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Update: Please restrict your answers to "tweets" that give more than just the statement of the result, and give also the essence (or a useful hint) of the argument/novelty.

I am looking for examples that the essence of a notable mathematical development fits a tweet (140 characters in English, no fancy formulas).

Background and motivation: This question was motivated by some discussion with Lior Pachter and Dave Bacon (over Twitter). Going over my own papers and blog posts I found very few such examples. (I give one answer.) There were also few developments that I do not know how to tweet their essence but I feel that a good tweet can be made by a person with deep understanding of the matter.

I think that a good list of answers can serve as a good educational window for some developments and areas in mathematics and it won't be overly big.

At most 140 characters per answer, single link is permitted. Tweeting one's own result is encouraged.

Update: I learned quite a few things, and Noam's tweet that I accepted is mind-boggling.

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    $\begingroup$ I have discovered a truly remarkable proof of this theorem which this Tweet is too small to contain. $\endgroup$
    – Glorfindel
    Commented Apr 26, 2017 at 8:26
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    $\begingroup$ I feel like most of the answers are mis-interpreting the question. This doesn't ask for a result whose statement is in 140 characters; that would be too broad: most paper titles fit in them. It asks for a result whose essence is tweetable: given the tweet alone, a mathematician with good knowledge of the field can fill in the details and complete a proof. So I am going to downvote almost all of them. $\endgroup$ Commented Apr 26, 2017 at 10:16
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    $\begingroup$ @FedericoPoloni: what is it, a kind of joke? Could you please indicate to me where exactly in the question it is written that the tweet should be such that "a mathematician with good knowledge of the field can fill in the details and complete a proof"? Honestly, I do not think that your personal interpretation of the locution "essence of a notable mathematical development" should be taken as a rule here. $\endgroup$ Commented Apr 26, 2017 at 10:21
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    $\begingroup$ Perhaps the next big MO question should be, "What is the essence of a mathematical result?" I myself lean toward Federico's interpretation- pithifying a theorem's statement does not necessarily clarify or illuminate the ideas at play. $\endgroup$
    – Neal
    Commented Apr 26, 2017 at 12:52
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    $\begingroup$ My initial intention was indeed that the "tweet" gives more than just the statement of the result but also the essence of the argument/novelty. To demand that a mathematician in the field can fill the details is too much to ask for. $\endgroup$
    – Gil Kalai
    Commented Apr 26, 2017 at 12:58

89 Answers 89

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Every non-constant complex polynomial has a complex root : If not the inverse is bounded analytic. Use Liouville. #FundamentalTheoremOfAlgebra

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Not a theorem but a good result on Mersenne integers using Lucas-Lehmer test (via @toorandom)

2^74207281-1 is #prime, they checked: Consider s_1=4, s_2=14,...,s_j=(s_{j-1}^2)-2... m=(2^n)-1 prime <=> s_{n-1} is multiple of m #mathchat

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Aubrey de Grey's strategies for finding the chromatic number of the plane: A) prolonging life to 1000 years and waiting for a solution B) Constructing a unit-distance planar graph that requires 5 colors!!! Based on SAT-solvers and a lot of Moser Spindles. link

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    $\begingroup$ The above far exceeds 140 characters. A more straightforward version works: “Aubrey de Grey constructed a unit-distance planar graph that requires 5 colors!!! Based on SAT-solvers and a lot of Moser Spindles.” $\endgroup$
    – user44143
    Commented Dec 30, 2022 at 3:00
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"The Magic Words Are Squeamish Ossifrage" - to factor a 129-digit semiprime took way less than 40 quadrillion years when early-90's era computers work together using early-90's era factoring algorithms over the early-90's era Internet!

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One cannot hear the shape of a drum. link

Proof via Sunada's Theorem.

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Lucas-Lehmer for Wagstaff. Let $p$ be an odd prime, $s_0=4,s_{n+1}=s_n^2-2$, then $N_p=(2^p+1)/3$ is prime implies $N_p$ divides $s_{p-1}-5-9\left( \frac{p}{3} \right) $.

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ZFC + "exists a Reinhardt cardinal" is inconsistent.

V = L implies that measurable cardinals do not exist.

