# Surprising identities / equations

What are some surprising equations/identities that you have seen, which you would not have expected?

This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc.

I'd request to avoid 'standard' / well-known results like $$e^{i \pi} + 1 = 0$$.

Please write a single identity (or group of identities) in each answer.

I found this list of Funny identities, in which there is some overlap.

• I really can't believe no one has posted this yet: xkcd.com/687 Sep 26, 2013 at 14:48
• This is not in line with what you are looking for, but as a child I discovered that 10million pi is the number of seconds in a year to 1/2% accuracy. This is useful for quick back of envelope calculations, where seconds are involved. Sep 26, 2013 at 19:48
• Pi seconds is a nanocentury! Oct 1, 2013 at 21:59
• The three trigonometric identities in the following exercises of my Wikibook: en.wikibooks.org/wiki/On_2D_Inverse_Problems/…
– DVD
Oct 12, 2015 at 2:51
• This shouldn't be closed. There are still so many nice equations.In spherical trigonometry $$\frac{sin(A)}{sin(a)}=\frac{sin(B)}{sin(b)}=\frac{sin(B)}{sin(b)}$$ where the capital letters are the angles and lowercase are the opposite sides.
– skan
Nov 26, 2016 at 17:12

This one by Ramanujan gives me the goosebumps:

$$\frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}.$$

P.S. Just to make this more intriguing, define the fundamental unit $U_{29} = \frac{5+\sqrt{29}}{2}$ and fundamental solutions to Pell equations,

$$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$

$$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$

$$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$

then we can see those integers all over the formula as,

$$\frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k} = \frac{1}{\pi}$$

Nice, eh?

• Ramanujan used to say a goddess revealed those identities to him in dreams en.wikipedia.org/wiki/… Sep 26, 2013 at 20:32
• What made Ramanujuan think like this? How did he arrive at this equation? Sep 27, 2013 at 12:02
• One needs to realize that $26390 =5 \times 7\times 13\times 58$ and $9801=99 \times 99$ and $396=4 \times99$. Then, it's obvious. ;-) Sep 27, 2013 at 19:09
• How is this formula not immediately obvious? Oct 2, 2013 at 10:47
• Can anyone tell me how one obtains the equation? I can't seem to figure out why is this so. Dec 10, 2013 at 20:34

${1\over 2} < \left\lfloor \mathrm{mod}\left(\left\lfloor {y \over 17} \right\rfloor 2^{-17 \lfloor x \rfloor - \mathrm{mod}(\lfloor y\rfloor, 17)},2\right)\right\rfloor$

The above is the most interesting inequality in mathematics. If you plot it so that areas satisfying the inequality are shaded, this is what you get: This is known as Tupper's self referential formula.

• @GottfriedHelms To be fair, a large portion of the miracle is hidden in the labelling of the $y$-axis. Sep 28, 2013 at 14:51
• "The formula itself is a general purpose method of decoding a bitmap stored in the constant $k$, so it could actually be used to draw any other image. When applied to the unbounded positive range $0 \le y$, the formula tiles a vertical swath of the plane with a pattern that contains all possible 17-pixel-tall bitmaps. One horizontal slice of that infinite bitmap depicts the drawing formula itself, but this is not remarkable since other slices depict all other possible formulae that might fit in a 17-pixel-tall bitmap." (Wikipedia) Sep 28, 2013 at 20:18
• c.f. “The Library of Babel” by Jorge Luis Borges arts.ucsb.edu/faculty/reese/classes/artistsbooks/… Jan 3, 2015 at 10:23
• Is there a derivation of how such function is constructed which graph itself ? Mar 15, 2021 at 1:16

$\mathrm{GCD}(F_{n},F_{m}) = F_{\mathrm{GCD}(n,m)}$ where $F_n$ is the $n$th Fibonacci number.

