Primes” Without “Pi”

You can’t spell “primes” without “pi”

David Patrick Art of Problem Solving aops.com

March 2017

David Patrick (AoPS) Primes March 2017 1 / 37 Over 260,000 members in our online community Over 5,300,000 messages posted Over 3,400,000 unique visitors in the past year Accredited by WASC as a supplementary education program

Art of Problem Solving History

www.artofproblemsolving.com Founded in 2003 Founder: Richard Rusczyk, 1989 USAMO Winner, co-author of Art of Problem Solving books (1993-94)

Created to provide resources and community for eager math students and their teachers and parents.

David Patrick (AoPS) Primes March 2017 2 / 37 Art of Problem Solving History

www.artofproblemsolving.com Founded in 2003 Founder: Richard Rusczyk, 1989 USAMO Winner, co-author of Art of Problem Solving books (1993-94)

Created to provide resources and community for eager math students and their teachers and parents.

Over 260,000 members in our online community Over 5,300,000 messages posted Over 3,400,000 unique visitors in the past year Accredited by WASC as a supplementary education program

David Patrick (AoPS) Primes March 2017 2 / 37 Standards for Mathematical Practice Common Core State Standards for Mathematics

1 Make sense of problems and persevere in solving them 2 Reason abstractly and quantitatively 3 Construct viable arguments and critique the reasoning of others 4 Model with mathematics 5 Use appropriate tools strategically 6 Attend to precision 7 Look for and make use of structure 8 Look for and express regularity in repeated reasoning

David Patrick (AoPS) Primes March 2017 3 / 37 Problem-solving perspective: new concepts introduced via challenging problems, not as unmotivated facts and tools. Problem-solving skills are explicitly taught. Being able to solve every problem means that the problems are too easy—students should learn that not every problem is easy, and “not solving a problem” is not the same as “failing”. Importance of peer group: Students of like interest and ability feed off of each other. They learn from each other. They challenge and inspire each other. Many students thrive on competition...but not all competitions and not all students.

Some Philosophies

There’s more to life than standardized tests or just racing to calculus.

David Patrick (AoPS) Primes March 2017 4 / 37 Being able to solve every problem means that the problems are too easy—students should learn that not every problem is easy, and “not solving a problem” is not the same as “failing”. Importance of peer group: Students of like interest and ability feed off of each other. They learn from each other. They challenge and inspire each other. Many students thrive on competition...but not all competitions and not all students.

Some Philosophies

David Patrick (AoPS) Primes March 2017 4 / 37 Importance of peer group: Students of like interest and ability feed off of each other. They learn from each other. They challenge and inspire each other. Many students thrive on competition...but not all competitions and not all students.

Some Philosophies

David Patrick (AoPS) Primes March 2017 4 / 37 Many students thrive on competition...but not all competitions and not all students.

Some Philosophies

There’s more to life than standardized tests or just racing to calculus. Problem-solving perspective: new concepts introduced via challenging problems, not as unmotivated facts and tools. Problem-solving skills are explicitly taught. Being able to solve every problem means that the problems are too easy—students should learn that not every problem is easy, and “not solving a problem” is not the same as “failing”. Importance of peer group: Students of like interest and ability feed off of each other. They learn from each other. They challenge and inspire each other.

David Patrick (AoPS) Primes March 2017 4 / 37 ..but not all competitions and not all students.

Some Philosophies

David Patrick (AoPS) Primes March 2017 4 / 37 Some Philosophies

David Patrick (AoPS) Primes March 2017 4 / 37 Art of Problem Solving Online Resources

Free! (some require free registration): Large discussion forum. Topics include math, problem solving, other Community subjects, fun & games. Supports online school and Alcumus. National & Local communities.

Adaptive learning system—customizes to student performance. Work at own pace. Alcumus Over 13,000 problems, many linked to video lessons. Teacher tools to track students’ progress.

MATHCOUNTS-style live interactive multiplayer For the Win! game.

David Patrick (AoPS) Primes March 2017 5 / 37 Art of Problem Solving Online Resources

Free! (some require free registration):

Articles and advice about problem solving, Articles mathematical writing, “proofs without words”, sermonettes, etc.

Over 300 videos, including “MATHCOUNTS Video lessons Minis” and video solutions to recent years’ hardest AMC/AIME problems.

Problem-solving wiki written by and for AoPSWiki students.

LATEX and Tutorials in mathematical writing Asymptote tutorials and graphics

David Patrick (AoPS) Primes March 2017 6 / 37 Online resources Cut-The-Knot (Alex Bogomolny) Nrich Maths (Univ. of Cambridge) mathcircles.org and mathteacherscircle.org

Lots of math circles for students and teachers in the Bay Area!

Other Resources

Online communities Math Forum (Drexel Univ. NCTM) → Math Stack Exchange (Q & A site) many forums for parents (WTM) or for research professionals (MathOverﬂow)

David Patrick (AoPS) Primes March 2017 7 / 37 Lots of math circles for students and teachers in the Bay Area!

Other Resources

Online communities Math Forum (Drexel Univ. NCTM) → Math Stack Exchange (Q & A site) many forums for parents (WTM) or for research professionals (MathOverﬂow) Online resources Cut-The-Knot (Alex Bogomolny) Nrich Maths (Univ. of Cambridge) mathcircles.org and mathteacherscircle.org

David Patrick (AoPS) Primes March 2017 7 / 37 Other Resources

Lots of math circles for students and teachers in the Bay Area!

David Patrick (AoPS) Primes March 2017 7 / 37 High School AMC 10/12 – amc.maa.org Who Wants To Be A Mathematician? – ams.org/wwtbam ARML (at Las Vegas) – arml.com Purple Comet – purplecomet.org Mandelbrot Competition – mandelbrot.org Planning to restart in 2017-18?

Contests

Middle School AMC 8 – amc.maa.org MATHCOUNTS – mathcounts.org MOEMS – moems.org

David Patrick (AoPS) Primes March 2017 8 / 37 Contests

Middle School AMC 8 – amc.maa.org MATHCOUNTS – mathcounts.org MOEMS – moems.org

High School AMC 10/12 – amc.maa.org Who Wants To Be A Mathematician? – ams.org/wwtbam ARML (at Las Vegas) – arml.com Purple Comet – purplecomet.org Mandelbrot Competition – mandelbrot.org Planning to restart in 2017-18?

