Tweetable Mathematics 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.  
 A: Not deep, but if [0,1]² is cut in N triangles of equal area, N is even. If not, extend 2-adic valuation on Q to R, tricolor plane and apply Sperner.
A: No surjection f from S to its powerset. If so, let T = {x in S | x notin f(x)}. Must be some y such that f(y)=T. Then y in T iff y notin T!#
(140 characters, nothing fancy)
A: This question is actually much older than what it appears to be. The famous fundamental anagram of calculus (Newton, 1676)

6accdae13eff7i3l9n4o4qrr4s8t12ux

is an anagram of the Latin

Data aequatione quotcunque fluentes quantitates involvente, fluxiones invenire; et vice versa

(less than 140 symbols). In modern English it means

Given an equation involving any number of fluent quantities to find the fluxions, and vice versa

or, in Arnold's interpretation

It is useful to solve differential equations

A: This is based on a real tweet of Ryan O'Donnell on the proof of MIP*=RE which disproves Connes' Embedding Conjecture from 1976.

MIP* = RE, by Zhengfeng Ji, Anand Natarajan, Thomas Vidick, John
  Wright, Henry Yuen: https://arxiv.org/abs/2001.04383 . There is a
  multiple-entagled-quantum-provers proof system for the Halting
  Problem, and Connes' Embedding Conjecture is false.

A: Chebyshev said, and I say it again,
there is always a prime between n and 2n.
#BertrandsPostulate
A: Integral of exp(-x²) dx over R = Γ(1/2) = √π.
Proof: square the Gaussian integral and use polar coordinates!
A: Gödel's first incompleteness theorem - any consistent formal system capable of basic arithmetic is incomplete. Link
A: If Ax = y underdetermined w unique sparse soln x0, minimizing L1(x) recovers x0 under mild conds. Shannon-Nyquist bad (or sick) thm! Link
A: Few things multiply nicely (division rings usually noncommutative) but few things multiply nicely (finite ones are) #Wedderburn
A: A rectangle tiled by rectangles with an integral side has an integral side because it has mass zero under a certain periodic signed measure.
(Tweeted here)
A: How about a proof that, given a set, there always exists a bigger set?

For every set A, there is a set that doesn't inject into A. Take the
  set B of ordinals that inject into A. B is an ordinal and is not in B.

I think one can also fit a proof of Sylvester's theorem in 137 characters.

Take n points not on a line. Let L be a line containing >1 points
  minimizing the distance to a point off L. It contains exactly 2
  points.

A: Littlewood's example of a 2 line dissertation fits here, I believe.
An integral function never 0 or 1 is a constant:
$\exp(i\Omega(f(z)))$ is a bounded integral function 
A: Cole 1903: 2^67-1=147,573,952,589,676,412,927=193,707,721 × 761,838,257,287  #MoreThanThreeYearsOfSundays
A: Every set of integers with divergent reciprocals contains 3-term arithmetic progressions. Erdos and Turan conjectured; Bloom and Sisask proved; Bateman and Katz's method for cap set used.  link.
#TWTMTH
A: Erdos: You can color edges between √2ᵏ vertices red/blue so no monochrome size k subgraphs (Pick a random coloring.  It probably works)
A: There is an algorithm to compute étale cohomology in finite time (but we don't know how long it takes) — link
A: 
Weyl's law: pack a domain with tiny squares. In the limit, hear the domain's volume.

and

Polya's theorem: pack a square with tiny domains. In the limit, the domain is always higher-pitched than Weyl's law thinks it should be.

