David Patrick Art of Problem Solving aops.com March 2017€¦ · Some Philosophies There’s more...

You can’t spell “primes” without “pi”

David PatrickArt of Problem Solving

aops.com

March 2017

David Patrick (AoPS) Primes March 2017 1 / 37

Art of Problem SolvingHistory

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

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

Over 260,000 members in our online communityOver 5,300,000 messages postedOver 3,400,000 unique visitors in the past yearAccredited by WASC as a supplementary education program


Art of Problem SolvingHistory

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

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

Over 260,000 members in our online communityOver 5,300,000 messages postedOver 3,400,000 unique visitors in the past yearAccredited by WASC as a supplementary education program


Standards for Mathematical PracticeCommon Core State Standards for Mathematics

1 Make sense of problems and persevere in solvingthem

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reasoningDavid Patrick (AoPS) Primes March 2017 3 / 37

Some Philosophies

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

Problem-solving perspective: new concepts introduced viachallenging problems, not as unmotivated facts and tools.Problem-solving skills are explicitly taught.Being able to solve every problem means that the problems aretoo 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 feedoff of each other. They learn from each other. They challenge andinspire each other.Many students thrive on competition...but not all competitions andnot all students.


Some Philosophies

There’s more to life than standardized tests or just racing tocalculus.Problem-solving perspective: new concepts introduced viachallenging problems, not as unmotivated facts and tools.Problem-solving skills are explicitly taught.

Being able to solve every problem means that the problems aretoo 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 feedoff of each other. They learn from each other. They challenge andinspire each other.Many students thrive on competition...but not all competitions andnot all students.


Some Philosophies

There’s more to life than standardized tests or just racing tocalculus.Problem-solving perspective: new concepts introduced viachallenging problems, not as unmotivated facts and tools.Problem-solving skills are explicitly taught.Being able to solve every problem means that the problems aretoo 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 feedoff of each other. They learn from each other. They challenge andinspire each other.Many students thrive on competition...but not all competitions andnot all students.


Some Philosophies

There’s more to life than standardized tests or just racing tocalculus.Problem-solving perspective: new concepts introduced viachallenging problems, not as unmotivated facts and tools.Problem-solving skills are explicitly taught.Being able to solve every problem means that the problems aretoo 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 feedoff of each other. They learn from each other. They challenge andinspire each other.

Many students thrive on competition...but not all competitions andnot all students.


Some Philosophies

There’s more to life than standardized tests or just racing tocalculus.Problem-solving perspective: new concepts introduced viachallenging problems, not as unmotivated facts and tools.Problem-solving skills are explicitly taught.Being able to solve every problem means that the problems aretoo 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 feedoff of each other. They learn from each other. They challenge andinspire each other.Many students thrive on competition.

..but not all competitions andnot all students.


Some Philosophies

There’s more to life than standardized tests or just racing tocalculus.Problem-solving perspective: new concepts introduced viachallenging problems, not as unmotivated facts and tools.Problem-solving skills are explicitly taught.Being able to solve every problem means that the problems aretoo 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 feedoff of each other. They learn from each other. They challenge andinspire each other.Many students thrive on competition...but not all competitions andnot all students.


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Contests

Middle SchoolAMC 8 – amc.maa.orgMATHCOUNTS – mathcounts.orgMOEMS – moems.org

High SchoolAMC 10/12 – amc.maa.orgWho Wants To Be A Mathematician? – ams.org/wwtbamARML (at Las Vegas) – arml.comPurple Comet – purplecomet.orgMandelbrot Competition – mandelbrot.orgPlanning to restart in 2017-18?


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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"0123456789

t uH 0 123 4

1 3 7 9

5 678 91 3 7 9

10 111213 141 3 7 9

15 161718 191 3 7 9

20 212223 241 3 7 9


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!




Infinitely Many!


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


(p1p2p3 · · · pN) + 1?





Infinitely Many!


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


(p1p2p3 · · · pN) + 1?





Infinitely Many!


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


(p1p2p3 · · · pN) + 1?





Infinitely Many!


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


(p1p2p3 · · · pN) + 1?




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 N10 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!




Meaning:


Some data:


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%)





Meaning:


Some data:


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%)





Meaning:


Some data:


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%)





Meaning:


Some data:


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%)





Meaning:


Some data:


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%)





Meaning:


Some data:


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%)





Meaning:


Some data:


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%)





Meaning:


Prime Number Theorem

ApproximatelyNlnN

of the first N positive integers are prime.

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




Meaning:


Prime Number Theorem

ApproximatelyNlnN

of the first N positive integers are prime.

