Have Yourself a Merry Little Christmas (Theorem)

Have Yourself a Merry Little Christmas(Theorem)

Kate Thompson

October 11, 2017

Personal Story Time:

I How I became a number theorist.

I Why I love number theory.

I I’m teaching number theory this spring! (even thoughtypically offered in fall only) 21-441

What odd primes can be written as a sum of two (integer)squares?

3? No.5? Yes.Data?

→ 3← 12 + 22 = 5 → 7←→ 11← 22 + 32 = 13 42 + 12 = 17→ 19← → 23← 52 + 22 = 29→ 31← 62 + 12 = 37 52 + 42 = 41

What’s going on here?


3?

No.5? Yes.Data?

→ 3← 12 + 22 = 5 → 7←→ 11← 22 + 32 = 13 42 + 12 = 17→ 19← → 23← 52 + 22 = 29→ 31← 62 + 12 = 37 52 + 42 = 41



3? No.

5? Yes.Data?

→ 3← 12 + 22 = 5 → 7←→ 11← 22 + 32 = 13 42 + 12 = 17→ 19← → 23← 52 + 22 = 29→ 31← 62 + 12 = 37 52 + 42 = 41



3? No.5?

Yes.Data?

→ 3← 12 + 22 = 5 → 7←→ 11← 22 + 32 = 13 42 + 12 = 17→ 19← → 23← 52 + 22 = 29→ 31← 62 + 12 = 37 52 + 42 = 41



3? No.5? Yes.

Data?

→ 3← 12 + 22 = 5 → 7←→ 11← 22 + 32 = 13 42 + 12 = 17→ 19← → 23← 52 + 22 = 29→ 31← 62 + 12 = 37 52 + 42 = 41



3? No.5? Yes.Data?

→ 3← 12 + 22 = 5 → 7←→ 11← 22 + 32 = 13 42 + 12 = 17→ 19← → 23← 52 + 22 = 29→ 31← 62 + 12 = 37 52 + 42 = 41



3? No.5? Yes.Data?

→ 3← 12 + 22 = 5 → 7←→ 11← 22 + 32 = 13 42 + 12 = 17→ 19← → 23← 52 + 22 = 29→ 31← 62 + 12 = 37 52 + 42 = 41


Theorem (Fermat’s Christmas Theorem, 1640)

Let p be an odd prime. There exist integers x and y such that

x2 + y2 = p

if and only if p ≡ 1 (mod 4); i.e., if and only if p = 4n + 1 forsome n ∈ N.

First proof?

I Euler, ≈ 1747.

I Lagrange (and Gauss), ≈ 1775− 1800

I Dedekind, ≈ 1875− 1890

I Minkowski, ≈ 1900

I Zagier, 1990

I A. David Christopher, 2016



x2 + y2 = p


First proof?

I Euler, ≈ 1747.


I Dedekind, ≈ 1875− 1890


I Zagier, 1990




x2 + y2 = p


First proof?

I Euler, ≈ 1747.


I Dedekind, ≈ 1875− 1890


I Zagier, 1990




x2 + y2 = p


First proof?

I Euler, ≈ 1747.


I Dedekind, ≈ 1875− 1890


I Zagier, 1990


One direction is “easy”.

(An odd prime p = x2 + y2) ⇒ (p ≡ 1 (mod 4)).

Proof.For n ∈ Z, n2 ≡ 0, 1 (mod 4). So, x2 + y2 ≡ 0, 1, 2 (mod 4).Thus, if p = x2 + y2 is odd, then p ≡ 1 (mod 4).

The other direction will not be a two line proof.













We will use one tool with some frequency:

If p ≡ 1 (mod 4), there is an integer 1 ≤ u ≤ p − 1 withu2 ≡ −1 (mod p).

Just so we’re clear:

I If p = 5, then u =2, 3.

I If p = 13, then u =5, 8

I If p = 101, then u = 10, 91

Not to get side-tracked, but for those interested this follows fromWilson’s Theorem.




I If p = 5, then u =2, 3.

I If p = 13, then u =5, 8

I If p = 101, then u = 10, 91





I If p = 5, then u =2, 3.

I If p = 13, then u =5, 8

I If p = 101, then u = 10, 91





I If p = 5, then u =

2, 3.

I If p = 13, then u =5, 8

I If p = 101, then u = 10, 91





I If p = 5, then u =2, 3.

I If p = 13, then u =5, 8

I If p = 101, then u = 10, 91





I If p = 5, then u =2, 3.


5, 8

I If p = 101, then u = 10, 91





I If p = 5, then u =2, 3.

