# Proofs without words

Can you give examples of proofs without words? In particular, can you give examples of proofs without words for non-trivial results?

(One could ask if this is of interest to mathematicians, and I would say yes, in so far as the kind of little gems that usually fall under the title of 'proofs without words' is quite capable of providing the aesthetic rush we all so professionally appreciate. That is why we will sometimes stubbornly stare at one of these mathematical autostereograms with determination until we joyously see it.)

(I'll provide an answer as an example of what I have in mind in a second)

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where possible could people also either note the image source or explain/provide a link to a "how to" for constructing the associated diagram? I think that such would also be helpful for folks` –  Carter Tazio Schonwald Dec 14 '09 at 23:57
I hope I am not alone in being (usually) unable to appreciate "proof by picture"... –  Suvrit Jul 8 '11 at 21:14
@Suvrit: I hope I am not alone in being most often unable to appreciate "proof by word" until I've read it at least twenty times and wrestled with it for many days per page! –  WetSavannaAnimal aka Rod Vance Jul 9 '11 at 12:11
I am actually quite fond of this question, David! I tend to make comments on answers that are not relevant, and they have a tendency to get deleted after that. –  Mariano Suárez-Alvarez Sep 16 '11 at 17:34
My opinion is that almost every proof-without-words is improved by a few well-chosen words. –  Joel David Hamkins Feb 12 '12 at 0:47

A proof of the identity $$1+2+\cdots + (n-1) = \binom{n}{2}$$

(Adapted from an entry I saw at Wolfram Demonstrations)

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Wow ! –  Dinakar Muthiah Dec 19 '09 at 22:56
@Johann, people who thing that mathematics is about deducing theorems from axioms have such a mistaken idea of what the mathematical activity is thar their judgment is more or less irrelevant :D –  Mariano Suárez-Alvarez Jun 29 '10 at 13:05
Am I the only one who doesn't understand this "proof" at all? –  mathreader Oct 17 '10 at 17:07
@mathreader - the yellow dots are the sum of the first n numbers. Choosing two of the n+1 blue dots uniquely specifies a yellow dot in a bijective fashion. –  Steven Gubkin Nov 11 '10 at 13:40
This beautiful proof warrants proper attribution. It was discovered by Loren Larson, professor emeritus at St. Olaf College. He included it along with a number of other, more standard, proofs, in "A Discrete Look at 1+2+...+n," published in 1985 in The College Mathematics Journal (vol. 16, no. 5, pp. 369-382). –  Barry Cipra Oct 15 '11 at 2:17

As you probably already know — there are loads of these in Proofs without Words (and II) by Roger Nelson.

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This is elementary as well, but one of my favorite ones :)

$1^2 + 2^2 + \dots + n^2 = \frac13n(n+1)(n+\frac12)$

(Author: Man-Keung Siu)

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There's an analogous proof that the integral of n^2 from 0 to x is x^3/3. It can be obtained from this proof by smoothing out the stepped pyramids into actual pyramids. –  Michael Lugo Dec 14 '09 at 16:47
I think very few people have enough spatial imagination to figure out what happens exactly in the area where the three pieces come together, or could easily depict the structure seen from the opposite end. For me the picture is not convincing at all (I'd rather say the formula convinces me the picture is correct than the other way round). However maybe playing with an actual model would be quite convincing. –  Marc van Leeuwen Dec 12 '11 at 13:31
@Mark - I think if you just think about the width of each step at each level, you will be able to see that they do all fit together. Just counting back along a given row or column shows you that it all fits. –  Steven Gubkin Feb 15 '12 at 15:10
A variant of Mike's construction for $\sum_{k=1}^n k^2$, easier to visualize (I'm going to try a proof-without-words, without pictures). Take $6$ copies of each parallelepiped of size $k \times k \times 1$. Glue them together so as to make the four lateral walls of a parallelepiped of (external) size $k \times (k+1) \times (2k+1)$. Do this for k from 1 to n, forming a collection of bracelets. Insert each one in the next, like matrioskas, getting a whole parallelepiped of size $n\times(n+1)\times(2n+1)$. –  Pietro Majer Apr 10 '13 at 10:48
• The first homotopy group of SO_3 has an order 2 element (that's a classic).