David Patrick (AoPS) Primes March 2017 8 / 37 Art of Problem Solving Contact

Our website is www.aops.com (Grades 2-5) Beast Academy: www.beastacademy.com

facebook.com/ArtofProblemSolving facebook.com/BeastAcademyMath Follow us on Twitter (@AoPSNews)

General inquiries: [email protected]

Contact me personally: [email protected]

David Patrick (AoPS) Primes March 2017 9 / 37 Prime Numbers

Primes are fundamental to the universe.

But a lot about primes is still very mysterious! David Patrick (AoPS) Primes March 2017 10 / 37 Prime Number Chart

Prime Numbers to 2500 !2,5

0 1 5 6 10 11 15 16 20 21 H 3 2 4 8 7 9 13 12 14 18 17 19 23 22 24 u 1 3 7 9 1 3 7 9 1 3 7 9 1 3 7 9 1 3 7 9 t ! " 0 1 2 3 4 5 6 7 8 9

David Patrick (AoPS) Primes March 2017 11 / 37 Inﬁnitely Many!

Suppose there were only ﬁnitely many:

p1, p2, p3,..., pN.

What can we say about the number

(p1p2p3 pN) + 1? ··· It cannot be the multiple of any prime. So it has no prime factors. But that’s impossible!

Questions about Primes

How many primes are there?

David Patrick (AoPS) Primes March 2017 12 / 37 Suppose there were only ﬁnitely many:

p1, p2, p3,..., pN.

What can we say about the number

(p1p2p3 pN) + 1? ··· It cannot be the multiple of any prime. So it has no prime factors. But that’s impossible!

Questions about Primes

How many primes are there?

Inﬁnitely Many!

David Patrick (AoPS) Primes March 2017 12 / 37 What can we say about the number

(p1p2p3 pN) + 1? ··· It cannot be the multiple of any prime. So it has no prime factors. But that’s impossible!

Questions about Primes

How many primes are there?

Inﬁnitely Many!

Suppose there were only ﬁnitely many:

p1, p2, p3,..., pN.

David Patrick (AoPS) Primes March 2017 12 / 37 It cannot be the multiple of any prime. So it has no prime factors. But that’s impossible!

Questions about Primes

How many primes are there?

Inﬁnitely Many!

Suppose there were only ﬁnitely many:

p1, p2, p3,..., pN.

What can we say about the number

(p1p2p3 pN) + 1? ···

David Patrick (AoPS) Primes March 2017 12 / 37 Questions about Primes

How many primes are there?

Inﬁnitely Many!

Suppose there were only ﬁnitely many:

p1, p2, p3,..., pN.

What can we say about the number

(p1p2p3 pN) + 1? ··· It cannot be the multiple of any prime. So it has no prime factors. But that’s impossible!

David Patrick (AoPS) Primes March 2017 12 / 37 Meaning:

About how many of the ﬁrst N positive integers are prime?

Some data: N number of primes less than N 10 4 (40%) 100 25 (25%) 1,000 168 (16.8%) 1,000,000 78,498 (7.85%) 1,000,000,000,000 37,607,912,018 (3.76%)

They seem to be getting less frequent!

Questions about Primes

How dense are the primes?

David Patrick (AoPS) Primes March 2017 13 / 37 Some data: N number of primes less than N 10 4 (40%) 100 25 (25%) 1,000 168 (16.8%) 1,000,000 78,498 (7.85%) 1,000,000,000,000 37,607,912,018 (3.76%)

They seem to be getting less frequent!

Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

David Patrick (AoPS) Primes March 2017 13 / 37 1,000,000 78,498 (7.85%) 1,000,000,000,000 37,607,912,018 (3.76%)

They seem to be getting less frequent!

Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

Some data: N number of primes less than N 10 4 (40%) 100 25 (25%) 1,000 168 (16.8%)

David Patrick (AoPS) Primes March 2017 13 / 37 78,498 (7.85%) 1,000,000,000,000 37,607,912,018 (3.76%)

They seem to be getting less frequent!

Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

Some data: N number of primes less than N 10 4 (40%) 100 25 (25%) 1,000 168 (16.8%) 1,000,000

David Patrick (AoPS) Primes March 2017 13 / 37 1,000,000,000,000 37,607,912,018 (3.76%)

They seem to be getting less frequent!

Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

Some data: N number of primes less than N 10 4 (40%) 100 25 (25%) 1,000 168 (16.8%) 1,000,000 78,498 (7.85%)

David Patrick (AoPS) Primes March 2017 13 / 37 37,607,912,018 (3.76%)

They seem to be getting less frequent!

Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

Some data: N number of primes less than N 10 4 (40%) 100 25 (25%) 1,000 168 (16.8%) 1,000,000 78,498 (7.85%) 1,000,000,000,000

David Patrick (AoPS) Primes March 2017 13 / 37 They seem to be getting less frequent!

Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

Some data: N number of primes less than N 10 4 (40%) 100 25 (25%) 1,000 168 (16.8%) 1,000,000 78,498 (7.85%) 1,000,000,000,000 37,607,912,018 (3.76%)

David Patrick (AoPS) Primes March 2017 13 / 37 Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

Some data: N number of primes less than N 10 4 (40%) 100 25 (25%) 1,000 168 (16.8%) 1,000,000 78,498 (7.85%) 1,000,000,000,000 37,607,912,018 (3.76%)

They seem to be getting less frequent!

David Patrick (AoPS) Primes March 2017 13 / 37 Example: if N = 1,000,000,000,000: Actual number of primes: 37,607,912,018 Predicted number of primes: 36,191,206,825 The prediction gets more and more accurate as N gets bigger.

Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

Prime Number Theorem N Approximately of the ﬁrst N positive integers are prime. ln N

David Patrick (AoPS) Primes March 2017 14 / 37 Questions about Primes

How dense are the primes?

Meaning:

About how many of the ﬁrst N positive integers are prime?