• Please provide a link to the proof! May 19, 2016 at 17:34
• @N.S.JOHN You only need basic number theory to prove this. Oct 7, 2017 at 20:03
• @N.S.JOHN Hint: consider the Fibonacci sequence mod $d$ for some $d$. Show that the zeros must be evenly spaced. Mar 24, 2019 at 9:05

Taken from the first question I posed upon joining M.SE:

Define a function $f(\alpha, \beta)$, $\alpha \in (-1,1)$, $\beta \in (-1,1)$ as

$$f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$

You can use, for example, the Residue Theorem to show that

$$f(\alpha, \beta) = \frac{\pi \sin{\pi \alpha \beta}}{ \sin{\pi \alpha} \sin{\pi \beta}}$$

Clearly, from this latter expression, $f(\alpha, \beta) = f(\beta, \alpha)$. But from where does such a symmetric result come? The integral itself does not lend itself to predicting any such symmetry so far as I (and many others so far) can see.

• I think you can transform the integral using complex numbers ($\mathrm{exp(iy)}$) into something more explicitly symmetrical... Feb 21, 2019 at 10:17

$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$

was surprising to me when I saw it for the first time.

• This is one of those times when you just have to accept Euler was a god among men. They had been trying to solve the sum of inverse squares for a very long time, and Euler just solved it matter-of-factly along with several other relations using roots of trigonometric taylor series and some simple substitutions. Sep 27, 2013 at 5:35
• How did he find these series summations? I know they are trivial once you have Fourier decomposition, but to get them without having Fourier series... Nov 19, 2013 at 14:46
• @finitud There is a nice article on this in the book Journey through Genius: The Great Theorems of Mathematics, Chapter 9: The Extraordinary Sums of Leonhard Euler. If you search the chapter title you can find several references that essentially tells the same story. It's a great read (and not lengthy)! Feb 24, 2014 at 10:41

$$10^2+11^2+12^2=13^2+14^2$$ I found that one stunning.

P.S. In general, for $n>0$, the sum of $n+1$ consecutive squares starting with $x_1 = 2n^2+n$ is equal to $n$ consecutive squares starting with $y_1 = x_1+(n+1)$. Hence,

$$3^2+4^2 = 5^2$$

$$10^2+11^2+12^2=13^2+14^2$$

$$21^2+22^2+23^2+24^2 = 25^2+26^2+27^2$$

and so on.

• One can go a step further and generalize: $\sum_{n=0}^a(2a^2+a+n)^2=\sum_{n=a+1}^{2a}(2a^2+a+n)^2$ Sep 27, 2013 at 2:55
• Stunning indeed. Because of $3^3+4^3+5^3=6^3$ (see other answer) can this be generalized even further? In the exponent? Sep 27, 2013 at 8:24
• Do you know of a way to prove this, other than "expand all the terms"? Sep 27, 2013 at 13:13
• @TobiMcNamobi Open your eyes and look at around your question! Oct 2, 2013 at 3:03
• abstrusegoose.com/63 Mar 19, 2014 at 19:50

By far my favorite identity: $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$

The fun part about this one (for me) is that it looks absolutely false at first glance. They both evaluate to $\pi$.

• +1 Oh gods, I can just imagine setting this as a question, and having students be frustrated with proving it via substitution. Let $x = y^2$ would be the start of most of their answers. Sep 27, 2013 at 13:45
• +1: Of course I wondered what the next powers would return and they returned all fractions of $\pi$ (starting with $p=1$) : $$1,1,\frac 34, \frac 23,\frac {115}{192},\frac {11}{10},\cdots$$ OEIS. Sep 29, 2013 at 9:08
• @RaymondManzoni Your last fraction is a typo: it should be $11 \over 20$. Sep 30, 2013 at 19:10
• You are right @MarkHurd but I can't correct that. Btw the denominators are here. Sep 30, 2013 at 20:19
• @skan: use integration by parts, starting from the right-hand side. Let $u=\sin^2(x)$, and $dv = x^{-2}dx$. Look at trig identities that involve the product $\sin(x)\cos(x)$. If you want to evaluate the integral, you will probably need complex analysis. Nov 27, 2016 at 10:43

This is slightly contrived, but consider a situation where you have two balls, of mass $M$ and $m$, where $M=16\times100^N\times m$ for some integer $N$. The balls are placed against a wall as shown: We push the heavy ball towards the lighter one and the wall. The balls are assumed to collide elastically with the wall and with each other. The smaller ball bounces off the larger ball, hits the wall and bounces back. At this point there are two possible solutions: the balls collide with each other infinitely many times until the larger ball reaches the wall (assume they have no size), or the collisions from the smaller ball eventually cause the larger ball to turn around and start heading in the other direction - away from the wall.