• The surface area of a quarter of the unit sphere is Pi via Gauss-Bonnet (My source is Ariel Shaqed - it should have been a classic, but no one I asked seems to knew it). The sphere is what you reach with a straight hand while standing still. Hold a Pencil in your hand, that's your tangent vector. Now parallel transport the pencil on a quarter sphere: it points in the opposite direction. QED

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For SO(3) has order 2 element: gregegan.customer.netspace.net.au/APPLETS/21/21.html –  Dan Piponi Dec 14 '09 at 15:29
Place a glass on the open palm of your hand. You can, with a bit of practice, rotate the glass twice (but not once) around the vertical axis without spilling any liquid from it, and return to your original position. Each part of your body goes through a loop in SO_3. Moving from the shoulder via the arm to the glass, you get a homotopy essentially proving the theorem. I have seen dancers from somewhere in south-east Asia incorporating this move into their dance. –  Harald Hanche-Olsen Dec 14 '09 at 21:21
Why are there so many words and so few pictures in this answer? –  David Eppstein Dec 14 '09 at 23:07
@David: well, you can think if this answer (or of Harald's comment, which gets my emphatic upvote) as a script for the choreography which, when acted out, is a proof without words :P –  Mariano Suárez-Alvarez Dec 15 '09 at 0:00
It doesn't feature Feynman, but here's a video of a human doing the plate trick (just after 1 minute in): youtube.com/watch?v=CYBqIRM8GiY –  Harrison Brown Dec 15 '09 at 2:42

The cover of Peter Winkler's first book is a great proof without words of a statement which I'll leave you to guess, regarding the combinatorics of tiling a heaxagon with rhombi.

EDIT: I think the guessing game isn't helpful. The statement is that when tiling a perfect hexagon with the appropriate kind of rhombi of various orientations, the number of tiles in each orientation is the same. The image is slightly misleading in its use of color; there ought to be just three colors, corresponding to the three orientations.

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I'd be more impressed by this if I knew what statement was supposedly being proven by this illustration. That rhombus tilings are in 1-1 correspondence with 3d orthogonal surfaces (Thurston 1990, dx.doi.org/10.2307/2324578)? –  David Eppstein Dec 14 '09 at 23:06
Also, there are equal numbers of rhombi of each orientation in any tiling, and in fact, any tiling can be obtained from any other one by rotating "unit" hexagons formed by three rhombi. –  Darsh Ranjan Dec 15 '09 at 2:35
What do the colors represent? In particular, there are two colors for "upward-facing" rhombi (red and light gray) and two colors for "right-facing" rhombi (brown and dark gray), and I don't see why. –  Michael Lugo Dec 15 '09 at 3:03

In an attempt to push the bar towards the non-trivial, I'll mention the proof that the boundary complex of every polytope is shellable. The proof is virtually word-free but requires an actual movie rather than a still image: imagine yourself in a spaceship, taking off in a straight line from one of the facets, away from the polytope. Every once in a while a new facet is visible to you; under assumptions of general position, this provides a shelling of the complex (obviously, you need to fly off to projective infinity and come back on the other side).

This was assumed by Euler but first proved only in 1970 by Brugesser and Mani, who said that the idea came to him in a dream. More details here (search for "shellability") or here.

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Why are there so many words and so few pictures in this answer? –  David Eppstein Dec 14 '09 at 23:07
Because I couldn't a way to draw this, let alone animate, in a reasonable time. I trust that the description is helpful in imagining what the actual wordless proof is. –  Alon Amit Dec 15 '09 at 5:17
I want a video! –  Emil Jan 16 '10 at 22:45

Wikipedia has a few nice proofs of the pythagorean theorem. Elementary, but elegant.

