It is not because things are difficult that we do not dare, it is because we do not dare that they are difficult. Lucius AnnæusSeneca (5 BC - AD 65) As our island of knowledge grows, so doestheshore of our ignorance. John Wheeler ( 1992)
![]() Tough questions mentioned elsewhere on this site:
Related articles on this site:
Mathematical Mystery Tour (BBC Horizon) |
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The Mertens conjecture (Stieltjes, 1885. Mertens, 1897) was yet another famous statement "stronger than RH" which seemed promising until it was disproved in 1985by 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+ ) for any > 0
The Mertens conjecture is the stronger (false) statement: |M(n)| < n½
The Zeta Function Ballad | |
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 until2016-09-14,when a pair of twin primes with 388342 digitswas found by Tom N. Greer, PrimeGrid, et al. :
{ 2996863034895 . 21290000 1 , 2996863034895 . 21290000 + 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).
The Twin Primes Conjecture says that thefollowing 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.
The weaker statement that the above holds for at least one nonzerovalue of K is equivalent to saying that the differencebetween consecutive primes doesn't tend to infinity. This was only proved recently:
![]() | In April 2013, Yitang Zhang established an upper-bound of 70 million for the least K verifying the above. Zhang was quoted as saying that the methods inhis 55-page papercould be perfected to pull this upper-bound downward. Quickly, Terry Tao initiated apolymathproject which reduced the upper-bound to 4680,using Zhang's own methods. In November 2013, James Maynard (b. 1987, 2002 Fields Medal) found a streamlined approach giving directly an upper-bound of 600. He joined Tao's group and they finally managed to reduce the bound to 246 (as of 2014-04-14). |
Assuming a generalized version of the Elliott-Halberstamconjecture (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.
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 construedas 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 sufficientlylarge odd numbers in 1923 by Hardy &Littlewood assuming the Riemann Hypothesis.
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 beenchecked by computerfor even numbers up to 4 18 or so.
There seems to be at least two primes between two consecutive squares.
Square | Primes | Square | Primes | Square | Primes |
---|---|---|---|---|---|
1 | 2, 3 | 4 | 5, 7 | 9 | 11, 13 |
16 | 17, 19, 23 | 25 | 29, 31 | 36 | 37, 41, 43, 47 |
49 | 53, 59, 61 | 64 | 67, 71, 73, 79 | 81 | 83, 89, 97 |
100 | 101, 103, 107, 109, 113 | 121 | 127, 131, 137, 139 | 144 | 149, 151, 157, 163, 167 |
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 failin the long run. That's a goodheuristic argumentbut it doesn't constitute a proof.
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 thisis not known for sure. This old questionwas popular enough in 1912 for Landau to include it in hislist of four unsolved problems related to primes.
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 isdefined 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 nonconstantpolynomials 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 ofthem 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.
A computational problem is said to be in the class Pofpolynomial time problems whenever there'sanalgorithm which can find a valid solution ina number of elementary steps which is always less than a certain polynomial function of the size of the input data (one measure of thissize could be the number of digits used in a reasonnably thrift encoding ofthe 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 allowedin the process which arrives at a solution. Such a nondeterministic process must still be such that only valid solutionsare produced... To put it in simpler words, a problem is in NP if and only if a solution ofit can be checked in polynomial time (an explicit nondeterministic algorithm would then be to guess a correct solutionand 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 thefollowing sense: Any NP problem can be reduced in polynomial time by a deterministic algorithmto the solution of an NP-complete problem whose data size is no more thana polynomial function of the original input data.
Therefore, ifany NP-complete problem could be solved deterministicallyin polynomial time (i.e., if it was a P problem) then all NP problemswould be in P and we would thus have P = NP.
Karp's original NP-complete problem (dubbed SAT)was thesatisfiability of boolean expressions: Is there a way to satisfy a givenboolean expression (i.e., make it "true") by assigning true/false valuesto the variables in it?
The SAT problem is clearly in NP (just guessa correct set of values and compute the booleanexpression to make sure it's true). Conversely,Stephen Cook (1939-) proved from scratch in 1971 that any problem in NP can be reduced in polynomialtime to a commensurable boolean satisfiability problem, thus establishingSAT to be NP-complete. This result is now known as theCook-Levin theorembecause 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 includeliterallyhundreds of "classical" examples.
The tantalizing thing is that many such NP-complete problems are very practicalproblems which, at first, look like they could be solved in polynomial time. Yet, nobody has ever "cracked" one of these in polynomial timeor proved that such a thing could not be done... Therefore we still don't know whether P=NP or not.
The problem is most commonly named after the German mathematician LotharCollatz (1910-1990) who formulated the conjecture in 1937 and shared it privately withStanislaw Ulam (1909-1984)andShizuo Kakutani (1911-2004)at theICM in 1950.