In fact, it is the second scenario which occurs: the larger ball eventually heads away from the wall. Denote by $p(N)$ the number of collisions between the two balls before the larger one changes direction, and gaze in astonishment at the values of $p(N)$ for various $N$:

\begin{align} p(0)&=3\\ p(1)&=31\\ p(2)&=314\\ p(3)&=3141\\ p(4)&=31415\\ p(5)&=314159\\ \end{align}

and so on. $p(N)$ is the first $N+1$ digits of $\pi$!

This can be made to work in other bases in the obvious way.

See 'Playing Pool with $\pi$' by Gregory Galperin.

• When the mass ratio gets large enough, the larger ball is going to have such a large radius that it will simply roll completely over the small ball.... unless you also allow infinite density, which I guess is a small gnat to swallow, given the camel of completely elastic collisions. Sep 26, 2013 at 18:22
• It must also be assumed that the balls do not roll, but rather slide without friction. (Of course, the point-mass idealization takes care of that too.) Sep 27, 2013 at 13:53
• As a rule of thumb, you will normally get less beautiful mathematical results if you start getting pedantic about what might be called real-world considerations. Sep 27, 2013 at 14:06
• Can you do a sample calculation of p(4) please?
– Neil
Apr 6, 2015 at 10:59

$3^3 + 4^3 + 5^3 = 6^3$.

Also,

$1/89 = 0.01 + 0.001 + 0.0002 + 0.00003 + 0.000005 + 0.0000008 + 0.00000013 + \cdots$.

Let $S = \sum \frac{F_n} {k^n}$. Then $S + kS = 1 + \sum \frac{ F_{n} + F_{n-1} } {k^n} = 1 + \sum \frac {F_{n+1}}{k^n} = 1 + k^2S -1 - k$

In particular, for $k=10$, we get $S = \frac{10}{89}$. Divide by 10 to get the second equation.

• +1 for the second one... wat? How does that even work? Does something similar work in other bases? Sep 26, 2013 at 3:00
• @fhyve Yes it does. $\frac{F_n}{k^n}$ is rational, which can be shown in the normal way. Sep 26, 2013 at 3:46
• @CalvinLin I don't think that's the formula here, though... otherwise the last and second-to-last term above would have equal numbers of leading zeros. Sep 26, 2013 at 3:47
• 89 is also a Fibonacci number. Sep 26, 2013 at 4:27
• The zeroes in your sum are off. It should be $...+0.00000013 + ...$. You have one too many. Sep 26, 2013 at 11:24

$$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac4{15}\right)}{\Gamma\left(\frac13\right)\Gamma\left(\frac2{15}\right)}=\frac{\sqrt2\,\,\sqrt3}{\sqrt5\,\sqrt{5-\frac{7}{\sqrt{5}}+\sqrt{6-\frac{6}{\sqrt{5}}}}}=\frac{\phi \,\, \sqrt3 \,\, \sqrt{\!\sqrt 3 \cdot \sqrt 5-\phi^{3/2}}}{\sqrt 2 \,\, \sqrt 5}$$

• This is the most unexpected identity I've seen in my life, and I would like to award the bounty to this answer. Oct 3, 2013 at 18:04
• In case you haven't seen this simplification of a similar Gamma ratio (admittedly with a simpler answer): math.stackexchange.com/questions/406200/… Oct 3, 2013 at 18:29
• Many thanks. Of course, one wonders how on earth your posted result is derived. Oct 3, 2013 at 19:02
• @RonGordon I wish it were my result, but actually it is from an amazing paper Raimundas Vidūnas, Expressions for values of the gamma function, which contains much more than that. Oct 3, 2013 at 19:10
• Whoa! Now that is a paper! Oct 3, 2013 at 19:18

Where $$\varphi = \frac{1 + \sqrt{5}}{2}$$ a golden ratio, $$\int_0^\infty\frac{1}{(1+x^\varphi)^\varphi}\mathrm dx = 1.$$

This follows immediately from the substitution $$t=[x^{\varphi}(1+x^{\varphi})^{-1}]^{\varphi}$$.