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Pythagoras' theorem is trivial? I had no idea … Seriously, I don't necessarily think that the existence of a very simple proof implies triviality. Such proofs are, after all, not so easily discovered. Anyway, this is my favourite proof of the theorem. –  Harald Hanche-Olsen Dec 14 '09 at 20:58
The 20th President of the US, James Garfield, independently discovered the proof obtained by halving the right-hand diagram along a diagonal of the square of side length c. It requires you to write down an equation, though. That's my favorite proof, but mostly because of the corollary that B. Obama isn't the first geeky POTUS. –  Harrison Brown Dec 15 '09 at 3:23
@HB: Um, Thomas Jefferson? –  Pete L. Clark Mar 6 '10 at 3:23
A typical fake proof --- a simple statement as Pythagorean theorem is proved using much more advanced theorem on existence of area... –  Anton Petrunin Nov 30 '10 at 20:26
A typical fake refutation. You don't need to define Lebesgue measure to do manipulations in geometry. All operations can be defined geometrically if I associate a number X with the segment of length X, and define $X \mapsto X^2$ as a function, mapping a segment to a square with such side. In fact, even many of infinite summations can be done geometrically, using the obvious topology and metric on shapes. Thanks to this formalistic tradition it took 100 years of pain to get from non-trivial Lebesgue construction to much more natural motivic integration. –  Anton Fetisov Nov 13 '11 at 10:38

The cardinality of the real number line is the same as a finite open interval of the real number line.

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I suppose this picture can also be adapted to obtain the stereographic projection proof that a sphere is a manifold? –  Kevin H. Lin Dec 14 '09 at 23:47
I usually use Inkscape for my vector-based needs, but this was just done with my Smartboard presentation software. –  Jason Dyer Dec 15 '09 at 14:21

There's a picture proof in the Princeton Companion, or alternatively on p. 340 of Hatcher, of the fact that the higher homotopy groups are abelian. Actually, here's a screenshot of the one in Hatcher (hopefully fair-use!):

Here $f$ and $g$ are mappings (with basepoint) of $S^n$ into some space for $n > 1$; the picture shows a homotopy between $f + g$ and $g + f$.

The above diagrams show an application of the interchange law, a more general expression of the Eckmann-Hilton argument, for double categories or groupoids. Here is a more general picture

which shows that the interchange law for a double groupoid implies the second rule $v^{-1}uv= u^{\delta v}$, where in the picture $a=\delta v$, for the crossed module associated to a double groupoid, taken from the book advertised here. There are many $2$-dimensional rewriting arguments which are essential to the results of this book.

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Page 340 of Hatcher's book: math.cornell.edu/~hatcher/AT/AT.pdf –  Dan Piponi Dec 14 '09 at 18:27
This is sometimes called the Eckmann-Hilton argument: en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument –  Kevin H. Lin Dec 14 '09 at 20:46
I've heard that term, but I've never quite understood how the diagram is supposed to prove the more general abstract nonsense theorem. But if you can explain it, that's what community wiki's for! :D –  Harrison Brown Dec 14 '09 at 20:53
There are lots of places on the web where this is explained nicely: youtube.com/watch?v=Rjdo-RWQVIY , math.ucr.edu/home/baez/week258.html , ncatlab.org/nlab/show/Eckmann-Hilton+argument , etc.... –  Kevin H. Lin Dec 14 '09 at 23:50

Rich Schwartz had on his site a great paper consisting of only a picture which proved that every right triangle admits a periodic billiard path. Unfortunately, he's since deleted it, so I can't post it here. (It shouldn't take too long for anyone interested to re-construct the proof, though.)

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I am guessing he did it by assembling four of the said right-triangles into a parallelogram. There is a path that bounces directly between the two longer sides. Mod out by the symmetry and you get a periodic path in the triangle. –  Willie Wong Mar 11 '10 at 17:04

Here is the very first piece of original mathematics I ever did, in high school:

The derivative of sine is cosine.