Prime Number Theorem N Approximately of the ﬁrst N positive integers are prime. ln N

Example: if N = 1,000,000,000,000: Actual number of primes: 37,607,912,018 Predicted number of primes: 36,191,206,825 The prediction gets more and more accurate as N gets bigger.

David Patrick (AoPS) Primes March 2017 14 / 37 Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

Hint: If two primes are at least N apart, then there has to be at least N 1 composite numbers in a row. How− can we construct N 1 composite numbers in a row? What numbers have lots of− factors? Factorials!!

Questions about Primes

Can primes be arbitrarily far apart?

David Patrick (AoPS) Primes March 2017 15 / 37 Hint: If two primes are at least N apart, then there has to be at least N 1 composite numbers in a row. How− can we construct N 1 composite numbers in a row? What numbers have lots of− factors? Factorials!!

Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

David Patrick (AoPS) Primes March 2017 15 / 37 How can we construct N 1 composite numbers in a row? What numbers have lots of− factors? Factorials!!

Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

Hint: If two primes are at least N apart, then there has to be at least N 1 composite numbers in a row. −

David Patrick (AoPS) Primes March 2017 15 / 37 What numbers have lots of factors? Factorials!!

Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

Hint: If two primes are at least N apart, then there has to be at least N 1 composite numbers in a row. How− can we construct N 1 composite numbers in a row? −

David Patrick (AoPS) Primes March 2017 15 / 37 Factorials!!

Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

Hint: If two primes are at least N apart, then there has to be at least N 1 composite numbers in a row. How− can we construct N 1 composite numbers in a row? What numbers have lots of− factors?

David Patrick (AoPS) Primes March 2017 15 / 37 Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

David Patrick (AoPS) Primes March 2017 15 / 37 N! + 3 is a multiple of 3 N! + 4 is a multiple of 4 . . N! + N is a multiple of N So that’s N 1 composite numbers in a row. Therefore, the nearest primes on either− side are at least N apart.

Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

N! + 2 is a multiple of 2

David Patrick (AoPS) Primes March 2017 16 / 37 N! + 4 is a multiple of 4 . . N! + N is a multiple of N So that’s N 1 composite numbers in a row. Therefore, the nearest primes on either− side are at least N apart.

Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

N! + 2 is a multiple of 2 N! + 3 is a multiple of 3

David Patrick (AoPS) Primes March 2017 16 / 37 . . N! + N is a multiple of N So that’s N 1 composite numbers in a row. Therefore, the nearest primes on either− side are at least N apart.

Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

N! + 2 is a multiple of 2 N! + 3 is a multiple of 3 N! + 4 is a multiple of 4

David Patrick (AoPS) Primes March 2017 16 / 37 So that’s N 1 composite numbers in a row. Therefore, the nearest primes on either− side are at least N apart.

Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

N! + 2 is a multiple of 2 N! + 3 is a multiple of 3 N! + 4 is a multiple of 4 . . N! + N is a multiple of N

David Patrick (AoPS) Primes March 2017 16 / 37 Questions about Primes

Can primes be arbitrarily far apart?

Meaning:

If I give you a positive integer N, are there two consecutive primes p and q such that q p + N? (In other words, there’s a “gap” of at least N between two consecutive≥ primes.)

N! + 2 is a multiple of 2 N! + 3 is a multiple of 3 N! + 4 is a multiple of 4 . . N! + N is a multiple of N So that’s N 1 composite numbers in a row. Therefore, the nearest primes on either− side are at least N apart.

David Patrick (AoPS) Primes March 2017 16 / 37 Meaning:

Are there three consecutive odd numbers, all of which are prime?

3,5,7 But that’s it. Why? If p > 3 is prime, then one of p + 2 or p + 4 is a multiple of 3.

Questions about Primes

Can we have three primes in a row?

David Patrick (AoPS) Primes March 2017 17 / 37 3,5,7 But that’s it. Why? If p > 3 is prime, then one of p + 2 or p + 4 is a multiple of 3.

Questions about Primes

Can we have three primes in a row?

Meaning:

Are there three consecutive odd numbers, all of which are prime?

David Patrick (AoPS) Primes March 2017 17 / 37 But that’s it. Why? If p > 3 is prime, then one of p + 2 or p + 4 is a multiple of 3.

Questions about Primes

Can we have three primes in a row?

Meaning:

Are there three consecutive odd numbers, all of which are prime?

3,5,7

David Patrick (AoPS) Primes March 2017 17 / 37 If p > 3 is prime, then one of p + 2 or p + 4 is a multiple of 3.

Questions about Primes

Can we have three primes in a row?

Meaning:

Are there three consecutive odd numbers, all of which are prime?

3,5,7 But that’s it. Why?

David Patrick (AoPS) Primes March 2017 17 / 37 Questions about Primes

Can we have three primes in a row?

Meaning:

Are there three consecutive odd numbers, all of which are prime?

3,5,7 But that’s it. Why? If p > 3 is prime, then one of p + 2 or p + 4 is a multiple of 3.

David Patrick (AoPS) Primes March 2017 17 / 37 3, 5 5, 7 11, 13 17, 19 29, 31 41, 43 How often does this occur? Twin Prime Conjecture: this occurs infinitely often Largest known pair (2016): 2996863034895 21290000 1 · ± (2013) Yitang Zhang: there are infinitely many pairs of primes that differ by at most 70 million. (2014) Polymath: there are infinitely many pairs of primes that differ by at most 246.

Questions about Primes

Can we have two primes in a row?

Meaning:

Are there two consecutive odd numbers, both of which are prime?

David Patrick (AoPS) Primes March 2017 18 / 37 Twin Prime Conjecture: this occurs infinitely often Largest known pair (2016): 2996863034895 21290000 1 · ± (2013) Yitang Zhang: there are infinitely many pairs of primes that differ by at most 70 million. (2014) Polymath: there are infinitely many pairs of primes that differ by at most 246.

Questions about Primes

Can we have two primes in a row?

Meaning:

Are there two consecutive odd numbers, both of which are prime?

3, 5 5, 7 11, 13 17, 19 29, 31 41, 43 How often does this occur?