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Elliptic curves produce a key exchange that may be safe against quantum computers. link

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  • $\begingroup$ They do? That is awesome. $\endgroup$
    – Asaf Karagila
    Commented Apr 27, 2017 at 19:35
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    $\begingroup$ "May be". That is, nobody knows of a sub-exponential quantum algorithm in the supersingular case, because the known sub-exponential algorithms for ordinary elliptic curves require the endomorphism ring to be commutative. But I see no reason for confidence that a sub-exponential quantum algorithm for this case does not exist. $\endgroup$ Commented Apr 27, 2017 at 20:40
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The smallest positive integer not definable in under sixty letters.

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f_N = (-d/dx)^N cos(x^½)
= A(y)f_0 + B(y)f_1 for y = 1/(4x); A,B∈Z[y], deg O(N)
~ N!/(2N)! as N → ∞ = o(ε^N)
f_1/f_0∈Q ⇒ y∉Q
x = π^2 ⇒ π^2∉Q

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e^x is sum of x^k/k!, k=0,1,...: Binomial theorem on (1+x/n)^n; coefficient of x^k is binom(n,k)/n^k=(1-1/n)...(1-(k-1)/n)/k!→1/k! as n→∞

(137 characters, uses fancy Unicode characters)

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First, my original tweet, on the essential content of IUT, i.e. how it intends to prove ABC, in 140 characters:

mochizuki: invented big deformation machine-tracks how deformed schemethry needs before HA-thry applies-amount of deformation IS Spziro ineq

Now, Twitter recently raised the character limit for all users to 280 characters, so with my whopping 140 extra characters, I will write a new style tweet of the same flavor:

Mochizuki: Invented deformation machine-IUT-which elim. obstruct'ns from applying fund.thm.of HAtheory to schemethry by deforming schemethry. By measuring distort. needed b4 FTHAT applies, ineq. appears-this is content of Spziro conj. hence ABC, modulo rigor check:IUT black box

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  • $\begingroup$ oh and I should say, HA theory is Hodge-Arakelov theory, Mochizuki's initial framework for solving ABC $\endgroup$
    – Samantha Y
    Commented Nov 17, 2017 at 20:21
  • $\begingroup$ This looks a good piece of ciphetext! $\endgroup$
    – vidyarthi
    Commented Dec 10, 2020 at 12:45
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Lattices with exponential kissing number discovered by Serge Vlăduţ. Another home run for algebraic-geometry codes. Link.

Actually, with the new 260 characters policy we can add:

Lattices with exponential kissing number discovered by Serge Vlăduţ. Another home run for algebraic-geometry codes. Will exponential improvement for Minkowski's 1905 lower bound for sphere packing be the next grand slam?

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The fundamental theorem of algebra: every polynomial splits in the field of complex numbers.

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Every consistent first-order theory has a model of countable size.

A set of sentences is consistent if and only if every finite subset is consistent.

A set of sentences has a model if and only if every finite subset of it has a model.

The first order logic is the only logic with a finitary syntax to possess the Löwenheim-Skolem property and be complete.

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Only nonbinary nontrivial perfect code is ternary w parity matrix rows 11122010000 11210201000 12101200100 12012100010 10221100001. Link

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You can always find a transversal line meeting a family of parallel line segments on the plane such that any 3 can be transversed.

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At a party with $4^n$ people and there are either $n$ people who all know each other, or $n$ who are all mutual strangers.

(I used crappy bounds so it could be easily stated. I know the upper bound for $R(n,n)$ is another answer.)

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    $\begingroup$ The upper bound, rather. $\endgroup$ Commented Apr 30, 2017 at 0:41
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    $\begingroup$ See also gilkalai.wordpress.com/2017/03/29/r55-%e2%89%a4-48 straight from OP's blog for the recent result that R(5,5) ≤ 48. The tweet could be "You can't have 48 people on a party such that no 5 are all acquainted and no 5 are all strangers." $\endgroup$ Commented Aug 27, 2017 at 19:15
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Erdös-Faber-Lovasz conjecture: If $n$ copies of $K_n$ have pairwise intersection of $\leq 1$, you can color all points with $n$ colors.

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  • $\begingroup$ I dont think it is already proved. Do you have a complete proof $\endgroup$
    – vidyarthi
    Commented Dec 10, 2020 at 12:43
  • $\begingroup$ That's right, this is an open problem. $\endgroup$ Commented Dec 11, 2020 at 13:19
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$\bar{\rho}$ irreducible Galois rep has finitely many lifts $\rho$ unramified outside of $S$. Proof: $(\rho_2^{-1}\rho_1\rho_2-\rho_1)/\mathcal{l}^r$ is a cocycle.

