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(2002年12月07日) RH = Riemann's Hypothesis (August 1859)
The real part of all nontrivial zeroes of Riemann's z function is ½.

I have temporarily put aside the search for a rigorous proof, after
some futile attempts, because it's not necessary for my immediate goal.
Bernhard Riemann, 1859.

As he solved the Basel problem in 1735, Leonhard Euler (1707-1783) introduced the following series, now universally called Zeta, focusing on integral values of s > 1 (for which the series converges). Leonhard Euler (1707-1783)

z(s) = 1 + 1 / 2^s + 1 / 3^s + 1 / 4^s + ... + 1 / n^s + ...

Using the fundamental theorem of arithmetic (which states that every positive integer has a unique factorization into primes) Euler observed (in 1737) that the term of rank n in the above series is obtained once and only once in the expansion of the product of all the following geometric series for any set of primes containing, at least, every prime divisor p of n.

1 + 1 / p^s + 1 / p^2s + 1 / p^3s + ... + 1 / p^ns + ... = 1 / ( 1 - 1 / p^s )

So, the sum of the whole series is equal to the product of all such things for all primes p. This yields the following celebrated Euler product formula :

z(s) = P _{p prime} ( 1 - p ^{- s} ) ^-1

Charles de la Vallee-Poussin (1866-1962) That formula characterizes the set of all prime numbers (it isn't true for any other set of integers). It's been the usual starting point of modern attacks on the set of primes, including the simultaneous proofs (1896) of the prime number theorem (PNT) by Jacques Hadamard (1865-1963) and Charles de la Vallée-Poussin (1866-1962).

Both the series and the infinite product converge for any complex number s in the half-plane where the real part of s is greater than 1 (Re(s) > 1).

Consider now the related Dirichlet eta function, defined by the following alternating series which converges on the right half-plane (Re(s) > 0).

h(s) = 1 - 1 / 2^s + 1 / 3^s - 1 / 4^s + ... + (-1)ⁿ⁺¹/ n^s + ...

At least when z(s) converges, that's equal to z(s) - 2 z(s) / 2^s . So:

z(s) = h(s) / (1 - 2^1-s )

Except for s = 1, this provides directly an analytic expression for z(s) in terms of an alternating series which converges when Re(s) > 0. Nice...

Dirichlet used the above to show that z has a simple pole of residue 1 at s = 1. (HINT: The numerator and the denominator are respectively asymptotic to Log 2 and (s-1) Log 2 as s goes to 1.)

In 1859, Bernhard Riemann (1826-1866) showed that the Zeta function can actually be extended to the entire complex plane, except at the pole s = 1, by analytic continuation (a concept invented by Weierstrass in 1842). establishing in the process a relation between values at (1-s) and s (which had been conjectured by Euler in 1749, in an equivalent form):

p^{-½ s} G( s ) z(s) = p^-½(1-s) G( 1-s ) z(1-s)

Vinculum Vinculum

2 2

Using the known properties of the Gamma function (whose reciprocal 1/G is an entire function with zeroes at all the nonpositive integers) this relation confirms the existence of a simple pole for the Zeta function at s = 1 and reveals trivial zeroes at negative even integers: -2, -4, -6...

Using the regularity of the Zeta function for Re(s) > 1 (due to the aforementioned convergence of the defining series in that domain) this same relation shows that nontrivial zeroes can only exist in the so-called critical strip (0 ≤ Re(s) ≤1). They could thus a priori be of two different types:

• Pairs of zeroes { s, s* } on the critical line (i.e., Re(s) = ½).
• Quadruplets of zeroes { s, s*, 1-s, 1-s* } off that critical line.

The famous Riemann Hypothesis (RH) is the conjecture, formulated by Bernhard Riemann in 1859, that there are no zeroes of the latter type:

Zeta shares its nontrivial zeros with the above convergent series: (Re(s)>0)

h(s) = 1 - 1 / 2^s + 1 / 3^s - 1 / 4^s + ... + (-1)ⁿ⁺¹/ n^s + ...

Listed below are the imaginary parts of the 29 smallest zeroes of the Zeta function located in the upper half-plane (conjugate zeroes exist in the lower half-plane whose imaginary parts are simply the opposites of these). The gigantic ZetaGrid distributed project of Sebastian Wedeniwski managed to compute more than 10 trillion zeros over the course of its lifetime (2001-2005) but they were scooped by Xavier Gourdon and Patrick Demichel who achieved that same goal earlier with modest means by using superior software based on an algorithm devised in 1988 by Andrew M. Odlyzko (1949-) and Arnold Schönhage (1934-)...