Proof (below) by filmor

\begin{align} \int_0^\infty\frac{1}{(1+x^\varphi)^\varphi}\mathrm dx &= \varphi^{-1}\int_0^\infty\frac{y^{\varphi^{-1} - 1}}{(1+y)^\varphi}\mathrm dy \\ &= \varphi^{-1}\int_0^\infty\frac{y^{\varphi - 2}}{(1+y)^\varphi}\mathrm dy \\ &= \varphi^{-1} B\bigl(\varphi - 1, 1\bigr) \\ &= \varphi^{-1}\frac{\Gamma(\varphi-1)\ \Gamma(1)}{\Gamma(\varphi)} \\ &= \varphi^{-1} \frac{1}{\varphi - 1} = 1 \end{align}

One more thing with golden ratio: by Ramanujan, $$r=\dfrac{e^{-2\pi/5}}{1 + \dfrac{e^{-2\pi}}{ 1 + \dfrac{e^{-4\pi}}{1 + \cdots}}} = \sqrt{ \sqrt{5}\varphi} - \varphi$$

and even more bizarrely (found based on the work of Vidunas), the hypergeometric function $$N=\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},1\big)$$ is a deg-80 algebraic number given by,

$$N=\frac{1}{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{1/20}}$$

• Could you add links to proofs of these results? Sep 26, 2013 at 4:30
• Sorry, I cannot. I know just the result. Is there anyone other than me? Sep 26, 2013 at 4:33
• Just added one :) Sep 26, 2013 at 11:59
• Your second equation involves the Rogers-Ramanujan continued fraction (it's a rearrangement of Eq.19 at that link). The paper The Rogers-Ramanujan continued fraction, by Berndt et al (pp 2-3) says the first proof of that equation was by G. N. Watson, in his Theorems stated by Ramanujan (VII): Theorems on continued fractions, J. London Math. Soc. 4 (1929), 39–48. A more general result is proved in a paper by Kang. Sep 27, 2013 at 3:36
• @filmor I added your proof to the post, someone delete it idk why... It's a really nice proof. Mar 3, 2021 at 0:04

If $A+B+C=180^\circ$ then $$\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$$

Here is a mathematical scherzo.

$$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3.$$

$\displaystyle\sum_{k=1}^{24} k^2=70^2$ is novel.

• such a nontrivial pair (24, 70) is unique. Sep 26, 2013 at 3:52
• It is also critical in the theory of the Leech lattice. See page 130, Theorem 4.5, in Lattices and Codes by Wolfgang Ebeling, second edition. Or see SPLAG, by Conway and Sloane, page 524 in Chapter 26, leading up to Theorem 3; chapter title Lorentzian Forms for the Leech Lattice. Sep 26, 2013 at 4:47
• The cannonball problem. Sep 26, 2013 at 14:55
• Related to sphere packing constant in 24-dimensional Euclidean space ie. the generalisation of the Kepler conjecture to dimension 24.
– Tom
Sep 26, 2019 at 19:13

$${a \over b} = {c \over d} \quad\Longrightarrow\quad {a + b\over a - b} = {c + d \over c - d}$$

