Cool theorems proved by undergraduates

Cool theorems proved by undergraduates Ken Ono Emory University

Child s play...

Child s play... Thanks to the,

Child s play... Thanks to the, undergrads and I play by...

Child s play... Thanks to the, undergrads and I play by...shooting off

Child s play... Thanks to the, undergrads and I play by...shooting off and getting in trouble,

Child s play... Thanks to the, undergrads and I play by...shooting off and getting in trouble,...and by proving theorems in number theory.

Child s play... Our toys include...

Child s play... Our toys include... (Prime numbers)

Child s play... Our toys include... (Prime numbers) (Partitions)

Child s play... Our toys include... (Prime numbers) (Partitions) (Numbers and Number Fields)

Prime Numbers Some theorems on primes

Prime Numbers Some theorems on primes Theorem (Euclid) There are infinitely many prime numbers.

Prime Numbers Some theorems on primes Theorem (Euclid) There are infinitely many prime numbers. Theorem (Prime Number Theorem) If π(x) := #{p x : prime}, then π(x) x ln x.

Prime Numbers

Prime Numbers Theorem (Dirichlet) If 0 r < t are integers for which gcd(r, t) = 1, then let π r,t (x) := #{p x : prime and p r (mod t)}.

Prime Numbers Theorem (Dirichlet) If 0 r < t are integers for which gcd(r, t) = 1, then let π r,t (x) := #{p x : prime and p r (mod t)}. Then in terms of Euler s phi-function, we have π r,t (x) 1 φ(t) x ln x. Example (Half of the primes are of the following forms:) 3n + 1, 3n + 2 4n + 1, 4n + 3 6n + 1, 6n + 5

Prime Numbers Arithmetic progressions of primes

Prime Numbers Arithmetic progressions of primes Theorem (van der Corput (1939)) There are ly many length 3 arithmetic progressions of primes.

Prime Numbers Arithmetic progressions of primes Theorem (van der Corput (1939)) There are ly many length 3 arithmetic progressions of primes. Example Here are examples of arithmetic progressions of length 3: (3, 3 + 2, 3 + 2 2) = (3, 5, 7), (5, 5 + 42, 5 + 2 42) = (5, 47, 89), (43, 43 + 30, 43 + 2 30) = (43, 73, 103).

Prime Numbers How long can these get?

Prime Numbers How long can these get? Theorem (Green-Tao (2000s)) For every k there are ly many length k AP of primes.

Prime Numbers How long can these get? Theorem (Green-Tao (2000s)) For every k there are ly many length k AP of primes. Example Length 10 example: 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089.

Prime Numbers Different problem Question Can you find k consecutive primes which end with the digit 1?

Prime Numbers Different problem Question Can you find k consecutive primes which end with the digit 1? Example We have the following consecutive primes 181, 191, 241, 251,. 4831, 4861, 4871.

Prime Numbers A great theorem...

Prime Numbers A great theorem... Theorem (Shiu (2000)) Let p 1 = 2, p 2 = 3,... be the primes in order. If gcd(r, t) = 1, then for every positive integer k there is an n for which p n p n+1 p n+2 p n+k r (mod t).

Prime Numbers A cool theorem... Theorem (Monks, Peluse, Ye) Special sets of primes have arbitrarily long sequences of primes in any arithmetic progression.

Prime Numbers A cool theorem... Theorem (Monks, Peluse, Ye) Special sets of primes have arbitrarily long sequences of primes in any arithmetic progression. Corollary Define the sequence of integers N π := { π, 2π, 3π,... }. Let p π (n) be the nth prime in this sequence, and so p π (1) = 3, p π (2) = 31, p π (3) = 37,....

Prime Numbers More on strings of prime

Prime Numbers More on strings of prime Example The first 6 consecutive primes 5 (mod 7) in N π is 26402437, 26402507, 26402591, 26402843, 26402899, 26402927.

Prime Numbers More on strings of prime Example The first 6 consecutive primes 5 (mod 7) in N π is 26402437, 26402507, 26402591, 26402843, 26402899, 26402927. Remark Special sets includes primes in the sets N α := { α, 2α, 3α,... } α irrational real alg. int., N := { n log log n : n = 1, 2, 3,... }.

Prime Numbers

Recurrence sequences The Fibonacci Sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377

Recurrence sequences The Fibonacci Sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377 Conjecture (Folklore) These are the only perfect powers!