14.1347251417346937904572519835624702707842571156992431756855674+ 21.0220396387715549926284795938969027773343405249027817546295204+ 25.0108575801456887632137909925628218186595496725579966724965420+ 30.4248761258595132103118975305840913201815600237154401809621460+ 32.9350615877391896906623689640749034888127156035170390092800034+ 37.5861781588256712572177634807053328214055973508307932183330011+ 40.9187190121474951873981269146332543957261659627772795361613037- 43.3270732809149995194961221654068057826456683718368714468788937- 48.0051508811671597279424727494275160416868440011444251177753125+ 49.7738324776723021819167846785637240577231782996766621007819558- 52.9703214777144606441472966088809900638250178888212247799007481+ 56.4462476970633948043677594767061275527822644717166318454509698+ 59.3470440026023530796536486749922190310987728064666696981224518- 60.8317785246098098442599018245240038029100904512191782571013488+ 65.1125440480816066608750542531837050293481492951667224059665011- 67.0798105294941737144788288965222167701071449517455588741966696- 69.5464017111739792529268575265547384430124742096025101573245400- 72.0671576744819075825221079698261683904809066214566970866833062- 75.7046906990839331683269167620303459228119035306974003016477753+ 77.1448400688748053726826648563046370157960324492344610417652315- 79.3373750202493679227635928771162281906132467431200308784387205- 82.9103808540860301831648374947706094975088805937821491465713063- 84.7354929805170501057353112068277414171066279342408187027355297- 87.4252746131252294065316678509192132521718864012690281864555579+ 88.8091112076344654236823480795093783954448934098186750421998716+ 92.4918992705584842962597252418106848787217940277306461750967505- 94.6513440405198869665979258152081539377280270156548520195924743- 95.8706342282453097587410292192467816952564612249879984205292817- 98.8311942181936922333244201386223278206580390634281961028193217+

Statements Equivalent to the Riemann Hypothesis :

Many statements have been shown to hold if RH is assumed to be true, and a number of them are known to imply RH, so they are actually equivalent to it.

In 1901, Helge von Koch (1870-1924) proved RH equivalent to a relation between the prime counting function p(x) and the logarithmic integral : Helge von Koch (1870-1924)

p(x) = li (x) + O ( x^½ Log x )

Beyond the Riemann Hypothesis :

Several nice statements have been made which seem true and are simpler and stronger than RH (each implies RH but the converse need not be true). This includes a conjecture made in the doctoral dissertation of Sebastian Wedeniwski (the aforementioned mastermind of ZetaGrid, 2001-2005) :

Wedeniwski's Property (2001)

Modulo p, there is a quadratic nonresidue below ³/₂ (Log p)²

The Mertens conjecture (Stieltjes, 1885. Mertens, 1897) was yet another famous statement "stronger than RH" which seemed promising until it was disproved in 1985 by Andrew Odlyzko (b.1949) and Herman te Riele (b.1947). It had been formulated in 1885 by Thomas Stieljes (1856-1894) after he thought he could prove the following weaker statement (actually equivalent to RH) about the asymptotics of the Mertens function M :

M(n) = O(n ^½+e ) for any e > 0

The Mertens conjecture is the stronger (false) statement: |M(n)| < n^½

Riemann Hypothesis at... Wikipedia | MathWorld | Prime Pages | Clay Mathematics Institute
How Euler discovered the zeta function by Keith Devlin (Devlin's Angle at MAA, The Math Guy at NPR, etc.)
Euler and the Zeta Function by Sumida-Takahashi Hiroki (Hiroshima University)
Zeta function | Eta function | Ten Trillion Zeta Zeros by Ed Pegg, Jr.

Videos : The Riemann Hypothesis (19:35) by James Grime (Singingbanana, 2014年01月17日).
The Riemann Hypothesis (1:05:40) by Alex V. Kontorovich (Math Mornings at Yale, 2012年12月02日).
Visualizing the Riemann hypothesis and analytic continuation (22:10) by Grant Sanderson (2016年12月09日).
The Key to the Riemann Hypothesis (12:37) by Jon Keating (2016年05月10日).