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



Can primes be arbitrarily far apart?

Meaning:

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

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




Meaning:






Meaning:


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

How can we construct N − 1 composite numbers in a row?What numbers have lots of factors?Factorials!!




Meaning:


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

What numbers have lots of factors?Factorials!!




Meaning:


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

Factorials!!




Meaning:






Meaning:


N! + 2 is a multiple of 2

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




Meaning:


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

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




Meaning:


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

...N! + N is a multiple of NSo that’s N − 1 composite numbers in a row. Therefore, the nearestprimes on either side are at least N apart.




Meaning:


N! + 2 is a multiple of 2N! + 3 is a multiple of 3N! + 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 nearestprimes on either side are at least N apart.




Meaning:


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



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.




Meaning:


3,5,7






Meaning:


3,5,7






Meaning:


3,5,7






Meaning:


3,5,7





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 oftenLargest known pair (2016):

2996863034895 · 21290000± 1

(2013) Yitang Zhang: there are infinitely many pairs of primes thatdiffer by at most 70 million.(2014) Polymath: there are infinitely many pairs of primes thatdiffer by at most 246.




Meaning:


3,5 5,7 11,13 17,19 29,31 41,43

How often does this occur?

Twin Prime Conjecture: this occurs infinitely oftenLargest known pair (2016):

2996863034895 · 21290000± 1





Meaning:


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





Meaning:


3,5 5,7 11,13 17,19 29,31 41,43


2996863034895 · 21290000± 1





Meaning:


3,5 5,7 11,13 17,19 29,31 41,43


2996863034895 · 21290000± 1

(2013) Yitang Zhang: there are infinitely many pairs of primes thatdiffer by at most 70 million.

(2014) Polymath: there are infinitely many pairs of primes thatdiffer by at most 246.




Meaning:


3,5 5,7 11,13 17,19 29,31 41,43


2996863034895 · 21290000± 1



An Amazing Fact

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

Some experimental data:

N Prob that two pos ints ≤ N are relatively prime10 0.63

100 0.60871000 0.608383

10000 0.60794971

Turns out that as N →∞, the probability is exactly:

6π2 = 0.607927101854 . . . .

WHERE THE HECK DID THAT π COME FROM???


An Amazing Fact




100 0.60871000 0.608383

10000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact



N Prob that two pos ints ≤ N are relatively prime

10 0.63100 0.6087

1000 0.60838310000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact




100 0.60871000 0.608383

10000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact




100 0.6087

1000 0.60838310000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact




100 0.60871000 0.608383

10000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact




100 0.60871000 0.608383

10000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact




100 0.60871000 0.608383

10000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact




100 0.60871000 0.608383

10000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact




100 0.60871000 0.608383

10000 0.60794971


6π2 = 0.607927101854 . . . .



An Amazing Fact


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

which happens with probability 1 −(

12

)2.

They’re both not multiples of 3. . .


13

)2.



15

)2.

And so on. . . we have to avoid both numbers being the multiple of thesame prime.


An Amazing Fact




12

)2.



13

)2.



15

)2.



An Amazing Fact




12

)2.



13

)2.



15

)2.



An Amazing Fact




12

)2.



13

)2.



15

)2.



An Amazing Fact




12

)2.



13

)2.



15

)2.



An Amazing Fact




12

)2.



13

)2.



15

)2.



An Amazing Fact




12

)2.



13

)2.



15

)2.



An Amazing Fact


So the probability is:(1 −

122

) (1 −

132

) (1 −

152

) (1 −

172

)· · ·

which can we written as: ∏p prime

(1 −

1p2

).



An Amazing Fact


So the probability is:(1 −

122

) (1 −

132

) (1 −

152

) (1 −

172

)· · ·

which can we written as: ∏p prime

(1 −

1p2

).



An Amazing Fact


Now we use the infinite geometric series:

1 +1p2 +

1p4 +

1p6 + · · · =

11 − 1

p2

.

So our probability is the reciprocal of(1 +

122 +

124 + · · ·

) (1 +

132 +

134 + · · ·

) (1 +

152 +

154 + · · ·

)· · · .



An Amazing Fact



1 +1p2 +

1p4 +

1p6 + · · · =

11 − 1

p2

.


122 +

124 + · · ·

) (1 +

132 +

134 + · · ·

) (1 +

152 +

154 + · · ·

)· · · .



An Amazing Fact



1 +1p2 +

1p4 +

1p6 + · · · =

11 − 1

p2

.