Recurrence sequences Perfect Powers in the Fibonacci Sequence Theorem (Siksek, Bugeaud, Mignotte, Annals of Math 2006) F 0 = 0 p, F 1 = 1 p, F 6 = 2 3, and F 12 = 12 2 are the only perfect powers in the Fibonacci sequence.

Recurrence sequences Lucas Sequences Definition (Lucas Sequence) A Lucas sequence u n is a nondegenerate integral linear binary recurrence relation defined by with u 0 = 0 and u 1 = 1. u n+2 = bu n+1 + cu n Definition (Companion Sequence) The companion sequence v n is defined by v n+2 = bv n+1 + cv n with v 0 = 2 and v 1 = b.

Recurrence sequences Lucas Sequences A Lucas sequence (b, c) has characteristic polynomial and roots g(z) = z 2 bz c, α, β = b ± b 2 + 4c. 2

Recurrence sequences Lucas Sequences A Lucas sequence (b, c) has characteristic polynomial and roots g(z) = z 2 bz c, α, β = b ± b 2 + 4c. 2 Fact 1. u n = αn β n α β v n = α n + β n 2. u 2k = u k v k 3. (b 2 + 4c)u 2 n = v 2 n 4( c) n

Recurrence sequences Specific Examples Theorem (Silliman, Vogt) For the following values of b and c: (b, c) = (3, 2), (5, 6), (7, 12), (17, 72), (9, 20) the Lucas sequence u n has no nontrivial pth powers, except u 2 = 3 2 in (9, 20).

Recurrence sequences General Bound Conjecture (Frey-Mazur) Let E, E be two elliptic curves defined over Q. If E[p] = E [p] for some p > 17, then E and E are isogenous.

Recurrence sequences General Bound Conjecture (Frey-Mazur) Let E, E be two elliptic curves defined over Q. If E[p] = E [p] for some p > 17, then E and E are isogenous. Theorem (Silliman, Vogt) Assume the Frey Mazur conjecture. Then a pth power u n = y p satisfies p Ψ(b, c). where Ψ(b, c) is an explicit constant.

Recurrence sequences More Examples Example The sequence (3, 1): 0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, We haven t found any nontrivial perfect powers...

Recurrence sequences More Examples Example The sequence (3, 1): 0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, We haven t found any nontrivial perfect powers... Theorem (Silliman, Vogt) Assume the Frey-Mazur Conjecture. There are no nontrivial powers in the sequences (b, c) = (3, 1), (5, 1), and (7, 1).

Recurrence sequences

Partitions Beautiful identities

Partitions Beautiful identities Euler proved that q (1 q 24n ) = q q 25 q 49 + q 121 + q 169 q 289. n=1

Partitions Beautiful identities Euler proved that q (1 q 24n ) = q q 25 q 49 + q 121 + q 169 q 289. n=1 Jacobi proved that q (1 q 8n ) 3 = q 3q 9 + 5q 25 7q 49 + 9q 81 11q 121 +. n=1

Partitions Beautiful identities Euler proved that q (1 q 24n ) = q q 25 q 49 + q 121 + q 169 q 289. n=1 Jacobi proved that q (1 q 8n ) 3 = q 3q 9 + 5q 25 7q 49 + 9q 81 11q 121 +. n=1 Gauss proved that q n=1 (1 q 16n ) 2 (1 q 8n ) = q+q 9 +q 25 +q 49 +q 81 +q 121 +q 169 +q 225 +

Partitions Such rare identities have been discovered by:

Partitions Such rare identities have been discovered by: Crazy combinatorial manipulation of power series. Higher identities such as Jacobi s n=1(1 q 2n )(1 + z 2 q 2n 1 )(1 + z 2 q 2n 1 ) = m Z z 2m q m2. Modular forms.

Partitions Nekrasov-Okounkov Theory

Partitions Nekrasov-Okounkov Theory Deeper structure for such identities.

Partitions Nekrasov-Okounkov Theory Deeper structure for such identities. One doesn t have to multiply out and combine terms.

Nekrasov-Okounkov Partitions Definition A nonincreasing sequence of positive integers summing to n is a partition of n.

Nekrasov-Okounkov Partitions Definition A nonincreasing sequence of positive integers summing to n is a partition of n. p(n) := # {partitions of n}.

Nekrasov-Okounkov Partitions Definition A nonincreasing sequence of positive integers summing to n is a partition of n. p(n) := # {partitions of n}. Example {Partitions of 4} = {4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1}

Nekrasov-Okounkov Generating function for p(n)

Nekrasov-Okounkov Generating function for p(n) Lemma We have that p(n)q n 1 = 1 q n. n=0 n=1

Nekrasov-Okounkov Wishful thinking

Nekrasov-Okounkov Wishful thinking Question Is there a combinatorial theory of infinite products where coefficient of q n = formula in partitions of n?