The Mertens conjecture (9:55) by Holly Krieger (Numberphile, 2020年01月23日).

The paper of Bernhard Riemann was published in November 1859 (Monatsberichte der Berliner Akademie):
Über die Anzahl der Primzahlen unter einer gegebenen Grösse (German original, 1859)
On the Number of Primes less than a Given Quantity (English translation, by David R. Wilkins, 1998)

Proposed Proofs of the Riemann Hypothesis, by Matthew R. Watkins.
The Seventh Millennium Talk (1:13:10) by Jeff Vaaler (2001年05月02日).
A Lecture on Primes and the Riemann Hypothesis by Barry Mazur (MSRI, 2014年04月08日).
The Riemann Hypothesis (42:00) by Michael Atiyah (2018年09月24日).

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(2013年07月02日) The Twin Primes Conjecture
Are there infinitely many pairs of primes whose difference is 2 ?

The first such pairs are: {3,5} {5,7} {11,13} {17,19} {29,31} ...

In December 2011, a large pair of twin primes (200700 digits) was discovered by Timothy D. Winslow, PrimeGrid, et al. :

{ 3756801695685 . 2⁶⁶⁶⁶⁶⁹ - 1 , 3756801695685 . 2⁶⁶⁶⁶⁶⁹ + 1 }

This remained the largest known until 2016年09月14日, when a pair of twin primes with 388342 digits was found by Tom N. Greer, PrimeGrid, et al. :

{ 2996863034895 . 2^1290000 - 1 , 2996863034895 . 2^1290000 + 1 }

Nobody knows for sure whether that sequence is infinite or not, although everybody's guessing that it is. That's one of the two oldest unsolved problems in mathematics (the other one pertains to odd perfect numbers).

In 1966, Chen Jingrun (1933-1996) proved that there are infinitely many primes p such that p+2 is either prime or semiprime (in which case p is called a Chen prime). A semiprime is defined as the product of two primes.

The Twin Primes Conjecture says that the following is true for K = 2:

There are infinitely many pairs of primes whose difference is K.

It's widely believed that the above statement holds for any even integer K. Polignac's coat-of-arms

Polignac's conjecture (1849) is the belief that there are infinitely many pairs of consecutive primes differing by K, for any even K.

The weaker statement that the above holds for at least one nonzero value of K is equivalent to saying that the difference between consecutive primes doesn't tend to infinity. This was only proved recently:

Assuming a generalized version of the Elliott-Halberstam conjecture (1968) the above lower-bound would be reduced down to 6 [Polymath, August 2014]. However, new methods seem needed for the ultimate reduction to 2.

"Bounded Gaps between Primes" by Yitang Zhang (Annals of Mathematics, 179, pp.1121-1174)

Zhang's Theorem on Bounded Gaps between Primes by Dan Goldston (1954-).
First proof that infinitely many primes come in pairs by Maggie McKee (Nature, 2013年05月14日)
Unheralded Mathematician Bridges the Prime Gap by Erica Klarreich (Simons Foundation, 2013年05月19日)
Philosophy behind Yitang Zhang's work (MathOverflow, 2013年05月20日)

Videos : Gaps between Primes & extra footage, in "Numberphile" by Brady Haran (2013年05月27日)
A conversation between Yitang Zhang and David Eisenbud (2013年09月13日).
Small Gaps between Prime Numbers by Yitang Zhang (Hong Kong, 2014年08月11日)
Terry Tao, Ph.D. Small and Large Gaps Between the Primes (UCLA, 2014年10月07日)
Counting From Infinity by George Csicsery (2015 documentary).
Yitang Zhang's "Counting From Infinity" Bonus Material : 1 | 2 | 3 | 4 | 5 | 6 | 7
Twin Prime Conjecture (17:41) by James Maynard (Number[hile, 2017年04月13日).

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(2016年01月29日) Goldbach Conjectures (1742)
Every even number greater than 2 is the sum of two primes.

That conjecture was first formulated by the talented recreational mathematician Christian Goldbach (1690-1764) who wrote to Euler about it in 1742.

An equivalent satement is obtained with odd primes by excluding the number 4 = 2+2 (which isn't the sum of two odd primes).

The weaker statement that odd numbers are sums of three primes can be construed as a corollary, from the remark that an odd number above 5 is 3 plus an even number above 2. That weaker statement is less formidable; it was shown to hold for sufficiently large odd numbers in 1923 by Hardy & Littlewood assuming the Riemann Hypothesis.