I If p = 13, then u =5, 8

I If p = 101, then u = 10, 91





I If p = 5, then u =2, 3.

I If p = 13, then u =5, 8


10, 91





I If p = 5, then u =2, 3.

I If p = 13, then u =5, 8

I If p = 101, then u = 10, 91





I If p = 5, then u =2, 3.

I If p = 13, then u =5, 8

I If p = 101, then u = 10, 91


So, in what time remains we will prove:

(p ≡ 1 (mod 4))⇒ (p = x2 + y2)

using

I Algebraic techniques (mimic-ing Dedekind)

I Geometric techniques (a la Minkowski)

I Analytic techniques (local-global principle)


(p ≡ 1 (mod 4))⇒ (p = x2 + y2)

using





(p ≡ 1 (mod 4))⇒ (p = x2 + y2)

using




Algebra Idea: Use Arithmetic in the Gaussians

The Gaussian integers are simply

Z[i ] = {a + bi : a, b ∈ Z, i2 = −1}.

Why should this have anything to do with the sum of two squares?

(x + iy)(x − iy) = x2 + y2

So essentially, saying that p = x2 + y2 is somehow equivalent tofactoring p in the Gaussians.



Z[i ] = {a + bi : a, b ∈ Z, i2 = −1}.


(x + iy)(x − iy) = x2 + y2




Z[i ] = {a + bi : a, b ∈ Z, i2 = −1}.


(x + iy)(x − iy) = x2 + y2




Z[i ] = {a + bi : a, b ∈ Z, i2 = −1}.


(x + iy)(x − iy) = x2 + y2


So suppose p ≡ 1 (mod 4). Then, by our fact, there is an integerm with m2 ≡ −1 (mod p). That means that m2 + 1 ≡ 0 (mod p)or p|(m2 + 1).

Factor m2 + 1 in the Gaussians as (m + i)(m − i). So what wehave now is

p|(m + i)(m − i).

One fun fact if p|ab and p is prime, then p|a or p|b. What we’regoing to first argue, then, is that p is not prime in Z[i ]. Thatmeans we’re going to show that while p|(m + i)(m − i), p - m + iand p - m − i .



p|(m + i)(m − i).




p|(m + i)(m − i).


Suppose that p|m + i . Then there would exist integers a, b so that

p(a + bi) = m + i .

Among other things, this means there’s an integer b so thatpb = 1. Einen. Kleinen. Problemo. So p - m + i .

Similarly, you can conclude that p - m − i and therefore p is notprime in the Gaussians.

Now we still have to show that p is a sum of two integer squares.


p(a + bi) = m + i .





p(a + bi) = m + i .




Since p composite, there exist nonzero integers x1, x2, y1, y2 with:

I gcd(x1, y1) = gcd(x2, y2) = 1

I |y2| ≤ |y1|such that

p = (x1 + y1i)(x2 + y2i)

= p + 0i = (x1x2 − y1y2) + i(x2y1 + x1y2)

y1 = −(x1y2)/x2, so x2|x1 or (x1/x2) ∈ Z.

p = x1x2 +

(x1y2x2

)y2

=x1x2

(x22 + y22 )

Since x22 , y22 ∈ N, we know x1 = x2, and we’re done.


I gcd(x1, y1) = gcd(x2, y2) = 1

I |y2| ≤ |y1|

such that

p = (x1 + y1i)(x2 + y2i)

= p + 0i = (x1x2 − y1y2) + i(x2y1 + x1y2)

y1 = −(x1y2)/x2, so x2|x1 or (x1/x2) ∈ Z.

p = x1x2 +

(x1y2x2

)y2

=x1x2

(x22 + y22 )



I gcd(x1, y1) = gcd(x2, y2) = 1


p = (x1 + y1i)(x2 + y2i)

= p + 0i = (x1x2 − y1y2) + i(x2y1 + x1y2)

y1 = −(x1y2)/x2, so x2|x1 or (x1/x2) ∈ Z.

p = x1x2 +

(x1y2x2

)y2

=x1x2

(x22 + y22 )



I gcd(x1, y1) = gcd(x2, y2) = 1


p = (x1 + y1i)(x2 + y2i)

= p + 0i = (x1x2 − y1y2) + i(x2y1 + x1y2)

y1 = −(x1y2)/x2, so x2|x1 or (x1/x2) ∈ Z.

p = x1x2 +

(x1y2x2

)y2

=x1x2

(x22 + y22 )



I gcd(x1, y1) = gcd(x2, y2) = 1


p = (x1 + y1i)(x2 + y2i)