A: Cap set problem solved: polynomial method, punchline going back to (a+b)²=a²+b²+2ab. link 
A: A finite(dimensional) domain has to be a field: all nonzero "multiply-by-an-element" maps are self-embeddings.
A: The n-th binary digit of π is calculable without calculating all the previous digits.
A: There is a way of pretending that any reduced ring is Noetherian and a field. Grothendieck's generic freeness lemma is then quite easy. link (Section 11.5)
A: **ΕΥΡΗΚΑ num=Δ+Δ+Δ
(Every positive integer can be written as the sum of three triangular numbers, Gauss, July 10, 1796)
A: $S$ unit sphere of an $\infty$-dim Banach isn't compact. Pf: for $H$ closed hyperplane, $\bigcap(S\cap H)=\varnothing$ (Hahn-Banach), but no finite subintersection is empty.
Note: $\infty$, $\bigcap$, $\cap$, $\neq$, and $\varnothing$ are unicode characters, so this is actually tweetable!
A: There are infinitely many primes p for which there is a prime between p+1 and p+246.
A: Euclid: there are infinitely many primes, if not, multiply all and add 1. Link
If only they had Twitter back then. They had so many tweetable proofs :)
A: 33 = 8866128975287528³ + (-8778405442862239)³ + (-2736111468807040)³. Andrew Booker! link.
Update 2019-09-05: $(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3 = 42$, https://people.maths.bris.ac.uk/~maarb/. This completes all the numbers less than $114$.
A: Stark-Heegner theorem: There are nine imaginary quadratic fields with class number one. Link
A: I can't resist. Birkhoff's theorem:
For an erg measure-pres trans, orbit-wise avgs of a fn agree with the spatial avgs.
A: Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos & generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.
Not in English, 268 characters so two tweets, @Glorfindel 's comment,  but I couldn't not post it.
A: Trichotomy: the rational points on an algebraic curve are parametrizable; form a finitely-generated abelian group; or form a finite set.
A: The square of the hypotenuse is equal to the sum of the squares of the other two sides.
A: Every non-constant complex polynomial has a complex root : If not the inverse is bounded analytic. Use Liouville. #FundamentalTheoremOfAlgebra
A: 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
A: 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
A: "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!
A: One cannot hear the shape of a drum. link
Proof via Sunada's Theorem.
A: 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) $.
A: The most beautifully useless theorem of mathematics IMO:
"This statement is a theorem (and moreover, I can prove it)" — link
A: $AB-BA=I$ has no solution in finite matrices. Pf: Trace it!
A: $n!$ divides $(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1})$.
Proof: $\mathfrak{S}_n$ embeds in $\mathrm{GL}_n(\mathbf{F}_2)$. $\quad \square$
(I have read this elegant justification on MO someday, yet I do not find it now.)
A: Halting problem - there is no computer program which can determine if an arbitrary computer program halts on a specified input. Link
A: ZFC + "exists a Reinhardt cardinal" is inconsistent.
V = L implies that measurable cardinals do not exist.
A: 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
A: Elliptic curves produce a key exchange that may be safe against quantum computers. link
A: The smallest positive integer not definable in under sixty letters.
A: The fundamental theorem of algebra: every polynomial splits in the field of complex numbers.
A: 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)
A: 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

A: 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?
A: You cannot comb a hairy ball, should you wish to.
A: 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.
A: Only nonbinary nontrivial perfect code is ternary w parity matrix rows 11122010000 11210201000 12101200100 12012100010 10221100001. Link
A: 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.
A: You can always find a transversal line meeting a family of parallel line segments on the plane such that any 3 can be transversed.
A: 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.)
A: Erdös-Faber-Lovasz conjecture: If $n$ copies of $K_n$ have pairwise intersection of $\leq 1$, you can color all points with $n$ colors.
A: 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?
A: 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).)
A: 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
A: Brooks' theorem and list-coloring variants can be proved using the combinatorial Nullstellensatz and a related theorem of Alon and Tarsi.
A: Sphere packing-The E8 lattice provides the optimal packing in eight-dimensional space. Link
A: Mihăilescu's theorem (ex Catalan conjecture) - $8$ and $9$ are the only consecutive proper powers of natural numbers. Link
A: Borsuk's conjecture - a counterexample for DIM>2000, the tensor product of the unit sphere with itself. Link 
A: Wedderburn's theorem-Every finite skew field is a field. Pf: class equation on centralizers and cyclotomic irreducibility. Link
A: Four color theorem: any planar graph can be colored with 4 colors. Only proof by computer. SAD.
A: Turing discovers problems that even computers cannot solve. Link
A: Cantor's theorem!