• Even Goldbach Conjecture : (Goldbach's conjecture)
  All even numbers above 2 are sums of two primes.
  All even numbers above 4 are sums of two odd primes.
• Odd Goldbach Conjecture : (Goldbach's weak conjecture)
  All odd numbers above 5 are sums of three primes.
  All odd numbers above 7 are sums of three odd primes.

In 2012 and 2013, Harald Helfgott (b. 1977) published two papers which provide a complete proof of the weak conjecture.

The strong conjecture remains an open question which has only been checked by computer for even numbers up to 4 10¹⁸ or so.

On 1752年11月18日, Goldbach also formulated the lesser-known conjecture that any odd number is twice a square plus a prime. Two counterexamples (5777 and 5993) were discovered in 1856 by Moritz A. Stern (of Stern-Brocot tree fame).

The Goldbach conjecture (9:58) by David Eisenbud (1947-) (Numberphile, 2017年05月24日).
210 is the most Goldbachy number (6:34) by Carl Pomerance (1944-) (Numberphile, 2017年05月28日).

Wikipedia: Goldbach's (strong) conjecture | Goldbach's weak conjecture

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(2017年07月08日) Legendre's Conjecture (Legendre, 1808)
Is there always at least one prime between two nonzero perfect squares?

There seems to be at least two primes between two consecutive squares.

A014085(n) is the number of primes between n² and (n+1)². Namely:

0, 2, 2, 2, 3, 2, 4, 3, 4, 3, 5, 4, 5, 5, 4, 6, 7, 5, 6, 6, 7, 7, 7, 6, 9, 8, 7, 8, 9, 8, 8, 10, 9, 10, 9, 10, 9, 9, 12, 11, 12, 11, 9, 12, 11, 13, 10, 13, 15, 10, 11, 15, 16, 12, 13, 11, 12, 17, 13, 16, 16, 13, 17, 15, 14, 16, 15, 15, 17, 13, 21, 15, 15...

Asymptotically, there should be as many primes between n² and (n+1)² as between 1 and n (roughly n / ln n, by the prime-number theorem). So, we're very confident that Legendre's conjecture won't fail in the long run. That's a good heuristic argument but it doesn't constitute a proof.

Legendre's conjecture | Adrien-Marie Legendre (1752-1833) | Landau's problems (1912)

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(2017年07月31日) Landau's Fourth Problem
Are there infinitely many primes following a square?

A005574 is the sequence of the numbers n for which n²+1 is prime:

1, 2, 4, 6, 10, 14, 16, 20, 24, 26, 36, 40, 54, 56, 66, 74, 84, 90, 94, 110, 116, 120, 124, 126, 130, 134, 146, 150, 156, 160, 170, 176, 180, 184, 204, 206, 210, 224, 230, 236, 240, 250, 256, 260, 264, 270, 280, 284, 300, 306, 314, 326, 340, 350, 384, 386, 396, 400, 406, 420, 430, 436, 440, 444, 464, 466, 470, 474, 490, 496, 536, 544, 556, 570, 576, 584, 594...

It seems likely that there are infinitely many primes of this form but this is not known for sure. This old question was popular enough in 1912 for Landau to include it in his list of four unsolved problems related to primes.

Landau's problems (1912)

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(2017年11月25日) Schinzel's Hypothesis H (Schinzel & Sierpinski, 1958)
A set of irreducible integer-valued polynomials without a fixed prime divisor are simultaneously prime infinitely often.

Note that, bt definition, an irreducible polynomial is non-constant.

Integer-valued polynomials need not have integer coefficients.
½ n² + ½ n is integer-valued (A000217) but it's reducible.

A fixed divisor of several integer-valued polynomials is defined to be an integer which divides their product at every point. A set of polynomials without a prime fixed divisor is said to verify Bunyakovsky's property.

Bunyakovsky's property is clearly necessary for nonconstant polynomials to be simultaneously prime infinitely often. Otherwise, there would be a fixed prime p dividing the product of the polynomials at every point...

In this case, at every point where all polynomials are prime, at least one of them must be equal to p (if a prime divides a product of primes, it's equal to one of them).