= p + 0i = (x1x2 − y1y2) + i(x2y1 + x1y2)

y1 = −(x1y2)/x2, so x2|x1 or (x1/x2) ∈ Z.

p = x1x2 +

(x1y2x2

)y2

=x1x2

(x22 + y22 )



I gcd(x1, y1) = gcd(x2, y2) = 1


p = (x1 + y1i)(x2 + y2i)

= p + 0i = (x1x2 − y1y2) + i(x2y1 + x1y2)

y1 = −(x1y2)/x2, so x2|x1 or (x1/x2) ∈ Z.

p = x1x2 +

(x1y2x2

)y2

=x1x2

(x22 + y22 )


What this problem really was using was:

I Unique factorization of elements into primes.

Believe it or not, this doesn’t hold in greater generality. And thisbecomes a delicacy in higher-level algebraic number theory.

As a parting example, consider instead Z[√−5]. Then

6 = 2 · 3 = (1 +√−5)(1−

√−5) are two “distinct” factorizations.

So if, say, you wanted to try to generalize and say what primes areof the form x2 + 5y2, you’d have problems using these techniques.

What this problem really was using was:

I Unique factorization of elements into primes.

Believe it or not, this doesn’t hold in greater generality. And thisbecomes a delicacy in higher-level algebraic number theory.

As a parting example, consider instead Z[√−5]. Then

6 = 2 · 3 = (1 +√−5)(1−

√−5) are two “distinct” factorizations.

So if, say, you wanted to try to generalize and say what primes areof the form x2 + 5y2, you’d have problems using these techniques.

Geometric Proof

The main tools here are:

I Lattices

I Minkowski’s Convex Body Theorem (MCBT)

A lattice formally-speaking is a discrete subset of Rn, and given agenerating set {v1, v2, ..., vn} any point in the lattice is someZ-linear combination of the vi ’s.

Informally, you can think of Zn ⊆ Rn, or any “graph paper onsteroids” series of dots in the plane.

Geometric Proof


I Lattices




Geometric Proof


I Lattices




A lattice tiles the plane (informally, think literally of graph paper).A fundamental domain is a basic tile. That is, given thegenerating set {v1, ..., vn} the fundamental domain is then-dimensional parallelogram whose sides are given by those vectors.

Example: If you’re thinking of Z2 ⊆ R2, a fundamental domaincould be a unit square.

Theorem (Minkowski’s Convex Body Theorem)

Let L ⊆ Rn be a lattice with fundamental domain D. Let B ⊆ Rn

be convex, and centrally-symmetric. If

vol(B) > 2nvol(D)

then B contains a nonzero lattice point in its interior.






vol(B) > 2nvol(D)







vol(B) > 2nvol(D)


What does this have to do with p ≡ 1 (mod 4)⇒ p = x2 + y2?

If you think about Q(x , y) = x2 + y2 and what the level sets are forthat function, they are circles centered at the origin. In particular,those are convex, centrally-symmetric bodies. And if you think ofyour lattice as containing integer points, then what you’d really liketo show is that x2 + y2 = p has an integer point on the boundary.There are just a few issues:

1. Z2 is too big of a lattice.

2. MCBT shows there’s a nonzero lattice point inside a region,not on the boundary.


If you think about Q(x , y) = x2 + y2 and what the level sets are forthat function, they are circles centered at the origin. In particular,those are convex, centrally-symmetric bodies. And if you think ofyour lattice as containing integer points, then what you’d really liketo show is that x2 + y2 = p has an integer point on the boundary.

There are just a few issues:















To fix these issues, we alter the lattice and do not consider Z2.Instead, we consider the lattice L generated by {(p, 0), (u, 1)},where 1 ≤ u ≤ p is an integer such that u2 ≡ −1 (mod p).

Consider any vector v ∈ L. Then

v = a(p, 0) + b(u, 1)

= (ap + bu, b)

|v | = p(a2p + 2abu) + b2(u2 + 1)

≡ 0 (mod p).

That is, every vector in this lattice has a norm divisible by p.



v = a(p, 0) + b(u, 1)

= (ap + bu, b)

|v | = p(a2p + 2abu) + b2(u2 + 1)

≡ 0 (mod p).




v = a(p, 0) + b(u, 1)

= (ap + bu, b)

|v | = p(a2p + 2abu) + b2(u2 + 1)

≡ 0 (mod p).


And now, in a second moment of brilliance, we don’t consider thelevel set associated to x2 + y2 = p. Instead, we look at the circlex2 + y2 = 2p.