David Patrick (AoPS) Primes March 2017 18 / 37 Largest known pair (2016): 2996863034895 21290000 1 · ± (2013) Yitang Zhang: there are inﬁnitely many pairs of primes that differ by at most 70 million. (2014) Polymath: there are inﬁnitely many pairs of primes that differ by at most 246.

Questions about Primes

Can we have two primes in a row?

Meaning:

Are there two consecutive odd numbers, both of which are prime?

3, 5 5, 7 11, 13 17, 19 29, 31 41, 43 How often does this occur? Twin Prime Conjecture: this occurs inﬁnitely often

David Patrick (AoPS) Primes March 2017 18 / 37 (2013) Yitang Zhang: there are inﬁnitely many pairs of primes that differ by at most 70 million. (2014) Polymath: there are inﬁnitely many pairs of primes that differ by at most 246.

Questions about Primes

Can we have two primes in a row?

Meaning:

Are there two consecutive odd numbers, both of which are prime?

3, 5 5, 7 11, 13 17, 19 29, 31 41, 43 How often does this occur? Twin Prime Conjecture: this occurs inﬁnitely often Largest known pair (2016): 2996863034895 21290000 1 · ±

David Patrick (AoPS) Primes March 2017 18 / 37 (2014) Polymath: there are inﬁnitely many pairs of primes that differ by at most 246.

Questions about Primes

Can we have two primes in a row?

Meaning:

Are there two consecutive odd numbers, both of which are prime?

David Patrick (AoPS) Primes March 2017 18 / 37 Questions about Primes

Can we have two primes in a row?

Meaning:

Are there two consecutive odd numbers, both of which are prime?

3, 5 5, 7 11, 13 17, 19 29, 31 41, 43 How often does this occur? Twin Prime Conjecture: this occurs infinitely often Largest known pair (2016): 2996863034895 21290000 1 · ± (2013) Yitang Zhang: there are infinitely many pairs of primes that differ by at most 70 million. (2014) Polymath: there are infinitely many pairs of primes that differ by at most 246. David Patrick (AoPS) Primes March 2017 18 / 37 Some experimental data: N Prob that two pos ints N are relatively prime 10 0.63 ≤ 100 0.6087 1000 0.608383 10000 0.60794971 Turns out that as N , the probability is exactly: → ∞ 6 = 0.607927101854 .... π2 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

David Patrick (AoPS) Primes March 2017 19 / 37 N Prob that two pos ints N are relatively prime 10 0.63 ≤ 100 0.6087 1000 0.608383 10000 0.60794971 Turns out that as N , the probability is exactly: → ∞ 6 = 0.607927101854 .... π2 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Some experimental data:

David Patrick (AoPS) Primes March 2017 19 / 37 10 0.63 100 0.6087 1000 0.608383 10000 0.60794971 Turns out that as N , the probability is exactly: → ∞ 6 = 0.607927101854 .... π2 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Some experimental data: N Prob that two pos ints N are relatively prime ≤

David Patrick (AoPS) Primes March 2017 19 / 37 100 0.6087 1000 0.608383 10000 0.60794971 Turns out that as N , the probability is exactly: → ∞ 6 = 0.607927101854 .... π2 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Some experimental data: N Prob that two pos ints N are relatively prime 10 0.63 ≤

David Patrick (AoPS) Primes March 2017 19 / 37 1000 0.608383 10000 0.60794971 Turns out that as N , the probability is exactly: → ∞ 6 = 0.607927101854 .... π2 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Some experimental data: N Prob that two pos ints N are relatively prime 10 0.63 ≤ 100 0.6087

David Patrick (AoPS) Primes March 2017 19 / 37 10000 0.60794971 Turns out that as N , the probability is exactly: → ∞ 6 = 0.607927101854 .... π2 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Some experimental data: N Prob that two pos ints N are relatively prime 10 0.63 ≤ 100 0.6087 1000 0.608383

David Patrick (AoPS) Primes March 2017 19 / 37 Turns out that as N , the probability is exactly: → ∞ 6 = 0.607927101854 .... π2 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Some experimental data: N Prob that two pos ints N are relatively prime 10 0.63 ≤ 100 0.6087 1000 0.608383 10000 0.60794971

David Patrick (AoPS) Primes March 2017 19 / 37 6 = 0.607927101854 .... π2 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Some experimental data: N Prob that two pos ints N are relatively prime 10 0.63 ≤ 100 0.6087 1000 0.608383 10000 0.60794971 Turns out that as N , the probability is exactly: → ∞

David Patrick (AoPS) Primes March 2017 19 / 37 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Some experimental data: N Prob that two pos ints N are relatively prime 10 0.63 ≤ 100 0.6087 1000 0.608383 10000 0.60794971 Turns out that as N , the probability is exactly: → ∞ 6 = 0.607927101854 .... π2

David Patrick (AoPS) Primes March 2017 19 / 37 An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

David Patrick (AoPS) Primes March 2017 19 / 37 1 2 which happens with probability 1 2 . They’re both not multiples of 3. . . − 1 2 which happens with probability 1 3 . They’re both not multiples of 5. . . − 1 2 which happens with probability 1 5 . And so on. . . we have to avoid both numbers being the multiple− of the same prime.

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Two numbers are relatively prime if: They’re both not multiples of 2. . .

David Patrick (AoPS) Primes March 2017 20 / 37 They’re both not multiples of 3. . . 1 2 which happens with probability 1 3 . They’re both not multiples of 5. . . − 1 2 which happens with probability 1 5 . And so on. . . we have to avoid both numbers being the multiple− of the same prime.

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Two numbers are relatively prime if: They’re both not multiples of 2. . . 2 which happens with probability 1 1 . − 2

David Patrick (AoPS) Primes March 2017 20 / 37 1 2 which happens with probability 1 3 . They’re both not multiples of 5. . . − 1 2 which happens with probability 1 5 . And so on. . . we have to avoid both numbers being the multiple− of the same prime.