This implies that at least one of the polynomials is equal to p infinitely many times, which can only happen if it's constant, which is ruled out. QED

Schinzel's hypothesis H | Andrzej Schinzel (1937-) | Viktor Bunyakovsky (1804-1889)

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(2002年11月17日) P = NP (?)
What's an NP-complete problem?

A computational problem is said to be in the class P of polynomial time problems whenever there's an algorithm which can find a valid solution in a number of elementary steps which is always less than a certain polynomial function of the size of the input data (one measure of this size could be the number of digits used in a reasonnably thrift encoding of the input data).

The class NP (nondeterministic polynomial time problems) consists of those problems which could be so solved nondeterministically, which is a fancy way to say that an unlimited number of lucky guesses are allowed in the process which arrives at a solution. Such a nondeterministic process must still be such that only valid solutions are produced... To put it in simpler words, a problem is in NP if and only if a solution of it can be checked in polynomial time (an explicit nondeterministic algorithm would then be to guess a correct solution and check it).

In 1972, Richard Karp (1935-) discovered that there are problems in NP which he dubbed "NP-complete" because they are at least as tough to solve as any other problem in NP, in the following sense: Any NP problem can be reduced in polynomial time by a deterministic algorithm to the solution of an NP-complete problem whose data size is no more than a polynomial function of the original input data.

Therefore, if any NP-complete problem could be solved deterministically in polynomial time (i.e., if it was a P problem) then all NP problems would be in P and we would thus have P = NP.

Karp's original NP-complete problem (dubbed SAT) was the satisfiability of boolean expressions: Is there a way to satisfy a given boolean expression (i.e., make it "true") by assigning true/false values to the variables in it?

The SAT problem is clearly in NP (just guess a correct set of values and compute the boolean expression to make sure it's true). Conversely, Stephen Cook (1939-) proved from scratch in 1971 that any problem in NP can be reduced in polynomial time to a commensurable boolean satisfiability problem, thus establishing SAT to be NP-complete. This result is now known as the Cook-Levin theorem because it was also obtained independently by Leonid Levin (1948-).

If a known NP-complete problem like SAT can be reduced polynomially to some NP problem Q, the problem Q is then established to be NP-complete. This way, from Karp's original NP-complete problem, the list of known NP-complete problems has grown to include literally hundreds of "classical" examples.

The tantalizing thing is that many such NP-complete problems are very practical problems which, at first, look like they could be solved in polynomial time. Yet, nobody has ever "cracked" one of these in polynomial time or proved that such a thing could not be done... Therefore we still don't know whether P=NP or not.

Videos: What computers can't do (1:04:05) with Q&A (10:40) by Kevin Buzzard (RI, 2017-01).

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(2002年12月21日) Collatz Sequences and the Syracuse Conjecture
Collatz sequences are sequences of integers where an even N is followed by N/2 and an odd N by 3N+1. Do they all land on 4, 2, 1, 4...?

The problem is most commonly named after the German mathematician Lothar Collatz (1910-1990) who formulated the conjecture in 1937 and shared it privately with Stanislaw Ulam (1909-1984) and Shizuo Kakutani (1911-2004) at the ICM in 1950.

This was first described in print in 1971, in a transcript of a talk given in 1970 by Harold Coxeter (1907-2003). In 1975, Helmut Hasse (1898-1979) coined the name Syracuse problem after Syracuse University.

It's also called the hailstone problem (because of its dynamics) and has been popularized under several other names like Kakutani's problem (Yale University, early 1960's) and Ulam's problem.

[画像: Come back later, we're still working on this one... ]

La conjecture de Syracuse (15:18) by ElJj.

On the Dirichlet inverse of the integers modulo 2 (A087003)

The sequence A087003 may be defined as the Dirichlet inverse of the character modulo 2. It was first defined by Labos E. as the sum of all the Möbius values found at the points of the Collatz trajectory until a "4" is found (2003年10月02日).

However, Marc Lebrun (2004年02月19日) has shown that either definition simply means that A087003 is equal to the Möbius function at odd points and vanishes at even points... All told, the Collatz trajectories turn out to be irrelevant !

Breakthrough in 2019

Almost all Collatz orbits attain almost bounded values.

This is about as close as anyone can get to the
Collatz comjecture without actually solving it.
Terry Tao (September 2019)

Almost all Collatz orbits attain almost bounded values by Terry Tao (2019年09月10日).