Nekrasov-Okounkov Hooklengths of partitions

Nekrasov-Okounkov Hooklengths of partitions Definition Hooks are the maximal. in the Ferrers board of λ.

Nekrasov-Okounkov Hooklengths of partitions Definition Hooks are the maximal. in the Ferrers board of λ. Hooklengths are their lengths, and H(λ) = { multiset of hooklengths of λ}

Nekrasov-Okounkov An example Example (λ = 5 + 3 + 2) 7 6 4 2 1 4 3 1 2 1

Nekrasov-Okounkov An example Example (λ = 5 + 3 + 2) 7 6 4 2 1 4 3 1 2 1 An so H(λ) = {1, 1, 1, 2, 2, 3, 4, 4, 6, 7}.

Nekrasov-Okounkov Nekrasov-Okounkov q-series Definition For z C, let O z (q) := λ q λ h H(λ) (1 z h 2 ).

Nekrasov-Okounkov Example (λ = 5 + 3 + 2) 7 6 4 2 1 4 3 1 2 1

Nekrasov-Okounkov Example (λ = 5 + 3 + 2) 7 6 4 2 1 4 3 1 2 1 The λ-contribution to O z (q) is q 10 (1 z ) h 2 h H(λ) = q 10 (1 z) 3 ( 1 z 4 ) 2 ( 1 z ) ( 1 z ) 2 ( 1 z ) ( 1 z ) 9 16 36 49

Nekrasov-Okounkov A famous identity revisited

Nekrasov-Okounkov A famous identity revisited Letting z = 4, we consider O 4 (q) = λ q λ h H(λ) (1 4h 2 ).

Nekrasov-Okounkov A famous identity revisited Letting z = 4, we consider O 4 (q) = λ q λ h H(λ) (1 4h 2 ). We only need λ where every hook h 2.

Nekrasov-Okounkov A famous identity revisited Letting z = 4, we consider O 4 (q) = λ q λ h H(λ) (1 4h 2 ). We only need λ where every hook h 2. = {Triangular partitions 1 + 2 + + k}

Nekrasov-Okounkov A famous identity revisited.

Nekrasov-Okounkov A famous identity revisited. λ 8 λ + 1 H(λ) (1 4/h 2 ) 0 1 φ 1 1 9 {1} 3 2 + 1 25 {1, 1, 3} 5 3 + 2 + 1 49 {1,..., 5} 7

Nekrasov-Okounkov A famous identity revisited. λ 8 λ + 1 H(λ) (1 4/h 2 ) 0 1 φ 1 1 9 {1} 3 2 + 1 25 {1, 1, 3} 5 3 + 2 + 1 49 {1,..., 5} 7 And so we have = qo 4 (q 8 ) = q 3q 9 + 5q 25 7q 49 +

Nekrasov-Okounkov A famous identity revisited. λ 8 λ + 1 H(λ) (1 4/h 2 ) 0 1 φ 1 1 9 {1} 3 2 + 1 25 {1, 1, 3} 5 3 + 2 + 1 49 {1,..., 5} 7 And so we have = qo 4 (q 8 ) = q 3q 9 + 5q 25 7q 49 + Jacobi? = q (1 q 8n ) 3 n=1

Nekrasov-Okounkov Big Identity

Nekrasov-Okounkov Big Identity Theorem (Nekrasov-Okounkov) If z is complex, then O z (q) := q λ (1 z ) h 2 = (1 q n ) z 1. λ n=1 h H(λ)

Nekrasov-Okounkov Big Identity Theorem (Nekrasov-Okounkov) If z is complex, then O z (q) := λ q λ (1 z ) h 2 = (1 q n ) z 1. h H(λ) n=1 Remark Letting z = 0 gives the generating function for p(n).

Nekrasov-Okounkov Euler, Gauss, and Jacobi-type identities

Nekrasov-Okounkov Euler, Gauss, and Jacobi-type identities Question 1 When do the sums below vanish? A(a, b; n) := λ =n h H a(λ) ( 1 ab ) h 2.

Nekrasov-Okounkov Euler, Gauss, and Jacobi-type identities Question 1 When do the sums below vanish? A(a, b; n) := λ =n h H a(λ) ( 1 ab ) h 2. 2 Are there more identities of Euler, Gauss and Jacobi-type?