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It looks like your image is no longer available... –  I. J. Kennedy Nov 16 '10 at 19:09

Sphere eversion

And here's a two-dimensional rendering of the sphere eversion:

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As pretty as it is, that is nowhere understandable as a proof. More as an illustration. –  Willie Wong Mar 11 '10 at 16:44
@Willie: Suppose someone wrote down the equations/formulas for the sphere eversion in that video. It seems to me that checking that the formulas indeed give a sphere eversion would be a rather difficult and tedious task, whereas a video animation is, although not a rigorous proof, much more immediately convincing. –  Kevin H. Lin Apr 6 '10 at 16:30
I just watched the video, which was excellent, but it had a lot of words in it. –  Patricia Hersh Aug 19 '12 at 0:23

Here's a proof of the inequality of the arithmetic and geometric means in the form $$\frac{x\_1^n}{n} + \cdots + \frac{x\_n^n}{n} \geq x\_1\cdots x\_n.$$

Proof for $n=3$:

The "figure" for general $n$ is similar, with $n$ right pyramids, one with an $(n-1)$-cube of side length $x\_k$ as its base and height $x\_k$ for each $k=1,\ldots,n$.

(I made this in Inkscape, a wonderful free-software vector drawing application. For the inequality and associated labels, I used the textext extension.)

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The box has volume xyz and is contained in the union of the three square pyramids, which respectively have volumes x^3/3, y^3/3, and z^3/3. Thus xyz <= x^3/3 + y^3/3 + z^3/3. –  Darsh Ranjan Nov 11 '10 at 3:41

Duality between $\ell^1$ and $\ell^\infty$ norms.

and the reverse animation

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I... don't quite get it. I think I need a few more words: What's the dot representing in each picture? –  Harrison Brown Dec 16 '09 at 15:01
The red line in xy-space satisfies the given equation. The dot gives the (a,b) coordinates of the same line in ab-space. The xy- and ab-spaces are linearly dual to each other. The resulting black and red shapes represent the unit balls in respective norms. –  Igor Khavkine Dec 16 '09 at 15:34

Q: Can you tile with ?

 

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I'm partial to the proof using Dandelin spheres that (certain) cross sections of cones are ellipses, where an ellipse is defined as the locus of points whose total distance to two foci is constant. It's particularly nice because it explains the foci geometrically, as well as the focus-directrix property with some more work.

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Yes, this one is beautiful. –  Andres Caicedo May 15 '10 at 18:47
@PatrickDaSilva: $PF1 = PP1$ because tangents to a circle/sphere have equal length. The total distance is thus equal to $PF1 + PF2 = PP1 + PP2 = P1P2$, which is constant. –  aorq Jul 29 '13 at 8:53

Have a look at this document from an MIT-instructor: http://mit.edu/18.098/book/extract2009-01-21.pdf

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There are a couple of Fibonacci identities, I think. For example

$F_0^2+F_1^2+\cdots+F_n^2=F_{n}F_{n+1}$, with $F_0=1$.

By puting together squares of side $F_n$, one at a time, you get a rectangle of dimension $F_nF_{n+1}$: The two squares of side 1, then the square of side 2, then the square of side 3 and so on.

Here is an image I found online

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fantastic ! –  Martin Brandenburg Apr 17 '10 at 23:30
I think that there is a nice pictorial proof for this fact, but I don't think this is it. It's a proof for a specific $n$. To make it a general proof, the inductive step needs to be illustrated. –  Max Mar 16 '11 at 14:08
@Max: The inductive step is easy to figure out, since the rectangle above contains the rectangles from previous steps. –  Daniel Litt Mar 16 '11 at 20:01

The sequence of pictures

proves the area formula for spherical triangles $A=\hat{ABC}+\hat{BCA}+\hat{CAB}-\pi$.