We want to use Minkowski, and we can see fairly easily the volumeof our ‘B’ is 2πp. But what is the volume of the fundamentaldomain? Since the lattice is generated by (p, 0) and (u, 1)

vol(D) = det

∣∣∣∣(p u0 1

)∣∣∣∣ = p.

And since 2πp > 22p, MCBT tells us that there is a nonzerolattice point inside our B.

And now, in a second moment of brilliance, we don’t consider thelevel set associated to x2 + y2 = p. Instead, we look at the circlex2 + y2 = 2p.We want to use Minkowski, and we can see fairly easily the volumeof our ‘B’ is 2πp. But what is the volume of the fundamentaldomain? Since the lattice is generated by (p, 0) and (u, 1)

vol(D) = det

∣∣∣∣(p u0 1

)∣∣∣∣ = p.


And now, in a second moment of brilliance, we don’t consider thelevel set associated to x2 + y2 = p. Instead, we look at the circlex2 + y2 = 2p.We want to use Minkowski, and we can see fairly easily the volumeof our ‘B’ is 2πp. But what is the volume of the fundamentaldomain? Since the lattice is generated by (p, 0) and (u, 1)

vol(D) = det

∣∣∣∣(p u0 1

)∣∣∣∣ = p.


But now we’re done. Every lattice point has norm equal to someinteger multiple of p. We’re looking in a region where all pointsinside have norm strictly less than 2p. So, our nonzero latticepoint has norm equal to some multiple of p bounded between 0pand 2p (exclusive). Therefore, we must have some element ofnorm p, which gives us a solution to

x2 + y2 = p.

There are a few restrictions to these geometric methods:

I The reliance on MCBT that the region B be convex andcentrally symmetric (let alone of some kind of finite volume).For instance, this would prevent you from using MCBT on thelevel sets associated to x2 − y2.

I Beyond that, in higher variables, generalizations of thesegeometric techniques only really work well when you have aneven number of variables. So, trying to use this even forx2 + y2 + z2 is more difficult that you would (want to) think.







Analytic Proof: Local-Global Theorem

The idea behind the local-global principle (aka the Hasseprinciple) is this: if you have a polynomial P(x1, x2, ..., xn) (overthe integers, with integer coefficients) satisfying

P(x1, x2, ..., xn) = 0

then ‘obviously’ you have a solution to

P(x1, x2, ..., xn) ≡ 0 (mod n)

for all n ∈ N.

But when can you move backwards? That is, when does asolution (mod n) for all n imply a solution in Z?

Analytic Proof: Local-Global Theorem

The idea behind the local-global principle (aka the Hasseprinciple) is this: if you have a polynomial P(x1, x2, ..., xn) (overthe integers, with integer coefficients) satisfying

P(x1, x2, ..., xn) = 0

then ‘obviously’ you have a solution to

P(x1, x2, ..., xn) ≡ 0 (mod n)

for all n ∈ N.

But when can you move backwards? That is, when does asolution (mod n) for all n imply a solution in Z?

This greatly depends upon the polynomial. Very famously, in 1951Selmer showed that while

3x3 + 4y3 + 5z3 = 0

has nontrivial solutions (mod n) for all n, the only solution overthe integers is the trivial x = y = z = 0.

Lucky for us, P(x , y) = x2 + y2 − p satisfies the Hasse principle.That is, if P(x , y) = 0 has a nontrivial solution (mod n) for all n,then it is guaranteed to have a solution in Z.

This greatly depends upon the polynomial. Very famously, in 1951Selmer showed that while

3x3 + 4y3 + 5z3 = 0

has nontrivial solutions (mod n) for all n, the only solution overthe integers is the trivial x = y = z = 0.

Lucky for us, P(x , y) = x2 + y2 − p satisfies the Hasse principle.That is, if P(x , y) = 0 has a nontrivial solution (mod n) for all n,then it is guaranteed to have a solution in Z.

Still, it’s a daunting task to check that there’s a solution (mod n)for all n. One thing we use to simplify life is the ChineseRemainder Theorem (CRT–typically covered in AlgebraicStructures once you’ve gotten to rings).

We won’t state it, but what it does for you is it tells you you onlyhave to look (mod pk) for all primes p and all powers k (sonot–strictly speaking– (mod n) for all n).

Still, it’s a daunting task to check that there’s a solution (mod n)for all n. One thing we use to simplify life is the ChineseRemainder Theorem (CRT–typically covered in AlgebraicStructures once you’ve gotten to rings).

We won’t state it, but what it does for you is it tells you you onlyhave to look (mod pk) for all primes p and all powers k (sonot–strictly speaking– (mod n) for all n).

