frobenius
doesn't want itself to be a quadratic or cubic residue. Sometimes you want to count a
slightly different thing than what we were just counting. When you do the point counting,
when it's avoiding a cubic, quadratic or cubic residue, then you end up having to go to
some other curves, not necessarily just that elliptic curve that we were looking at. And it
ends up you can break up K for K prime up into different cases mod 36. And you can
actually work it out but the calculations were so messy we actually had the computer do
the computations in the end.
All the simplifying and stuff. And if K splits, of course you have to do the two cases
separately and combine. If it isn't prime you have even more complicated stuff to look at.
So I won't ->>: The relationship between these curves has to get started with?
>> Katherine Stange: So there isn't really. The only reason I'm bringing this curve into it
is because this is the numerator of a cubic residue and I want it to be a cube. So when
you want it to be a cube but you also don't want this to be a quadratic residue or a cubic
residue, then you can count the number of points on that other curve and then take away
the ones that would fit on this with N equals 2.
And so, okay, but basically we turn it into a number theory problem and then ask that
problem naively and forget the original curves to do with the pairs.
So those computations are a challenge. And now I'm putting up the guts of the actual
answer in case anybody wants to actually see it written down. So basically instead of the
MK which was just -- there's four cases here. Here's the second slide. The last case
here, where K is congruent to 3 mod 4, or K, the prime factors of K are not congruent to
plus or minus 1. This means each prime factor. Then the MK you want is just the
original one. So just the fact where those are invertible in the residue. And then for
these other ones you have cases where it might not be a cubic residue, might not be
allowed to be a cubic residue or allowed to be a quadratic residue. It just depends on the
behavior of K mod 4 and mod 9.
And then again MK star is just all the things in MK, but where that's a cube. So that's the
like technical details to writing this out. And then once you have those definitions we
expect the limit going back to the amicable pairs question we expect this limit to be a
quarter of the limit of those counting functions.
And that doesn't even deal with the case when K is not co-prime to 6. So here's what it
comes out to when you actually count all these things. Okay. It took pages and pages in
the paper. And then it simplifies to these simple polynomials in K. So when you do the
computations you're counting points on all of these curves, on these Jacobeans, these
crazy curves, you do all this crazy stuff, traces of Frobenius are showing up, residue are
showing up. They cancel. Somehow you get polynomial K in the end. So when we
count these things, we get one-sixth plus a half of whatever polynomials in the different
cases.
And so in the one case that I showed that worked out when K was 2 mod 3 N prime, it
wasn't maybe surprising it came out to be a polynomial in K. But for these ones it really
is because the computation, there's no reason to expect to have all that stuff simplify. I
feel like there should be some real good reason for this that I'm just not, I don't know
what it is. And it's a purely -- you can state this problem purely in the case of as a
number theory problem, and if the answers are so simple I feel there should be a good
reason. And obviously there's not a good reason.
So finally here's some data for this conjecture. So we went and computed these, what
we expect the conjectural limit to be. And for each K it's just some rational number. So
here are those rational numbers for each K. And these two columns can bear conjecture
and experiment. And I think it's pretty convincing that we seem to have discovered that
the Frobenius is ranging through at least those allowable classes according to the law,
according to the way described.
So it seems plausible. And fortunately with these curves we can get lots of pairs. This is
the number of pairs. That gives you an idea how robust the data is. So that's up to 10 to
the 8. So seems reasonable. So these are the final suggestions, ideas, questions, I
don't know, that might come out of this project.
So, first of all, this first remark is just what I was saying a second ago. These things I'm
computing, the number of classes where this thing is a cube, are coming out to these
very simple answers, and I don't know why. So one could ask that question more
generally. So we had M times 1 minus M, but what happens if you have some
polynomial in terms of M up here.
I don't know what's known about that question in general. And so I mean I'm new to
Sage, but it seems like we could ask about what Sage can already do with respect to
these computations we didn't use them in Sage we used the computer PARI to do the
computations because they were so messy and could doing these sorts of computation
provide any insight and give some nicer answer to why these things came out nicely, or is
it a special case for M times 1 minus M somehow.
Another question that comes out of this is to think about this as a dynamical system. So
going back to the pairs, it's sort of like taking reduction modulo P is something you could
iterate. So you take reduction mod P with Q and you take it with Q, get back to P and
you have a two-cycle. We'd have to define it more generally, because what if you reduce
mod P and you get some composite number. What's the next step? What's the function
you want to iterate. That's reduction mod P for primes. So one idea is to define, instead
of your trace of Frobenius only for P to define it for composite numbers in terms of the L
series. And a student of Joe Silverman, Sahinoglu, is working on this. So we shouldn't
steal her problem. But there might be other ideas for how to think about this as a
dynamical system. This might not be the right dynamical system.