This was first described in print in 1971, in a transcript of atalk given in 1970 byHarold Coxeter (1907-2003). In 1975,Helmut Hasse (1898-1979) coined the name Syracuse problem after Syracuse University.
The sequence A087003 may be defined as the Dirichlet inverse of thecharacter modulo 2. It was first defined by Labos E. as the sum of all the Möbius values found at the points of the Collatz trajectoryuntil a "4" is found (2003-10-02).
However, Marc Lebrun (2004-02-19) has shown that either definition simplymeans that A087003 is equal to the Möbius functionat odd points and vanishes at even points... All told, the Collatz trajectories turn out to be irrelevant !
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)
In 1904,Henri Poincaré had conjectured that:
The 3-sphere is the only closed 3-manifold in which
every loopcan be continuously tightened to a point.
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 Pythagoreantriples. They are fairly easy to enumerate systematically, starting withx=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 forn = 4 (LiberQuadratorum, 1225).
The radical (rad) of a positive integer n is theinteger whose prime factorization consist of thesame 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) has infinitely many exceptions when = 0 but finitely many when > 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...
The "nontrivial" qualifier indicates that we're considering only families containing at least onenonempty set. A family of sets is said to be union-closed when itcontains anyunion of its members.
The conjecture clearly holds for families containing at least one singleton.
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).
This is one of the 7 Millenium Problems on which the Clay Mathematical Institute has placed a bounty of one million dollars.
François Proth (in 1878) andNormal L. Gilbreath(in 1958) independently considered that a sequence can be obtained from another as the absolutevalue of the differences between consecutive terms. When we start with the sequence of the prime numbers and apply thatprocess iteratively, we obtain the following intriguing table:
2 | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 29 | 31 | 37 | 41 | 43 | 47 | 53 | 59 | 61 | 67 | 71 | 73 | 79 | 83 | |||||||||||||||||||||||
1 | 2 | 2 | 4 | 2 | 4 | 2 | 4 | 6 | 2 | 6 | 4 | 2 | 4 | 6 | 6 | 2 | 6 | 4 | 2 | 6 | 4 | 6 | |||||||||||||||||||||||
1 | 0 | 2 | 2 | 2 | 2 | 2 | 2 | 4 | 4 | 2 | 2 | 2 | 2 | 0 | 4 | 4 | 2 | 2 | 4 | 2 | 2 | ||||||||||||||||||||||||
1 | 2 | 0 | 0 | 0 | 0 | 0 | 2 | 0 | 2 | 0 | 0 | 0 | 2 | 4 | 0 | 2 | 0 | 2 | 2 | 0 | 0 | ||||||||||||||||||||||||
1 | 2 | 0 | 0 | 0 | 0 | 2 | 2 | 2 | 2 | 0 | 0 | 2 | 2 | 4 | 2 | 2 | 2 | 0 | 2 | 0 | |||||||||||||||||||||||||
1 | 2 | 0 | 0 | 0 | 2 | 0 | 0 | 0 | 2 | 0 | 2 | 0 | 2 | 2 | 0 | 0 | 2 | 2 | 2 | 2 | |||||||||||||||||||||||||
1 | 2 | 0 | 0 | 2 | 2 | 0 | 0 | 2 | 2 | 2 | 2 | 2 | 0 | 2 | 0 | 2 | 0 | 0 | 0 | ||||||||||||||||||||||||||
1 | 2 | 0 | 2 | 0 | 2 | 0 | 2 | 0 | 0 | 0 | 0 | 2 | 2 | 2 | 2 | 2 | 0 | 0 | 0 | ||||||||||||||||||||||||||
1 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 0 | 0 | 0 | 2 | 0 | 0 | 0 | 0 | 2 | 0 | 0 | |||||||||||||||||||||||||||
1 | 0 | 0 | 0 | 0 | 0 | 0 | 2 | 0 | 0 | 2 | 2 | 0 | 0 | 0 | 2 | 2 | 0 | 0 | |||||||||||||||||||||||||||
1 | 0 | 0 | 0 | 0 | 0 | 2 | 2 | 0 | 2 | 0 | 2 | 0 | 0 | 2 | 0 | 2 | 0 | ||||||||||||||||||||||||||||
1 | 0 | 0 | 0 | 0 | 2 | 0 | 2 | 2 | 2 | 2 | 2 | 0 | 2 | 2 | 2 | 2 | 0 | ||||||||||||||||||||||||||||
1 | 0 | 0 | 0 | 2 | 2 | 2 | 0 | 0 | 0 | 0 | 2 | 2 | 0 | 0 | 0 | 2 |
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.
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 Cr-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) remainscompletely 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 indynamical systems,as it meets the premises of Denjoy's theorem, which wasestablished in 1932 by Arnaud Denjoy (1884-1974),the thesis advisor of the bourbakist Gustave Choquet (1915-2006).
: Let G = Diff+1+bv( S1)
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.