122 +

124 + · · ·

) (1 +

132 +

134 + · · ·

) (1 +

152 +

154 + · · ·

)· · · .



An Amazing Fact



122 +

124 + · · ·

) (1 +

132 +

134 + · · ·

) (1 +

152 +

154 + · · ·

)· · ·

= 1 +122 +

132 +

142 +

152 +

162 + · · · .

That is, our probability is the reciprocal of the sum of the reciprocals ofall the squares.



An Amazing Fact



122 +

124 + · · ·

) (1 +

132 +

134 + · · ·

) (1 +

152 +

154 + · · ·

)· · ·

= 1 +122 +

132 +

142 +

152 +

162 + · · · .




An Amazing Fact



122 +

124 + · · ·

) (1 +

132 +

134 + · · ·

) (1 +

152 +

154 + · · ·

)· · ·

= 1 +122 +

132 +

142 +

152 +

162 + · · · .




An Amazing Fact


In 1735, Euler computed that

1 +122 +

132 +

142 +

152 +

162 + · · · =

π2

6.

He used techniques from calculus to show that

sin(x) = x −x3

3!+

x5

5!−

x7

7!+ · · · .

But he also considered the function

f(x) = x(1 −

x2

π2

) (1 −

x2

22π2

) (1 −

x2

32π2

)· · · .

What do these functions have in common?


An Amazing Fact



1 +122 +

132 +

142 +

152 +

162 + · · · =

π2

6.


sin(x) = x −x3

3!+

x5

5!−

x7

7!+ · · · .


f(x) = x(1 −

x2

π2

) (1 −

x2

22π2

) (1 −

x2

32π2

)· · · .

What do these functions have in common?


An Amazing Fact



1 +122 +

132 +

142 +

152 +

162 + · · · =

π2

6.


sin(x) = x −x3

3!+

x5

5!−

x7

7!+ · · · .


f(x) = x(1 −

x2

π2

) (1 −

x2

22π2

) (1 −

x2

32π2

)· · · .

What do these functions have in common?David Patrick (AoPS) Primes March 2017 24 / 37

An Amazing Fact

sin(x) = x −x3

3!+

x5

5!−

x7

7!+ · · · ,

f(x) = x(1 −

x2

π2

) (1 −

x2

22π2

) (1 −

x2

32π2

)· · · .

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

Two polynomial functions with the same roots — they’re the samefunction!

Now compare the x3 terms:

−16= −

( 1π2 +

122π2 +

132π2 +

142π2 + · · ·

).

This simplifies to Euler’s formula!


An Amazing Fact

sin(x) = x −x3

3!+

x5

5!−

x7

7!+ · · · ,

f(x) = x(1 −

x2

π2

) (1 −

x2

22π2

) (1 −

x2

32π2

)· · · .




−16= −

( 1π2 +

122π2 +

132π2 +

142π2 + · · ·

).

This simplifies to Euler’s formula!


An Amazing Fact

sin(x) = x −x3

3!+

x5

5!−

x7

7!+ · · · ,

f(x) = x(1 −

x2

π2

) (1 −

x2

22π2

) (1 −

x2

32π2

)· · · .




−16= −

( 1π2 +

122π2 +

132π2 +

142π2 + · · ·

).

This simplifies to Euler’s formula!David Patrick (AoPS) Primes March 2017 25 / 37

An Amazing Fact

And thus. . .


The answer is6π2 .

This is also equivalent to the fact that

ζ(2) =π2

6,

where ζ is the Riemann Zeta Function, the key object of study in thefamous Riemann Hypothesis.

But that’s a topic for another day!


An Amazing Fact

And thus. . .


The answer is6π2 .


ζ(2) =π2

6,




An Amazing Fact

And thus. . .


The answer is6π2 .


ζ(2) =π2

6,




abc Conjecture

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

a + b = c.

We’re only interested in minimal solutions in which a,b , c have nocommon prime factors. (In other words, divide out by as much as youcan first.)

This still seems profoundly unexciting.


abc Conjecture


a + b = c.




abc Conjecture


a + b = c.




abc Conjecture

DefinitionThe radical of a number n, denoted rad(n), is the product of all theprime factors of n.

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

Example: a = 5, b = 7, c = 12rad(abc) = 5 · 7 · 2 · 3 = 210

Example: a = 8, b = 9, c = 17rad(abc) = 2 · 3 · 17 = 102

Example: a = 1, b = 4, c = 5rad(abc) = 2 · 5 = 10

Notice that in all these examples, c < rad(abc).


abc Conjecture


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





abc Conjecture


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





abc Conjecture



Example: a = 8, b = 9, c = 17

rad(abc) = 2 · 3 · 17 = 102




abc Conjecture







abc Conjecture




Example: a = 1, b = 4, c = 5

rad(abc) = 2 · 5 = 10



abc Conjecture







abc Conjecture







abc Conjecture


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).)


abc Conjecture




a + b = c


How about c = rad(abc)?