Nekrasov-Okounkov A cool theorem

Nekrasov-Okounkov A cool theorem Theorem (Clader, Kemper, Wage) The list of all pairs (a, b) for which almost all the A(a, b; n) vanish are (1, 2), (1, 3), (1, 4), (1, 5), (1, 7), (1, 9), (1, 11), (1, 15), (1, 27), (2, 2), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 9), (4, 5), (4, 7), (7, 9), (7, 15).

Nekrasov-Okounkov

Number Fields Number fields Wrong Definition This is a number field.

Number Fields Number fields Wrong Definition This is a number field..

Number Fields Number fields Wrong Definition This is a number field.. Definition A finite dimensional field extension of Q is called a number field.

Number Fields An invariant Remark The discriminant K Z \ {0} of a number field K does:

Number Fields An invariant Remark The discriminant K Z \ {0} of a number field K does: Measures the volume of algebraic integers.

Number Fields An invariant Remark The discriminant K Z \ {0} of a number field K does: Measures the volume of algebraic integers. Controls some properties of primes (ramification).

Number Fields Quadratic fields If D is square-free and K := Q( D), then { D if D 1 (mod 4), K := 4D otherwise.

Number Fields Quadratic fields If D is square-free and K := Q( D), then { D if D 1 (mod 4), K := 4D otherwise. Example For example, we have Q(i) = = 4 Q( 2) = = 8.

Number Fields Distribution of quadratic fields Notation. Let N 2 (X ) := #{quad fields with X }.

Number Fields Distribution of quadratic fields Notation. Let N 2 (X ) := #{quad fields with X }. X N 2 (X ) N 2 (X )/X 10 2 61 0.6100... 10 4 6086 0.6086... 10 6 607925 0.6079...

Number Fields Distribution of quadratic fields Notation. Let N 2 (X ) := #{quad fields with X }. X N 2 (X ) N 2 (X )/X 10 2 61 0.6100... 10 4 6086 0.6086... 10 6 607925 0.6079... Theorem (Easy) We have that N 2 (X ) lim = 6 X + X π 2 = 0.607927....

Number Fields General Number Fields Conjecture (Linnik) If we let N n (X ) := #{degree n fields with X }, then N n (X ) c n X.

Number Fields General Number Fields Conjecture (Linnik) If we let N n (X ) := #{degree n fields with X }, then N n (X ) c n X. Theorem (Easy, Davenport-Heilbronn (70s), Bhargava (2000s)) Linnik s Conjecture is true for n = 2, 3, 4, 5.

Number Fields General results

Number Fields General results Theorem (Schmidt (1995), Ellenberg-Venkatesh (2006)) We have that X n+2 4 if 6 n 84393, N n (X ) n X exp(c log n) if n > 84393.

Number Fields Related Functions Notation. If G is a finite group, then let N n (G; X ) := number of deg n fields with X whose Galois closure has Galois group G

Number Fields Related Functions Notation. If G is a finite group, then let N n (G; X ) := number of deg n fields with X whose Galois closure has Galois group G Remark Estimating N n (G; X ) is easy only for n = 2 and G = Z/2Z.

Number Fields Prime cyclic cases Theorem (Wright (1989)) If p is an odd prime, then there is a constant c(p) for which N p (Z/pZ; X ) = c(p) X 1 p 1 + O(X 1/p ).

Number Fields Prime cyclic cases Theorem (Wright (1989)) If p is an odd prime, then there is a constant c(p) for which N p (Z/pZ; X ) = c(p) X 1 p 1 + O(X 1/p ). Conjecture (Cohen, Diaz y Diaz, Olivier (2006)) We have that N 3 (Z/3Z; X ) = c(3) X 1 2 + O(X 1/6 ).

Number Fields A cool theorem...

Number Fields A cool theorem... Theorem (Lee-Oh) Assuming the GRH, if p is an odd prime, then N p (Z/pZ; X ) = c(p) X 1 p 1 + X 1 3(p 1) R p (log X ) + O(X 1 4(p 1) ), where R p (x) is a polynomial of degree p(p 2)/3 1.

Number Fields Linear regression when p = 3: Slope 1 6

Number Fields Linear regression when p = 3: Slope 1 6 3 2 1 5 10 15 20 25 1 2 x axis = log 10 (X ) y axis = log 10 (N 3 (Z/3Z; X ) c(3) X 1 2 )

In conclusion... Our summer toys include...

In conclusion... Our summer toys include... (Prime numbers) (Partitions) (Numbers and Number Fields)

In conclusion... So, we basically...

In conclusion... So, we basically......prove theorems, and...explode stuff...