• I have to admit I was way too surprised when I learnt this simple identity. Oct 13, 2013 at 11:35
• This is true iff $a\neq b$. More generally, $\frac{a}{b}=\frac{c}{d}\implies \frac{a+kb}{a-kb}=\frac{c+kd}{c-kd}$, which is true iff $a\neq kb$. Apr 3, 2015 at 12:24
• $\frac{a+b}{a-b}=\frac{a/b+1}{a/b-1}=\frac{c/d+1}{c/d-1}=\frac{c+d}{c-d}$ Aug 17, 2015 at 20:12
• Even more generally, $\frac{a}{b}=\frac{c}{d}$ implies $\frac{f(a,b)}{g(a,b)}=\frac{f(c,d)}{g(c,d)}$ where $f$ and $g$ are any two homogeneous polynomials of the same degree. Aug 9, 2017 at 6:24
• @RounakSarkar the only condition is that $g(a,b) \ne 0$. In fact the condition that they're polynomials can be lifted, they just need to be homogenous. Proof is simple, suppose $f,g$ have degree of homogeneity $n$. Since $\frac{a}{b}=\frac{c}{d}$ we can write $a=kc$ and $b=kd$ and make $\frac{f(a,b)}{g(a,b)} = \frac{f(kc,kd)}{g(kc,kd)} = \frac{k^nf(c,d)}{k^ng(c,d)}= \frac{f(c,d)}{g(c,d)}$. Jan 21, 2022 at 11:27

$$\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^\pi}dx=\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^e}dx$$

• Wow. Is this one of those things that generalize to any exponent or is it only $\pi$ and $e$ ? Nov 22, 2014 at 17:45
• It is true for all exponents for which the integral exists. Mar 15, 2015 at 1:18
• School students will say $\pi = e$ :P Jan 19, 2017 at 0:30
• What technique do you use to show this? Contour integration, series expansion, or some nifty substitution? @Vladimir Reshetnikov Feb 16, 2022 at 15:41
• @FShrike Substitute $t = \tan x$ and $t' = \frac{\pi}{2} - t$. Change the cotangent to tangents and then you'll be able to show that $I = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} dt$.
– Vue
Apr 10, 2022 at 16:12

When I began my serious encounter with number theory and looked at properties of prominent combinatorical matrices I found this identity. This impressed me so much (even a bit philosophically) that I wanted to printed it on a t-shirt (but the white-on-black printing was then too expensive). The german phrase means "the exponential of the counting is the binomial"

Here is, how it looked asymptotically: • Why on earth would you want that on a shirt? Oct 22, 2013 at 15:58
• @Phap - well, why? Perhaps it was something, which I had myself discovered, and it had a much philosophical resembling: the human ability of counting, ... the step to the (otherwise) ubiquituous binomial numbers and the again ubiquituous exponential function - that was something really magic to me. Oct 22, 2013 at 17:28
• @GottfriedHelms: So it won't be expensive, why not black print on a white shirt? :) Jan 8, 2014 at 5:50
• Make a kickstarter and I will donate to help you fulfill your dream of having that shirt :) But don't be cheap with it also add in costs for matching pants, maybe a suit and a tie also :)
– Neil
Apr 6, 2015 at 5:33
• This is an interesting identity. It has a natural interpretation after you identify vectors with coefficients of polynomials/power series. Then the transpose of the left corresponds to differentiation D, and the transpose of the right corresponds to the "shift" map T : f(x) -> f(x+1), and the identity becomes e^D = T; i.e. Taylor's theorem Feb 7, 2017 at 8:03
• how do you derive that? Jul 8, 2014 at 9:58
• @athos the Wikipedia page on the topic has multiple proofs Jan 23, 2017 at 15:46

This bit of notational juggling may cause one do double take...

$$\huge \sqrt[\sqrt{2}]{2} = \sqrt{2}^\sqrt{2}$$

By definition, the LHS is the number $x$ such that $x^\sqrt{2} = 2$. It is simple to check that the RHS also has this property.

• At first sight. I thought that the LHS is referring to tetration. Oct 18, 2021 at 3:58

Easy geometric series but I found this one charming when I found out:

1/7 = 0,142857...
= 0,14 +
0,0028 +
0,000056 +
0,00000112 +
0,0000000224 + ... (double the value and shift it by two spaces)

• Quite generalizable, 7 * 14 = 100 - 2, therefore you start with 14, divide by 100 and multiply by 2. You could also use 7 * 143 = 1000 - -1 for an even simpler series. Oct 2, 2013 at 11:35
• It would read like this: $$\frac{1}{7}=7\sum_{k=0}^{\infty}\frac{2k}{10^{2k}}$$ Or, as I prefer it $$\frac{1}{49}=\sum_{k=0}^{\infty}\frac{2k}{10^{2k}}$$ Jun 24, 2018 at 12:29

Do logic answers count? I like the Drinker Paradox, which isn't really a paradox but actually a theorem of logic:

$\exists x.\ [D(x) \rightarrow \forall y.\ D(y)]$

For every bar there is a person for whom, if that person is drinking, then everyone is drinking.