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Two numbers are relatively prime if: They’re both not multiples of 2. . . 1 2 which happens with probability 1 2 . They’re both not multiples of 3. . . −

David Patrick (AoPS) Primes March 2017 20 / 37 They’re both not multiples of 5. . . 1 2 which happens with probability 1 5 . And so on. . . we have to avoid both numbers being the multiple− of the same prime.

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Two numbers are relatively prime if: They’re both not multiples of 2. . . 1 2 which happens with probability 1 2 . They’re both not multiples of 3. . . − 2 which happens with probability 1 1 . − 3

David Patrick (AoPS) Primes March 2017 20 / 37 1 2 which happens with probability 1 5 . And so on. . . we have to avoid both numbers being the multiple− of the same prime.

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

David Patrick (AoPS) Primes March 2017 20 / 37 And so on. . . we have to avoid both numbers being the multiple of the same prime.

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Two numbers are relatively prime if: They’re both not multiples of 2. . . 1 2 which happens with probability 1 2 . They’re both not multiples of 3. . . − 1 2 which happens with probability 1 3 . They’re both not multiples of 5. . . − 2 which happens with probability 1 1 . − 5

David Patrick (AoPS) Primes March 2017 20 / 37 An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

David Patrick (AoPS) Primes March 2017 20 / 37 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

So the probability is:

1 1 1 1 1 1 1 1 − 22 − 32 − 52 − 72 ··· which can we written as: ! Y 1 1 . − p2 p prime

David Patrick (AoPS) Primes March 2017 21 / 37 An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

So the probability is:

1 1 1 1 1 1 1 1 − 22 − 32 − 52 − 72 ··· which can we written as: ! Y 1 1 . − p2 p prime

WHERE THE HECK DID THAT π COME FROM???

David Patrick (AoPS) Primes March 2017 21 / 37 So our probability is the reciprocal of

1 1 1 1 1 1 1 + + + 1 + + + 1 + + + . 22 24 ··· 32 34 ··· 52 54 ··· ···

WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Now we use the inﬁnite geometric series:

1 1 1 1 1 + + + + = . p2 p4 p6 ··· 1 1 − p2

David Patrick (AoPS) Primes March 2017 22 / 37 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Now we use the inﬁnite geometric series:

1 1 1 1 1 + + + + = . p2 p4 p6 ··· 1 1 − p2 So our probability is the reciprocal of

1 1 1 1 1 1 1 + + + 1 + + + 1 + + + . 22 24 ··· 32 34 ··· 52 54 ··· ···

David Patrick (AoPS) Primes March 2017 22 / 37 An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

Now we use the inﬁnite geometric series:

1 1 1 1 1 + + + + = . p2 p4 p6 ··· 1 1 − p2 So our probability is the reciprocal of

1 1 1 1 1 1 1 + + + 1 + + + 1 + + + . 22 24 ··· 32 34 ··· 52 54 ··· ···

WHERE THE HECK DID THAT π COME FROM???

David Patrick (AoPS) Primes March 2017 22 / 37 That is, our probability is the reciprocal of the sum of the reciprocals of all the squares. WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

So our probability is the reciprocal of

1 1 1 1 1 1 1 + + + 1 + + + 1 + + + 22 24 ··· 32 34 ··· 52 54 ··· ··· 1 1 1 1 1 = 1 + + + + + + . 22 32 42 52 62 ···

David Patrick (AoPS) Primes March 2017 23 / 37 WHERE THE HECK DID THAT π COME FROM???

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

So our probability is the reciprocal of

1 1 1 1 1 1 1 + + + 1 + + + 1 + + + 22 24 ··· 32 34 ··· 52 54 ··· ··· 1 1 1 1 1 = 1 + + + + + + . 22 32 42 52 62 ··· That is, our probability is the reciprocal of the sum of the reciprocals of all the squares.

David Patrick (AoPS) Primes March 2017 23 / 37 An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

So our probability is the reciprocal of

1 1 1 1 1 1 1 + + + 1 + + + 1 + + + 22 24 ··· 32 34 ··· 52 54 ··· ··· 1 1 1 1 1 = 1 + + + + + + . 22 32 42 52 62 ··· That is, our probability is the reciprocal of the sum of the reciprocals of all the squares. WHERE THE HECK DID THAT π COME FROM???

David Patrick (AoPS) Primes March 2017 23 / 37 He used techniques from calculus to show that

x3 x5 x7 sin(x) = x + + . − 3! 5! − 7! ··· But he also considered the function ! ! ! x2 x2 x2 f(x) = x 1 1 1 . − π2 − 22π2 − 32π2 ··· What do these functions have in common?

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

In 1735, Euler computed that

1 1 1 1 1 π2 1 + + + + + + = . 22 32 42 52 62 ··· 6

David Patrick (AoPS) Primes March 2017 24 / 37 But he also considered the function ! ! ! x2 x2 x2 f(x) = x 1 1 1 . − π2 − 22π2 − 32π2 ··· What do these functions have in common?

An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

In 1735, Euler computed that

1 1 1 1 1 π2 1 + + + + + + = . 22 32 42 52 62 ··· 6 He used techniques from calculus to show that

x3 x5 x7 sin(x) = x + + . − 3! 5! − 7! ···

David Patrick (AoPS) Primes March 2017 24 / 37 An Amazing Fact

What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)?

In 1735, Euler computed that

1 1 1 1 1 π2 1 + + + + + + = . 22 32 42 52 62 ··· 6 He used techniques from calculus to show that

x3 x5 x7 sin(x) = x + + . − 3! 5! − 7! ··· But he also considered the function ! ! ! x2 x2 x2 f(x) = x 1 1 1 . − π2 − 22π2 − 32π2 ··· What do these functions have in common? David Patrick (AoPS) Primes March 2017 24 / 37 Two polynomial functions with the same roots — they’re the same function! Now compare the x3 terms: 1 1 1 1 1 = + + + + . −6 − π2 22π2 32π2 42π2 ··· This simpliﬁes to Euler’s formula!

An Amazing Fact

x3 x5 x7 sin(x) = x + + , − 3! 5! − 7! ··· ! ! ! x2 x2 x2 f(x) = x 1 1 1 . − π2 − 22π2 − 32π2 ···

These functions have the same roots: all integer multiples of π make both of these functions 0, and no other values do.