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Thomas Harriot first proved this formula in 1603, apparently by a similar argument, though I have not seen his picture(s). –  John Stillwell Feb 22 '10 at 22:31
Haha, I'm happy to see these illustrations useful to someone! I created them some years ago, mainly to crystalize what I saw in my minds eye after finding some simple proofs of this identity online. The words accompanying these images can be found at planetmath.org/encyclopedia/AreaOfASphericalTriangle.html Also, original MetaPost source can be obtained from this unfortunately obscure link: images.planetmath.org:8080/cache/objects/5841/src/sph-tri.mp –  Igor Khavkine Apr 26 '10 at 20:55
There is an analogous proof using the fact that although the hyperbolic plane has infinite area, a triply asymptotic triangle has finite area, so once you pick one of the two triply asymptotic triangles containing your triangle, you're in business. The relevant picture's in my answer posted separately (I posted it before I had the reputation to leave comments): mathoverflow.net/questions/8846/proofs-without-words/… –  Vaughn Climenhaga May 18 '10 at 19:04

There a proof of Erd˝os-Mordell Inequality 'without words' is an impressive one. Please follow the link http://forumgeom.fau.edu/FG2007volume7/FG200711.pdf

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Also elementary, but here is a proof that

$C_n = \binom{2n}{n} - \binom{2n}{n+1} = \frac{\binom{2n}{n}}{n+1},$

where $C_n$ is the $n$th Catalan number.

http://utdallas.edu/~hagge/images/Catalan.pdf

Sorry for the link; new users may not use image tags.

Here's the image:

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Do you have an explanation for the picture? I looked at it, and looked at it, and don't get it. –  Willie Wong Mar 11 '10 at 16:38
Sorry for not noticing your question (much) earlier. The differences between adjacent terms in Pascal's triangle form another triangle which obeys the same generation rules. In my picture of that triangle, the yellow squares count some of the downward paths on a square grid which has been rotated $45^\circ$, namely those that never fall to the left of the top square. One definition of $C_n$ is that it is the number of such paths which terminate at the bottom corner of an $n \times n$ grid. –  Tobias Hagge Oct 26 '10 at 5:46

This other answer shows that an 8x8 board with opposite squares removed cannot be tiled with dominoes, as they are of the same "colour". But what if two squares of opposite colours are removed? Ralph E. Gomory showed that it is always possible, no matter where the two removed squares are, and this is his proof.

(Imagine A and B are the squares removed.) The image is from Honsberger's Mathematical Gems I.

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Because I think proof by picture is potentially dangerous, I'll present a link to the standard proof that 32.5 = 31.5:

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I think it is just as easy to introduce some kind of logical gap in a written proof as in a graphical one. –  Steven Gubkin Mar 7 '10 at 23:41
@Steven: I think there is some truth to your claim, but I don't agree fully. First, we may notice that most proofs rely much more on writing than on pictures, and so mathematicians have developed a better radar for "written gaps". Second, there is a very strong sense in which written proofs may be formalized and checked by computer. Picture proofs, unless they share quite a bit of the "discrete" character of written proofs, usually are not amenable to such treatment. (And the notions of discreteness I can think of pretty much ensure that the picture proof could be turned into words.) –  Pietro KC May 15 '10 at 20:22
+1 for "the standard proof that $32.5 = 31.5$." Made me laugh. :) –  Quadrescence Oct 15 '10 at 20:10
@Pietro: “there is a very strong sense in which written proofs may be formalised”? Formalisation is a highly non-trivial task, and typically depends on quite a lot of mathematical background. What affects the difficulty is not whether the proof is written or graphical, but whether it’s detailed or highly abstracted. Formalising a good proof-by-picture is no harder than formalising a high-level written proof. Insofar as there’s a difference, I’d say it’s just that written proofs can be made detailed enough that formalising them is straightforward, whereas picture proofs perhaps can’t. –  Peter LeFanu Lumsdaine Nov 29 '10 at 1:04

The idea is to prove things in ways that are obvious to different parts of your brain, right? Anyone found any "auditory proofs"? Some candidates -

1. Nyquist sampling theorem?

2. sin[a] + sin[b] = 2sin[(a+b)/2]cos[(a-b)/2]. If you use at and bt instead of a and b, you can translate that to show how the addition of two sine tones close in frequency can also be perceived as a modulation or "vibrato" around the centre frequency. The factor of 2 might be hard, though you can add a gain instead of 2 and show that the difference is silence when the gain is 2 :)

3. Sampling in frequency domain (comb filter) is periodicity in time domain?

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Are there more details on 1? 2 and 3 don't seem like proofs so much as examples, though maybe you are just putting them forth as challenges. –  j.c. Mar 7 '10 at 11:54

This is apparently not was intended, but I think it qualifies. From Principia Mathematica: the proof of 1+1=2 (I can't include the image bc I'm a new user, but perhaps an experienced user can edit this answer for me.)