The cardinality of the reals is greater than that of the natural numbers.

Also, various incompleteness results along the lines of the Mathematical T-Rex:

The continuum hypothesis is neither provable nor disprovable from ZFC, unless ZFC is inconsistent to begin with!

(The above can be extended to include a myriad of choice-related statements and ZF, for example.)
A: One more:
CFSG - The largest sporadic simple group has order 808017424794512875886459904961710757005754368000000000. 
A: Breaking News: A mathematician blew up 6 points on a plane! You won't believe what happened next...
A: Of course, every finite group of odd order is solvable.
A: Lindemann's theorem - You need more than a compass and a ruler to square a circle. Link
Edit Or if we really want the essence of the argument, on which a whole lot of stronger results have been patterned since:
Exploiting the properties of the exponential function, if $\pi$ was algebraic there'd be rational integer strictly between 0 and 1.
A: Invert 2×2 matrix A?  Easy: switch diagonal entries, negate off-diags,
& divide by Δ=det(A).
Pf: divide A²-tA+ΔI=0 by A, solve for 1/A !
MO link
A: One among $\pi+e$ and $\pi e$ is irrational. Proof: If not, then $(x+e)(x+\pi) \in \mathbb{Q}[x]$ and hence $e$ and $\pi$ would be algebraic.
NOTE: It is still open whether either of $\pi+e$ and $\pi e$ is irrational.
A: This is actually a real tweet by Ryan O'Donnell on Huang's proof of the sensitivity conjecture. 

Hao Huang@Emory: 
Ex.1: ∃edge-signing of n-cube with 2^{n-1} eigs each of +/-sqrt(n)
Interlacing=>Any induced subgraph with >2^{n-1} vtcs has max eig >=
  sqrt(n)
Ex.2: In subgraph, max eig <= max valency, even with signs
Hence [GL92] the Sensitivity Conj, s(f) >= sqrt(deg(f))

A: det(exp(X)) = exp(tr(X)). Remarkably, the RHS does not involve off-diagonal elements of X.
A: The prime numbers contain arbitrarily long arithmetic progressions.
A: 1_000_000 $ problem solved: every simply connected, closed 3-manifold is homeomorphic to the 3-sphere. Hint: use the Ricci flow.
A: Lagrange: every integer is sum of 4 squares. Pf: 4-squares identity + every real prime splits in the ring of Hurwitz quaternions. Link
A: 27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵. Nice try, Euler.  link
#Counterexamples
A: Banach–Tarski paradox: a solid sphere can be divided in a finite # of parts which can be joined to form 2 spheres identical to the original.
A: Here are three.
Wiles theorem: Every elliptic curve over Q is parameterized by modular functions.
Faltings Theorem/Mordell Conjecture: A curve of genus at least 2 has only finitely many rational points.
Faltings Theorem/Shafarevich Conj: There are only finitely many abelian varieties with good reduction
outside a given finite set of primes.
A: Borsuk–Ulam theorem:
If you crumple a ball and run a steam roller over it there are at least two antipodal points touching.
A: The sentence "The first positive integer that cannot be specified in a 140 character tweet" doesn't specify a well defined integer. 
A: Every rational r is xyz(x+y+z) for some rational x,y,z.
Proof: Euler (1749) found x(r),y(r),z(r).  Nobody knows how.
I have a guess.
https://people.math.harvard.edu/~elkies/euler_14t.pdf
A: Graham's number is so big, that its digits contain more information than can be contained within the volume of a human brain
A: There is no smooth surjection from $S^5$ to $S^6$. #Sard
A: $\bar{\rho}$ 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.
A: 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
A:  Cogito ergo sum #ReneDescartes 