David Patrick (AoPS) Primes March 2017 25 / 37 Now compare the x3 terms: 1 1 1 1 1 = + + + + . −6 − π2 22π2 32π2 42π2 ··· This simpliﬁes to Euler’s formula!

An Amazing Fact

x3 x5 x7 sin(x) = x + + , − 3! 5! − 7! ··· ! ! ! x2 x2 x2 f(x) = x 1 1 1 . − π2 − 22π2 − 32π2 ···

These functions have the same roots: all integer multiples of π make both of these functions 0, and no other values do. Two polynomial functions with the same roots — they’re the same function!

David Patrick (AoPS) Primes March 2017 25 / 37 An Amazing Fact

x3 x5 x7 sin(x) = x + + , − 3! 5! − 7! ··· ! ! ! x2 x2 x2 f(x) = x 1 1 1 . − π2 − 22π2 − 32π2 ···

These functions have the same roots: all integer multiples of π make both of these functions 0, and no other values do. Two polynomial functions with the same roots — they’re the same function! Now compare the x3 terms: 1 1 1 1 1 = + + + + . −6 − π2 22π2 32π2 42π2 ··· This simpliﬁes to Euler’s formula!

David Patrick (AoPS) Primes March 2017 25 / 37 This is also equivalent to the fact that

π2 ζ(2) = , 6 where ζ is the Riemann Zeta Function, the key object of study in the famous Riemann Hypothesis. But that’s a topic for another day!

An Amazing Fact

And thus. . . What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)? 6 The answer is . π2

David Patrick (AoPS) Primes March 2017 26 / 37 But that’s a topic for another day!

An Amazing Fact

And thus. . . What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)? 6 The answer is . π2 This is also equivalent to the fact that

π2 ζ(2) = , 6 where ζ is the Riemann Zeta Function, the key object of study in the famous Riemann Hypothesis.

David Patrick (AoPS) Primes March 2017 26 / 37 An Amazing Fact

And thus. . . What is the probability that two randomly chosen positive integers are relatively prime (have no common factors)? 6 The answer is . π2 This is also equivalent to the fact that

π2 ζ(2) = , 6 where ζ is the Riemann Zeta Function, the key object of study in the famous Riemann Hypothesis. But that’s a topic for another day!

David Patrick (AoPS) Primes March 2017 26 / 37 We’re only interested in minimal solutions in which a, b, c have no common prime factors. (In other words, divide out by as much as you can ﬁrst.)

This still seems profoundly unexciting.

abc Conjecture

The abc Conjecture is a statement about positive integer solutions to the highly complicated equation

a + b = c.

David Patrick (AoPS) Primes March 2017 27 / 37 This still seems profoundly unexciting.

abc Conjecture

The abc Conjecture is a statement about positive integer solutions to the highly complicated equation

a + b = c.

We’re only interested in minimal solutions in which a, b, c have no common prime factors. (In other words, divide out by as much as you can ﬁrst.)

David Patrick (AoPS) Primes March 2017 27 / 37 abc Conjecture

The abc Conjecture is a statement about positive integer solutions to the highly complicated equation

a + b = c.

We’re only interested in minimal solutions in which a, b, c have no common prime factors. (In other words, divide out by as much as you can ﬁrst.)

This still seems profoundly unexciting.

David Patrick (AoPS) Primes March 2017 27 / 37 Example: a = 5, b = 7, c = 12 rad(abc) = 5 7 2 3 = 210 · · · Example: a = 8, b = 9, c = 17 rad(abc) = 2 3 17 = 102 · · Example: a = 1, b = 4, c = 5 rad(abc) = 2 5 = 10 · Notice that in all these examples, c < rad(abc).

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

We’re interested in rad(abc) for solutions to a + b = c.

David Patrick (AoPS) Primes March 2017 28 / 37 rad(abc) = 5 7 2 3 = 210 · · · Example: a = 8, b = 9, c = 17 rad(abc) = 2 3 17 = 102 · · Example: a = 1, b = 4, c = 5 rad(abc) = 2 5 = 10 · Notice that in all these examples, c < rad(abc).

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

We’re interested in rad(abc) for solutions to a + b = c. Example: a = 5, b = 7, c = 12

David Patrick (AoPS) Primes March 2017 28 / 37 Example: a = 8, b = 9, c = 17 rad(abc) = 2 3 17 = 102 · · Example: a = 1, b = 4, c = 5 rad(abc) = 2 5 = 10 · Notice that in all these examples, c < rad(abc).

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

We’re interested in rad(abc) for solutions to a + b = c. Example: a = 5, b = 7, c = 12 rad(abc) = 5 7 2 3 = 210 · · ·

David Patrick (AoPS) Primes March 2017 28 / 37 rad(abc) = 2 3 17 = 102 · · Example: a = 1, b = 4, c = 5 rad(abc) = 2 5 = 10 · Notice that in all these examples, c < rad(abc).

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

We’re interested in rad(abc) for solutions to a + b = c. Example: a = 5, b = 7, c = 12 rad(abc) = 5 7 2 3 = 210 · · · Example: a = 8, b = 9, c = 17

David Patrick (AoPS) Primes March 2017 28 / 37 Example: a = 1, b = 4, c = 5 rad(abc) = 2 5 = 10 · Notice that in all these examples, c < rad(abc).

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

We’re interested in rad(abc) for solutions to a + b = c. Example: a = 5, b = 7, c = 12 rad(abc) = 5 7 2 3 = 210 · · · Example: a = 8, b = 9, c = 17 rad(abc) = 2 3 17 = 102 · ·

David Patrick (AoPS) Primes March 2017 28 / 37 rad(abc) = 2 5 = 10 · Notice that in all these examples, c < rad(abc).

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

We’re interested in rad(abc) for solutions to a + b = c. Example: a = 5, b = 7, c = 12 rad(abc) = 5 7 2 3 = 210 · · · Example: a = 8, b = 9, c = 17 rad(abc) = 2 3 17 = 102 · · Example: a = 1, b = 4, c = 5

David Patrick (AoPS) Primes March 2017 28 / 37 Notice that in all these examples, c < rad(abc).