• I've seen this in the context of the Riemann Hypothesis: there exists a real number $c>0$ such that if $\zeta$ has no roots $\rho$ off of the critical line with $\vert \mathrm{Im} \rho \vert < c$, then the Riemann Hypothesis is true. Sep 27, 2013 at 5:22
• well, "In every bar there is something for ..." (we should at least say "someone"). Saying "there is a person" is too strong, since the bar might be empty at the moment (FOL supposes that domain is non-empty, but predicates such as "x is as person" are allowed to have an empty extension) Sep 27, 2013 at 8:04
• @LukaMikec In every populated bar…? You can't very well say something is drinking. That just sounds creepy. Sep 27, 2013 at 15:13
• It's ok if it's creepy, being creepy shouldn't be too strange to mathematicians :) But we could say something like "In every bar there is something or someone for which it holds that if that something or someone is a person who drinks, then..." (or something shorter, but the point is that the "x" in the formula above might not be a person, it might be a number or whatever) Sep 27, 2013 at 19:14
• If you take the statement to mean "there exists a person in the bar" then it does need to be a populated bar. If you take the statement to mean "there exists a person somewhere in the universe for which..." then it doesn't need to be a populated bar. Perhaps I should have said "for every bar", I'll change that now. Sep 28, 2013 at 7:48

$$\ \ \ \ \ 2592=2^59^2\ \ \ \ \$$

$$\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}dx=\dfrac{\pi e}{4!}$$

• I'm not too surprised that this would come out to something like that (though I would never guess that in particular). It is a fairly complicated integral, it seems like it was specifically crafted to evaluate to a certain value Sep 26, 2013 at 2:58
• Could you add a link to a method to evaluate this? Sep 26, 2013 at 4:29
• It would be more attractive if you wrote $4!$ instead of $24$ Sep 26, 2013 at 5:19
• This doesn't strike me as remarkable at all. The $\pi$ probably comes from the $\sin$, the $e$ probably comes from the $x^x$. Sep 26, 2013 at 11:27
• What do you have if you integrate $\sin\pi(1-x) \times \textrm{the integrand}$? Sep 26, 2013 at 17:20

$\tan 10^\circ = \tan 20^\circ \times \tan 30^\circ \times \tan 40^\circ$.
$\tan 80^\circ = \tan 70^\circ \times \tan 60^\circ \times \tan 50^\circ$.

• That can be easily shown using $\tan 3x = \tan(60+x) \times \tan(x) \times \tan(60-x)$ Sep 30, 2013 at 1:16

$$\int_0^1 \frac{\ln(1+t^{4+\sqrt{15}})}{1+t}dt= -\frac{\pi^2}{12}(\sqrt{15}-2)+\ln 2\cdot \ln(\sqrt{3}+\sqrt{5})+\ln\frac{1+\sqrt{5}}{2}\cdot \ln(2+\sqrt{3})$$

For references, see http://ega-math.narod.ru/Chowla/index.htm (there is a scan of a paper of Herglotz where it is proved).

• I don't get it: how is this striking? I must be missing something... Sep 30, 2013 at 6:30
• How would one even approach such an integral? How does one evaluate integrals with quadratic irrationalities in the exponent? (Note that there is no formula for the same expression with arbitrary $p$ in place of $4+\sqrt{15}$). Sep 30, 2013 at 13:18
• This is an amazing integral! Are there other real non-rational algebraic exponents such that the integral can be expressed in an elementary closed form? Oct 3, 2013 at 18:08

The square root of 2 is also the only real number other than 1 whose infinite tetrate is equal to its square...