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A group G≠A_n,S_n with a core-free maximal subgroup of index n∈{266,506,759,1045,1288,1463,3795} is sporadic. Proof by GAP. Any other index?

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If $S^n$ x $S^n$ minus diag & antidiag self deforms and each (x,y) → (y,x) then n = 1, 3, 7, 15, 31, 63 or 127 (Kervaire invariant).

(Stolen from the epigraph of I. M. James, The topology of Stiefel manifolds (1976).)

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When playing poker with a quantum decks of cards, you can only look at one card at a time.

On first sight you might find three aces of hearts, and two of spades, but when you reveal your hand to claim the pot you suddenly have nothing but a pair of twos.


EDIT: More here: https://arxiv.org/abs/2104.02817

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  • $\begingroup$ This does not fit within 140 characters $\endgroup$
    – user44143
    Commented Dec 30, 2022 at 2:32
  • $\begingroup$ @MattF. question out of date. twitter.com/JP_Mc_C/status/… $\endgroup$ Commented Dec 30, 2022 at 8:58
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Sz(q) has two orbits more than PSL(2,q) under the action of its automorphism group - see https://doi.org/10.1081/AGB-120004501, Thm. 3.4.

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  • $\begingroup$ Just to follow the encouragement to post a result of one's own! $\endgroup$
    – Stefan Kohl
    Commented Apr 26, 2017 at 14:17
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Brooks' theorem and list-coloring variants can be proved using the combinatorial Nullstellensatz and a related theorem of Alon and Tarsi.

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Graham's number is so big, that its digits contain more information than can be contained within the volume of a human brain

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    $\begingroup$ Understatement of the century... $\endgroup$ Commented Apr 26, 2017 at 22:19
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    $\begingroup$ And of the last century. $\endgroup$ Commented Apr 27, 2017 at 1:15
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    $\begingroup$ But does it have a chance for the understatement of the millennium? $\endgroup$
    – Asaf Karagila
    Commented Apr 27, 2017 at 10:45
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    $\begingroup$ It's less of an understatement when you realize the word "volume" plays an important role: "It's a number so insanely, absurdly huge that storing all the digits of Graham's number in the brain could create a black hole, said John Baez, a mathematical physicist at the University of California, Irvine, who is researching big numbers. (Only so much information can be stored in a given amount of space, and trying to squish more matter into that space creates a black hole, he said.)" livescience.com/26870-ginormous-numbers-boggle-the-mind.html $\endgroup$ Commented Apr 29, 2017 at 23:24
  • $\begingroup$ this is perhaps three years too late but I wonder if John Baez is aware that he's "researching big numbers" $\endgroup$ Commented Dec 10, 2020 at 23:02
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There is no smooth surjection from $S^5$ to $S^6$. #Sard

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Let $u_0 \ge 2$ be a rational, and $u_{n+1}=⌊u_n⌋(u_n - ⌊u_n⌋ + 1)$.
Does the sequence $(u_n)$ reach an integer? Link.
#AlternativeToContinuedFraction

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  • $\begingroup$ Cool Tweetable Puzzle $\endgroup$ Commented Dec 10, 2020 at 11:56
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Cogito ergo sum #ReneDescartes

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  • $\begingroup$ I find it hilarious that my "tweet" was too short for MO. Andere Städtchen, andere Mädchen, I suppose? $\endgroup$ Commented Apr 26, 2017 at 21:14
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    $\begingroup$ Isn't that more in the area of Tweetable Philosophy? $\endgroup$ Commented Apr 26, 2017 at 21:54
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    $\begingroup$ Perhaps. Or metamathematics, as I would interpret it. Arguably, that sentence is also at the root of modern scientific method. To those downvoting this answer, I have another one for you: $$\ $$ You are not fun anymore! #MonthyPython $\endgroup$ Commented Apr 27, 2017 at 1:48
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    $\begingroup$ @Victor Protsak Wait, so "another roof, another proof" [attrib. Erdős] is a play on "andere Städtchen, andere Mädchen"? $\endgroup$ Commented May 1, 2017 at 18:49
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