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

We’re interested in rad(abc) for solutions to a + b = c. Example: a = 5, b = 7, c = 12 rad(abc) = 5 7 2 3 = 210 · · · Example: a = 8, b = 9, c = 17 rad(abc) = 2 3 17 = 102 · · Example: a = 1, b = 4, c = 5 rad(abc) = 2 5 = 10 ·

David Patrick (AoPS) Primes March 2017 28 / 37 abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

David Patrick (AoPS) Primes March 2017 28 / 37 How about c = rad(abc)? Only 1 + 1 = 2. (There are no primes left for a or b if c = rad(abc).)

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

So here’s the game:

Are there solutions in relatively prime positive integers to

a + b = c

for which c rad(abc)? ≥

David Patrick (AoPS) Primes March 2017 29 / 37 Only 1 + 1 = 2. (There are no primes left for a or b if c = rad(abc).)

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

So here’s the game:

Are there solutions in relatively prime positive integers to

a + b = c

for which c rad(abc)? ≥ How about c = rad(abc)?

David Patrick (AoPS) Primes March 2017 29 / 37 abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

So here’s the game:

Are there solutions in relatively prime positive integers to

a + b = c

for which c rad(abc)? ≥ How about c = rad(abc)? Only 1 + 1 = 2. (There are no primes left for a or b if c = rad(abc).)

David Patrick (AoPS) Primes March 2017 29 / 37 1 + 8 = 9 is the smallest rad(72) = 2 3 = 6. · 5 + 27 = 32 is the smallest with all numbers greater than 1 rad(5 27 32) = 5 3 2 = 30. · · · ·

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

Are there solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

David Patrick (AoPS) Primes March 2017 30 / 37 5 + 27 = 32 is the smallest with all numbers greater than 1 rad(5 27 32) = 5 3 2 = 30. · · · ·

abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

Are there solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

1 + 8 = 9 is the smallest rad(72) = 2 3 = 6. ·

David Patrick (AoPS) Primes March 2017 30 / 37 abc Conjecture

Deﬁnition The radical of a number n, denoted rad(n), is the product of all the prime factors of n.

Are there solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

1 + 8 = 9 is the smallest rad(72) = 2 3 = 6. · 5 + 27 = 32 is the smallest with all numbers greater than 1 rad(5 27 32) = 5 3 2 = 30. · · · ·

David Patrick (AoPS) Primes March 2017 30 / 37 32 + 49 = 81 (radical is 42)

Are there inﬁnitely many solutions?

Computer search has found over 23 million solutions!

abc Conjecture

Are there solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

Four others with c < 100: 1 + 48 = 49 (radical is 42) 1 + 63 = 64 (radical is 42) 1 + 80 = 81 (radical is 30)

David Patrick (AoPS) Primes March 2017 31 / 37 Are there inﬁnitely many solutions?

Computer search has found over 23 million solutions!

abc Conjecture

Are there solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

Four others with c < 100: 1 + 48 = 49 (radical is 42) 1 + 63 = 64 (radical is 42) 1 + 80 = 81 (radical is 30) 32 + 49 = 81 (radical is 42)

David Patrick (AoPS) Primes March 2017 31 / 37 Computer search has found over 23 million solutions!

abc Conjecture

Are there solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

Four others with c < 100: 1 + 48 = 49 (radical is 42) 1 + 63 = 64 (radical is 42) 1 + 80 = 81 (radical is 30) 32 + 49 = 81 (radical is 42)

Are there inﬁnitely many solutions?

David Patrick (AoPS) Primes March 2017 31 / 37 abc Conjecture

Are there solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

Four others with c < 100: 1 + 48 = 49 (radical is 42) 1 + 63 = 64 (radical is 42) 1 + 80 = 81 (radical is 30) 32 + 49 = 81 (radical is 42)

Are there inﬁnitely many solutions?

Computer search has found over 23 million solutions!

David Patrick (AoPS) Primes March 2017 31 / 37 No: there are inﬁnitely many solutions.

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

David Patrick (AoPS) Primes March 2017 32 / 37 1 + (26n 1) = 26n for n 1. − ≥ Let b = 26n 1 = 64n 1 and notice that b is a multiple of 9. − b− This means rad(b) 3 . 2b ≤ So rad(abc) 3 < b < b + 1 = c. Examples: ≤ 1 + 63 = 64, rad(63 64) = 42 · 1 + 4095 = 4096, rad(4095 4096) = 3 5 7 13 2 = 2730 · · · · ·

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

No: there are inﬁnitely many solutions.

David Patrick (AoPS) Primes March 2017 32 / 37 Let b = 26n 1 = 64n 1 and notice that b is a multiple of 9. − b− This means rad(b) 3 . 2b ≤ So rad(abc) 3 < b < b + 1 = c. Examples: ≤ 1 + 63 = 64, rad(63 64) = 42 · 1 + 4095 = 4096, rad(4095 4096) = 3 5 7 13 2 = 2730 · · · · ·

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

No: there are inﬁnitely many solutions.

1 + (26n 1) = 26n for n 1. − ≥

David Patrick (AoPS) Primes March 2017 32 / 37 b This means rad(b) 3 . 2b ≤ So rad(abc) 3 < b < b + 1 = c. Examples: ≤ 1 + 63 = 64, rad(63 64) = 42 · 1 + 4095 = 4096, rad(4095 4096) = 3 5 7 13 2 = 2730 · · · · ·

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

No: there are inﬁnitely many solutions.

1 + (26n 1) = 26n for n 1. − ≥ Let b = 26n 1 = 64n 1 and notice that b is a multiple of 9. − −

David Patrick (AoPS) Primes March 2017 32 / 37 2b So rad(abc) 3 < b < b + 1 = c. Examples: ≤ 1 + 63 = 64, rad(63 64) = 42 · 1 + 4095 = 4096, rad(4095 4096) = 3 5 7 13 2 = 2730 · · · · ·

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

No: there are inﬁnitely many solutions.