Progress on the Collatz conjecture by John D. Cook (2019年09月10日).

Videos: Norman J. Wildberger (2013年01月14日)
Marc Chamberland , Grinnell College : Part 1 | Part 2 (2013年04月12日).
David Eisenbud , MSRI : Part 1 | Part 2 (2016年08月08日).
The Simplest Math Problem No One Can Solve (22:08) by Derek Muller (Veritassium, 2021年07月30日).

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Yes (2006年08月29日) The Poincaré Conjecture (1904)
This was proved in 2002 by Grigori Yakovlevich Perelman, in the generalized form proposed by Thurston (geometrization conjecture).

In 1904, Henri Poincaré had conjectured that:

The 3-sphere is the only closed 3-manifold in which
every loop can be continuously tightened to a point.

[画像: Come back later, we're still working on this one... ]

Poincaré conjecture | Geometrization conjecture of Bill Thurston (1946-2012)

Grisha Perelman (1966-) | 2006年08月22日

Geometry in 2, 3, and 4 Dimensions (2010年06月08日) by Michael Atiyah (1929-, Fields Medal 1966).
History of the Poincaré Conjecture (2010年06月08日) by John Morgan (1946-).
Post-Perelman Problems in Topology (2010年06月08日) by Stephen Smale (1930-, Fields Medal 1966).

Ricci flow (14:40) by James Isenberg (Numberphile, 2014年04月23日).

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Yes (2010年10月15日) Fermat's Last Theorem (Fermat 1637, Wiles 1995)
xⁿ + yⁿ = zⁿ has no solutions in positive integers for n > 2.

Pierre de Fermat (1601-1665) Hanc marginis exiguitas non caperet.
This margin is too small to contain [my proof].
Pierre de Fermat (1601-1665)

Die Gleichung aⁿ=bⁿ+cⁿ für n>2 in ganzen Zahlen [ist] nicht auflösbar.
A. Ernest Wendt (1894)

Solutions for n = 2 are called Pythagorean triples. They are fairly easy to enumerate systematically, starting with x=3, y=4, z=5. Many special cases known in ancient times were recorded on Chaldean clay tablets.

In the Middle Ages, Leonardo Fibonacci proved that there was no solutions for n = 4 (Liber Quadratorum, 1225).

[画像: Come back later, we're still working on this one... ]

Fermat's Last Theorem, a film by Simon Singh & John Lynch (1996)

Nicholas M. Katz

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(2013年01月09日) The Oesterlé-Masser ABC Conjecture
Moshizuki has proposed a 500-page proof whose philosophy is obscure.

The radical (rad) of a positive integer n is the integer whose prime factorization consist of the same primes as that of n with multiplicity 1. The function rad is a multiplicative function.

The ABC conjecture says that the inequality rad ( a b ) > (a+b)^e has infinitely many exceptions when e = 0 but finitely many when e > 0.

The conjecture was formulated in 1985 by the French Bourbakist Joseph Oesterlé (b. 1954) and the British mathematician David Masser (b. 1948).

That's arguably the most important open Diophantine statement today.

On August 30, 2012, Shinichi Mochizuki, from Kyoto, released...

[画像: Come back later, we're still working on this one... ]

Proof of the ABC conjecture? (6:43) by James Grime (Numberphile, 2012年10月12日).
The ABC conjecture (55:29) by Jeff Vaaler (2013年02月18日).
Introduction to the ABC conjecture (53:43) by Héctor Pastén Vásquez (IAS, 2016年03月21日).

The Mochizuki Theorem? by Jeremy Teitelbaum, University of Connecticut (2012年10月11日).
An ABC proof too tough to check by By Kevin Hartnett, Boston Globe (2012年11月03日).

Wikipedia : abc conjecture | David Masser (1948-) | Joseph Oesterlé (1954-)

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(2016年05月30日) Union-closed sets conjecture (Péter Frankl, 1979)
In a nontrivial finite union-closed family of finite sets,
is there always an element that belongs to at least half of the sets?

The "nontrivial" qualifier indicates that we're considering only families containing at least one nonempty set. A family of sets is said to be union-closed when it contains any union of its members.

The conjecture clearly holds for families containing at least one singleton.

Wikipedia : Union-closed sets conjecture (1979) | Péter Frankl (1953-)

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(2019年07月18日) Hadwiger conjecture in graph theory (1943)
Connection between chromatic number and graph minors.