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If we have 3 circles on the plane with tangent lines, we can notice they have colinear intersection!

To prove it, we can visualize the same configuration in 3D, the balls lay on a surface and rather than tangent lines we take cones: The colinearity comes from the fact that if we lay a plane ontop of this configuration it will intersect the table in a line!

This is from 'curious and interesting geometry' and the proof is attributed to John Edson Sweet. I really like this proof because it gives a vivid example of the general idea that sometimes, to solve a problem in the most simple way you need to view it as a part of some bigger whole.

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You need to draw a 3D picture of this to get rid of the words! –  Ian Agol Jun 28 '11 at 16:39
In this pretty solution there is another pretty geometric problem: Given three spheres there is a plane which is tangent to all three. –  Rogelio Fernández-Alonso Jan 29 '12 at 16:51
Where is the picture? –  Patrick Da Silva Jul 28 '13 at 18:29
This was one of my favorite proofs in this list... it's a shame that imageshack took this picture off to promote their site. –  KalEl Nov 7 '14 at 22:36

This should really be a comment on Marco Radeschi's answer from Feb 22 involving the area formula for spherical triangles, but since I'm new here I don't have the reputation to leave comments yet.

In reply to Igor's comment (on Marco's answer) wondering about an analogous proof for the area formula of hyperbolic triangles: there is one along similar lines, and you're rescued from non-compactness by the fact that asymptotic triangles have finite area. In particular, the proof in the spherical case relies on the fact that the area of a double wedge with angle $\alpha$ is proportional to $\alpha$; in the hyperbolic case, you need to replace the double wedge with a doubly asymptotic triangle (one vertex in the hyperbolic plane and two vertices on the ideal boundary) and show that if the angle at the finite vertex is $\alpha$, then the area is proportional to $\pi - \alpha$. That follows from similar arguments to those in the spherical case (show that the area function depends affinely on $\alpha$ and use what you know about the cases $\alpha=0,\pi$).

Once you have that, then everything follows from the picture below, since you know the area of the triply asymptotic triangle and of the three (yellow, red, blue) doubly asymptotic triangles.

(That picture is slightly modified from p. 221 of this book, which has the whole proof in more detail.)

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Means inequalities:

The image was sent to me by James M. Lawrence, grazie! See also page 53 of "Proofs without words: exercises in visual thinking, Volume 2" for a very different layout of the same 4 inequalities.

Another one exists involving the sum $$1^3 + 2^3 + \cdots + n^3:$$

The second image is due to Brian Sears

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I used the second proof (involving sum of cubes) in my class today after proving it by induction. A few were quite inspired by it! –  Somnath Basu Feb 24 '12 at 18:42
2nd proof: It would be nicer if the small strips were above and to the left of the big square. –  Günter Rote Feb 25 '13 at 22:56

It's a long list of wonderful answers already, but I can't resist...

Question: Is it possible to find six points on a square lattice that form the vertices of a regular hexagon?

Proof without words:

Hint: A square lattice is invariant under rotation by π/2 around any lattice point. Use reductio ad absurdum.

Credit: I learned that proof from György Elekes during the Conjecture and Proof course in the Budapest Semesters in Mathematics, after constructing a proof of my own that used entirely too many words and made very laboured use of the fact that $\sqrt{3}$ is irrational. The picture here is my own creation (using Asymptote).

Follow-up: Can you find four points on a hexagonal lattice that form the vertices of a square? The proof is similar but not immediate.

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Why would you resist? –  Mariano Suárez-Alvarez May 20 '10 at 17:41
+1 for the "Conjecture & Proof" shout-out. Best, course, ever! –  Kevin O'Bryant Nov 10 '10 at 23:18
Igen, nagyon jó. –  Douglas Zare Feb 7 '11 at 5:08
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