>>: I think it's weird. N plus 1 minus N seems a little suspicious. Shouldn't it be like -- NF
times Q, seems to be P minus 1 AP, Q plus 1 minus AQ, the cardinality of either ->> Katherine Stange: Yeah, I have.
>>: Mod NC.
>> Katherine Stange: I have the same thought that this is suspicious. Yeah. It's fishy.
Certainly there are other alternatives. It seems more natural to me. I also thought
somehow ->>: Like the cardinality of E -- the cardinality of EZ mod.
>> Katherine Stange: Exactly. But then what do you do -- well, yeah, so that's definitely
something to ponder.
Also one reason that we found this interesting was because if a prime is anomalous, it's
actually a very special case. If you have an elliptic curve over FP where P is anomalous,
it's susceptible to an attack, elliptic curve discrete problem. If you have an amicable pair
on your elliptic curve, is there anything you get from that? We couldn't get anything, but
it's an interesting question.
And of course, I don't know, we could figure out other ways to construct pairs, cycles or
better ways to search for them. We really did exhaustive searching just by reducing and
checking what the order was.
So, yeah, so, thank you.
[applause]
>>: Does anybody have any questions?
>>: Sometimes when we look at these distributions, compare them with the certain matrix
groups, how often those matrices are finalized, the probabilities end up matching up with
that space?
>> Katherine Stange: You're talking about this question here?
>>: You were saying about that your probabilities where you have one-sixth plus
something in K. Now you're saying is there any explanation for that. I was wondering,
one place to look for that might be to look in to see whether any matrix groups associated
with your problem and then trying to analyze the probability that those matrices have a
certain value and eigenvalue. And so you think that the eigenvalue 1, for example,
should occur about 1 over P time. It's actually a little bit off from that. So sometimes
those -- sometimes that can give an explanation for something that ->> Katherine Stange: Oh, that's fascinating. Yeah, I'd like to find out more about that. I
haven't thought about that.
>>: So [inaudible] density theorem didn't pop up anywhere in your talk. Somehow feels
like it should if you're trying to prove one of these cases but it kind of doesn't matter if
we're just trying to find the conjecture. Why is that?
>> Katherine Stange: Well, I think we did think about it a little bit. But I'm not sure if I
have anything coherent to say about it.
>>: And so in every possible case you have kind of a conjecture now but you don't have
a proof of any cases at all?
>> Katherine Stange: Right. All we've got is the proof that it's sort of restricted in the CM
case.
>>: You've given a conjecture in every case which is really nice.
>> Katherine Stange: I think except for K not co-prime with 6.
>>: Few cases J0.
>> Katherine Stange: Yeah.
>>: Are you going to ask the same questions over number fields as well, they cost
everybody for the one CM curves, for example?
>> Katherine Stange: Yeah, we did sort of note that the similar proof will work for CM
over -- but of course there could be way more complications of this type that you get for
the J equals 0 case. So saying something works means the same sort of proof works but
the actual working it out, what the many cases are depending on the units and stuff, you
know ->>: What about the non-CM case over certain number fields?
>> Katherine Stange: There we only have a conjecture. We don't really have anything ->>: Conjecture there, easier, because it was easier to make if conjecture in the non-CM
case.
>> Katherine Stange: Yes. So I guess we should -- I don't think.
>>: Make it easy.
>> Katherine Stange: Yeah, I haven't had time to write it down. The paper was already
like -- with all this computing for the J equals 0 case, 50 pages, we're like who is going to
want this. The first referee came back and said it's interesting they don't prove anything.
It's 50 pages long.
[laughter]
Which was slightly unfair. It's a lot of figuring out how to write down the right conjectures.
>>: Regarding small jobs, so I've been talking some with [inaudible] into Sage. So that
could be one of the projects. I don't think it would be too difficult to include in Sage. And
then at least you might be able to replicate the tables you may be using -- he may be
using small junk [inaudible].
>> Katherine Stange: Yeah. He sent us more data than I have, that's up here, but he
was just using it, he thought it was good for testing.
>>: Which is fast.
>> Katherine Stange: Yeah, he got farther than we managed to do. We only got to here.
>>: 13th to the ninth and small jobs a little farther, works in parallel. So it's really nice.
>> Katherine Stange: Yeah. So it was really nice.
>>>: Any other questions? Okay. Let's thank our speaker.
[applause]