[画像: Come back later, we're still working on this one... ]

Hadwiger conjecture in graph theory (1943) | Hugo Hadwiger (1908-1981)

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(2016年05月30日) The Egyptian conjecture (Straus & Erdös, 1948)
For n≥2, is 4/n always the sum of three unit fractions?

Besides 2/3 and 3/4, ancient Egyptians only allowed fractions with numerator 1 (unit fractions). They represent other fractions as sums of those (without repetitions).

Paul Erdös (1913-1996) | Ernst G. Straus (1922-1983) | Wikipedia : Egyptian conjecture (1948)

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(2017年07月28日) Birch and Swinnerton-Dyer Conjecture (1960)
The algebraic rank of any elliptic curve is equal to its analytic rank.

This is one of the 7 Millenium Problems on which the Clay Mathematical Institute has placed a bounty of one million dollars.

The Birch and Swinnerton-Dyer conjecture (1:03:08) by Fernando Rodriguez-Villegas (2001年02月21日).

Recent results about the Birch and Swinnerton-Dyer conjecture (1:15:55) by Manjul Bhargava (2015年10月23日).
What's the Birch and Swinnerton-Dyer conjecture? (1:07:44) by Manjul Bhargava (2016).
What's known about the Birch and Swinnerton-Dyer conjecture (1:02:22) by Manjul Bhargava (Oslo, 2016年05月25日).

Wikipedia : Birch and Swinnerton-Dyer conjecture | Bryan Birch (1931-) | Peter Swinnerton-Dyer (1927-)

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(2017年11月13日) Gilbreath's Conjecture (Proth 1878, Gilbreath 1958).
It's true of "sufficiently random" sequences. Is it true of primes ?

François Proth (in 1878) and Normal L. Gilbreath (in 1958) independently considered that a sequence can be obtained from another as the absolute value of the differences between consecutive terms. When we start with the sequence of the prime numbers and apply that process iteratively, we obtain the following intriguing table:

This far, all the successive sequences so tabulated start with a leftmost 1. The Proth-Gilbreath conjecture is the unproved statement that it's always so.

In 1878, Proth gave an erroneous proof.

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Wikipedia | The Prime Glossary

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(2019年05月11日) The g-conjecture (McMullen, 1970)
Proved by Karim Adiprasito (1988-) in December 2018.

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Simplicial sphere | Peter McMullen (1942-) | Karim Adiprasito (1988-)

g-conjecture (22:10) by June Huh (Numberphile, 2018年05月21日).

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(2020年06月14日) Is Diff ² ( S¹ ) a simple group?

In the 1970s, Bill Thurston (1946-2012) and John Mather (1942-2017) proved a highly nontrivial result:

In the group Diff ^r (M) of the C^r-diffeomorphisms of a compact manifold M, the connected component of the identity is a simple subgroup, in most cases... The result need not hold when r is equal to dim(M)+1.

In particular, the case of Diff ² ( S¹ ) remains completely open.

Do the diffeomorphisms of class C² over the circle form a simple group?

The related group denoted Diff+^1+bv( S¹ ) is not simple, as pointed out by the late John Mather. This one is a well-studied group consisting of the orientation-preserving diffeomorphisms f of the circle, which are C1 (i.e., the first derivative f ' is a continuous function) with the added condion that Log f ' is of bounded variation (French: à variation bornée). That group is fundamental in dynamical systems, as it meets the premises of Denjoy's theorem, which was established in 1932 by Arnaud Denjoy (1884-1974), the thesis advisor of the bourbakist Gustave Choquet (1915-2006).

Proof (John N. Mather) : Let G = Diff₊^1+bv ( S¹ )

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My favorite groups (22:42) by Etienne Ghys (CIRCM, 2014年01月21日).

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(2019年12月18日) Hadwiger-Nelson Problem (Nelson, 1950)
Color the plane so two points one unit apart are never the same color.

The problem was made popular by Martin Gardner in 1960. Since 2018, we know that the least number of colors needed is 5, 6 or 7.

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Hadwiger-Nelson Problem | Hugo Hadwiger (1908-1981) | Edward Nelson (1932-2014)

The chromatic number if the plane is at least 5 by Aubrey D.N.J. de Grey (2018).

A Colorful Unsolved Problem (9:38) by James Grime (Numberphile, 2019年02月27日).