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


abc Conjecture




a + b = c


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


abc Conjecture



a + b = c

for which c > rad(abc)?

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

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


abc Conjecture



a + b = c





abc Conjecture



a + b = c





abc Conjecture


a + b = c


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!


abc Conjecture


a + b = c


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)




abc Conjecture


a + b = c






abc Conjecture


a + b = c






abc Conjecture

Are there finitely many solutions in relatively prime positive integers to

a + b = c


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

3 < b < b + 1 = c.Examples:1 + 63 = 64, rad(63 · 64) = 421 + 4095 = 4096, rad(4095 · 4096) = 3 · 5 · 7 · 13 · 2 = 2730


abc Conjecture


a + b = c



1 + (26n− 1) = 26n for n ≥ 1.

Let b = 26n− 1 = 64n





abc Conjecture


a + b = c



1 + (26n− 1) = 26n for n ≥ 1.

Let b = 26n− 1 = 64n





abc Conjecture


a + b = c



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) ≤ b3 .

So rad(abc) ≤ 2b3 < b < b + 1 = c.

Examples:1 + 63 = 64, rad(63 · 64) = 421 + 4095 = 4096, rad(4095 · 4096) = 3 · 5 · 7 · 13 · 2 = 2730


abc Conjecture


a + b = c



1 + (26n− 1) = 26n for n ≥ 1.

Let b = 26n− 1 = 64n


3 .

So rad(abc) ≤ 2b3 < b < b + 1 = c.

Examples:1 + 63 = 64, rad(63 · 64) = 421 + 4095 = 4096, rad(4095 · 4096) = 3 · 5 · 7 · 13 · 2 = 2730


abc Conjecture


a + b = c



1 + (26n− 1) = 26n for n ≥ 1.

Let b = 26n− 1 = 64n



3 < b < b + 1 = c.

Examples:1 + 63 = 64, rad(63 · 64) = 421 + 4095 = 4096, rad(4095 · 4096) = 3 · 5 · 7 · 13 · 2 = 2730


abc Conjecture


a + b = c



1 + (26n− 1) = 26n for n ≥ 1.

Let b = 26n− 1 = 64n





abc Conjecture

So we modify the question a little. . .


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

q = lograd(abc) c =log c

log rad(abc).

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


abc Conjecture



a + b = c




log rad(abc).



abc Conjecture



a + b = c




log rad(abc).



abc Conjecture


a + b = c


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

. . . but it’s unknown whether this is true or not!


abc Conjecture


a + b = c





abc Conjecture


a + b = c





abc Conjecture


a + b = c


Highest known quality:

2 + 6436341 = 6436343

2 + 310· 109 = 235

rad = 2 · 3 · 23 · 109 = 15042

150421.62991168... = 6436343


abc Conjecture


a + b = c



2 + 6436341 = 6436343

2 + 310· 109 = 235

rad = 2 · 3 · 23 · 109 = 15042

150421.62991168... = 6436343


abc Conjecture


a + b = c



2 + 6436341 = 6436343

2 + 310· 109 = 235

rad = 2 · 3 · 23 · 109 = 15042

150421.62991168... = 6436343


abc Conjecture


a + b = c



2 + 6436341 = 6436343

2 + 310· 109 = 235

rad = 2 · 3 · 23 · 109 = 15042

150421.62991168... = 6436343


abc Conjecture


a + b = c


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.


abc Conjecture


a + b = c


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.


abc Conjecture


a + b = c


In August, 2012, the Japanese mathematician Shinichi Mochizukipublished on his website a 500-page series of papers that he claimedproved the abc conjecture.

. . . but nobody understands the proof yet!

9/17/12 New York TimesAt 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 iscorrect or not. There have been a series of conferences in Kyoto,Japan to discuss it.


abc Conjecture


a + b = c


In August, 2012, the Japanese mathematician Shinichi Mochizukipublished on his website a 500-page series of papers that he claimedproved the abc conjecture. . . . but nobody understands the proof yet!




abc Conjecture


a + b = c


In August, 2012, the Japanese mathematician Shinichi Mochizukipublished on his website a 500-page series of papers that he claimedproved the abc conjecture. . . . but nobody understands the proof yet!




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