1 + (26n 1) = 26n for n 1. − ≥ Let b = 26n 1 = 64n 1 and notice that b is a multiple of 9. This means− rad(b) b−. ≤ 3

David Patrick (AoPS) Primes March 2017 32 / 37 Examples: 1 + 63 = 64, rad(63 64) = 42 · 1 + 4095 = 4096, rad(4095 4096) = 3 5 7 13 2 = 2730 · · · · ·

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

No: there are inﬁnitely many solutions.

1 + (26n 1) = 26n for n 1. − ≥ Let b = 26n 1 = 64n 1 and notice that b is a multiple of 9. This means− rad(b) b−. ≤ 3 So rad(abc) 2b < b < b + 1 = c. ≤ 3

David Patrick (AoPS) Primes March 2017 32 / 37 abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)?

No: there are inﬁnitely many solutions.

1 + (26n 1) = 26n for n 1. − ≥ Let b = 26n 1 = 64n 1 and notice that b is a multiple of 9. − b− This means rad(b) 3 . 2b ≤ So rad(abc) 3 < b < b + 1 = c. Examples: ≤ 1 + 63 = 64, rad(63 64) = 42 · 1 + 4095 = 4096, rad(4095 4096) = 3 5 7 13 2 = 2730 · · · · · David Patrick (AoPS) Primes March 2017 32 / 37 Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

The value q such that c = rad(abc)q is called the quality of the triple (a, b, c). If you know logarithms, this is just log c q = log c = . rad(abc) log rad(abc)

So the question is: if we ﬁx the quality Q that we want, are there ﬁnitely many solutions that have at least that quality (that is, that have q > Q)?

abc Conjecture

So we modify the question a little. . .

David Patrick (AoPS) Primes March 2017 33 / 37 The value q such that c = rad(abc)q is called the quality of the triple (a, b, c). If you know logarithms, this is just log c q = log c = . rad(abc) log rad(abc)

So the question is: if we ﬁx the quality Q that we want, are there ﬁnitely many solutions that have at least that quality (that is, that have q > Q)?

abc Conjecture

So we modify the question a little. . .

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

David Patrick (AoPS) Primes March 2017 33 / 37 abc Conjecture

So we modify the question a little. . .

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

The value q such that c = rad(abc)q is called the quality of the triple (a, b, c). If you know logarithms, this is just log c q = log c = . rad(abc) log rad(abc)

So the question is: if we ﬁx the quality Q that we want, are there ﬁnitely many solutions that have at least that quality (that is, that have q > Q)?

David Patrick (AoPS) Primes March 2017 33 / 37 The conjecture is YES: if we ﬁx a value of Q, then there are only ﬁnitely many solutions with quality q > Q. . . . but it’s unknown whether this is true or not!

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

David Patrick (AoPS) Primes March 2017 34 / 37 . . . but it’s unknown whether this is true or not!

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

The conjecture is YES: if we ﬁx a value of Q, then there are only ﬁnitely many solutions with quality q > Q.

David Patrick (AoPS) Primes March 2017 34 / 37 abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

The conjecture is YES: if we ﬁx a value of Q, then there are only ﬁnitely many solutions with quality q > Q. . . . but it’s unknown whether this is true or not!

David Patrick (AoPS) Primes March 2017 34 / 37 2 + 310 109 = 235 ·

rad = 2 3 23 109 = 15042 · · ·

150421.62991168... = 6436343

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

Highest known quality:

2 + 6436341 = 6436343

David Patrick (AoPS) Primes March 2017 35 / 37 rad = 2 3 23 109 = 15042 · · ·

150421.62991168... = 6436343

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

Highest known quality:

2 + 6436341 = 6436343

2 + 310 109 = 235 ·

David Patrick (AoPS) Primes March 2017 35 / 37 150421.62991168... = 6436343

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

Highest known quality:

2 + 6436341 = 6436343

2 + 310 109 = 235 ·

rad = 2 3 23 109 = 15042 · · ·

David Patrick (AoPS) Primes March 2017 35 / 37 abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

Highest known quality:

2 + 6436341 = 6436343

2 + 310 109 = 235 ·

rad = 2 3 23 109 = 15042 · · ·

150421.62991168... = 6436343

David Patrick (AoPS) Primes March 2017 35 / 37 The right-hand side of this

238841709663649705652770167283,

a 30-digit number.

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

There are 239 known triples with quality q 1.4. The largest one is: ≥ 23731291093 + 513131529391 = 7231117933458711.

David Patrick (AoPS) Primes March 2017 36 / 37 abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

There are 239 known triples with quality q 1.4. The largest one is: ≥ 23731291093 + 513131529391 = 7231117933458711.

The right-hand side of this

238841709663649705652770167283,

a 30-digit number.

David Patrick (AoPS) Primes March 2017 36 / 37 . . . but nobody understands the proof yet! 9/17/12 New York Times At ﬁrst glance, it feels like you’re reading something from outer space. – Jordan Ellenberg, math professor at Univ. of Wisconsin

The mathematical community is divided as to whether the proof is correct or not. There have been a series of conferences in Kyoto, Japan to discuss it.

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

In August, 2012, the Japanese mathematician Shinichi Mochizuki published on his website a 500-page series of papers that he claimed proved the abc conjecture.

David Patrick (AoPS) Primes March 2017 37 / 37 The mathematical community is divided as to whether the proof is correct or not. There have been a series of conferences in Kyoto, Japan to discuss it.

abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

In August, 2012, the Japanese mathematician Shinichi Mochizuki published on his website a 500-page series of papers that he claimed proved the abc conjecture. . . . but nobody understands the proof yet! 9/17/12 New York Times At ﬁrst glance, it feels like you’re reading something from outer space. – Jordan Ellenberg, math professor at Univ. of Wisconsin

David Patrick (AoPS) Primes March 2017 37 / 37 abc Conjecture

Are there ﬁnitely many solutions in relatively prime positive integers to

a + b = c

for which c > rad(abc)Q for a ﬁxed Q > 1?

The mathematical community is divided as to whether the proof is correct or not. There have been a series of conferences in Kyoto, Japan to discuss it.

David Patrick (AoPS) Primes March 2017 37 / 37