Again, though...there are infinitely many primes! And infinitelymany prime powers!

Consider the following: a solution (mod pk) implies a solution(mod p`) for all ` ≤ k (concretely, for instance, a solution(mod 9) implies a solution (mod 3)). Again, though, it’d beGREAT to work backwards. Wouldn’t it be amazing if for somepower of k , a solution (mod pk) implied a solution (mod p`) for` ≥ k?

Again, we luck out. The most important tool in computationalnumber theory: Hensel’s Lemma (think of it as a numbertheorist’s Newton’s method).







Applying Hensel just to x2 + y2 − p = 0 for p ≡ 1 (mod 4) primesays to guarantee a solution (mod qk) for all primes q it sufficesto show a solution exists:

I (mod q3) for q = 2.Not bad: p ≡ 1 (mod 4)⇒ p ≡ 1, 5 (mod 8). 02 + 12 ≡ 1(mod 8) and 22 + 12 ≡ 5 (mod 8).

I (mod q) for q odd.Slightly more delicate, but it’s one of CMU’s favorite kinds ofproofs: contradiction.


I (mod q3) for q = 2.

Not bad: p ≡ 1 (mod 4)⇒ p ≡ 1, 5 (mod 8). 02 + 12 ≡ 1(mod 8) and 22 + 12 ≡ 5 (mod 8).







I (mod q) for q odd.

Slightly more delicate, but it’s one of CMU’s favorite kinds ofproofs: contradiction.




And that basically finishes the analytic proof.

1. The Hasse principle applies to x2 + y2 − p = 0, so if solutionsexist (mod n) for all n, then a solution exists in Z.

2. The Chinese Remainder Theorem says instead of looking(mod n) for all n, you can just look at prime powers.

3. Hensel’s Lemma tells you the prime powers you really careabout are 8 (where there’s a solution) and q (for q odd, whereagain there’s a solution).

4. QED





4. QED





4. QED





4. QED

“Obviously” the downside to the analytic techniques is that they’rehard! But good news/bad news: they generalize most easily (atleast to other polynomials, and to higher numbers of variables).

For those really intrigued, in higher dimensions the buzz-word is“modular forms”.

“Obviously” the downside to the analytic techniques is that they’rehard! But good news/bad news: they generalize most easily (atleast to other polynomials, and to higher numbers of variables).

For those really intrigued, in higher dimensions the buzz-word is“modular forms”.

Where Do We Go From Here?

It may be surprising in the first place that a theorem firstannounced in 1640 had a new proof published as late as 2016.This is one increasingly “big deal” in number theory: showing thatcertain techniques work is crucial–especially for those trying tosolve still-open problems.

But still, what generalizations could there possibly be of this 1640result?

Where Do We Go From Here?

It may be surprising in the first place that a theorem firstannounced in 1640 had a new proof published as late as 2016.This is one increasingly “big deal” in number theory: showing thatcertain techniques work is crucial–especially for those trying tosolve still-open problems.

But still, what generalizations could there possibly be of this 1640result?

You could see the Christmas Theorem as saying “Withfinitely-many exceptions (p = 2), all the primes represented byx2 + y2 can be determined by congruence conditions.” Viewingthings that way, a natural question to ask would be “Does thishold for polynomials beyond x2 + y2?”

Theorem (Clark, Hicks, Parshall, T., 2013)

There are precisely 2779 polynomials of the form ax2 + bxy + cy2

(where b2 − 4ac < 0) for which–with finitely many (explicit)exceptions–the primes represented can be determined completelyby congruence conditions (also made explicit).

You could see the Christmas Theorem as saying “Withfinitely-many exceptions (p = 2), all the primes represented byx2 + y2 can be determined by congruence conditions.” Viewingthings that way, a natural question to ask would be “Does thishold for polynomials beyond x2 + y2?”

Theorem (Clark, Hicks, Parshall, T., 2013)

There are precisely 2779 polynomials of the form ax2 + bxy + cy2

(where b2 − 4ac < 0) for which–with finitely many (explicit)exceptions–the primes represented can be determined completelyby congruence conditions (also made explicit).

Where else could you go?

I Up the number of variables.

I Can you say how often a prime would be represented?

I Forget about primes. Look at full congruences classes. Lookat integers to avoid as well as integers to represent. Look atarithmetic progressions.

I Change the playing field (almost literally). Instead ofconsidering polynomials over Z, what about polynomials inZ[i ] (or other rings of integers)? What about over polynomialsrings (going meta and looking at polynomials of polynomials)?
















Thank you.

Documents

Have Yourself a Merry Little Christmas (Theorem)