$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}=2.$$

• Hardly remarkable. It's just $2=\sqrt{2}^2$. Nov 19, 2013 at 2:47
• Proof is pretty trivial: $m = {{x^x}^x}^{\dots}$. Then $x^m = m$ and $x = \sqrt[m]{m}$. So you could also say that $\sqrt{3}$ is the only real number other than $1$ whose infinite tetrate is equal to its cube.
– MCT
Feb 26, 2014 at 4:33
• @Soke Be careful! Your "proof" also works when $m=4$. Any correct proof must address converge. Feb 8, 2016 at 16:55
• @BorisBukh Convergence is trivial. $z = \sqrt{2}^\sqrt{2}$ is less than $2$, so inductively $(\sqrt{2})^z$ is less than $2$ since $\sqrt{2}^2 = 2$ and $z < 2$, repeat....
– MCT
Feb 8, 2016 at 22:55
• @Soke I was pointing out that your original "trivial" proof was flawed as (if taken literally) it implied that 2=4. The new proof is almost complete --- you show that the sequence is bounded. It remains to show that the sequence is increasing (which is not too hard). Feb 9, 2016 at 1:08

I find this identity due to Euler particularly striking (and not obvious at all): $$\prod_{n=1}^\infty (1-x^n) = \sum_{k=-\infty}^\infty (-1)^k\,x^{p(k)}$$

where the $p(k) = \dfrac{k(3k-1)}{2}$ are the generalized pentagonal numbers. This is what these numbers look like us for $1 \leq k \leq 5$, [image created by Aldoaldoz]

• Euler formula for partition function $p(k)=\sum_{d=1}^{\infty}(-1)^{d+1}\left(p\left(k-\frac{d(3d-1)}{2}\right)+ p\left(k-\frac{d(3d+1)}{2}\right)\right)$ Sep 28, 2013 at 13:00
• I think the RHS is prettier if you write it as $1-x^1-x^2+x^{2+3}+x^{3+4}-x^{3+4+5}-x^{4+5+6}+x^{4+5+6+7}+\cdots$
– bof
Sep 30, 2013 at 5:25
• @AdiDani And likewise Euler's similar recurrence for the sum-of-divisors function $\sigma(n)$.
– bof
Oct 1, 2013 at 0:42

Tetration :

consider the tower of taking infinite powers : $x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}}$ .

At first its seems big mystery and undefined one for lots of real numbers.

Surprising fact is its indeed converges in an closed interval which is bounded by the fancy real numbers $e^{-e}$, $e^\frac{1}{e}$

So $x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}}$ converges for $x \in [e^{-e}, e^\frac{1}{e} ]$

If $a,b,c,d$ are in arithmetical progression, then $$\frac{d^2-a^2}{c^2-b^2}=3.$$

• Do you mean 9 instead? Sep 17, 2014 at 12:29
• No… e.g. $1,2,3,4$ are in arithmetical progression, and $(4^2-1^2)/(3^2-2^2)= 15/5 = 3$. Sep 17, 2014 at 13:32
• Write $(a,b,c,d) = (a,a+\delta,a+2\delta,a+3\delta)$, and by substitution \begin{align} \frac{(d^2 - a^2)}{(c^2 - b^2)} = \frac{(d-a)(d+a)}{(c-b)(c+b)} &= \frac{(3\delta)(2a+3\delta)}{(\delta)(2a+3\delta)} = 3. \end{align} Sep 17, 2014 at 13:37
• @CalvinLin: Surprising, yes? =) Sep 17, 2014 at 20:56
• Oh sorry, I completely misread it as (d-a)^2 / (c-b)^2. Not sure why. Yes, this is initially surprising :) Sep 17, 2014 at 21:35

$3^3 + 4^4 + 3^3 + 5^5 = 3435$

$1^1=1$ is the only other such number.

• Oct 26, 2013 at 1:46
• Thanks to the numberphile $\text{You}\,\color{red}{\boxed{\color{black}{\text{Tube}}}}$ Channel.
– user93957
Dec 12, 2013 at 19:21
• It looks interesting. In other words: $^{2}3 + ^{2}4 + ^{2}3 + ^{2}5 = 3435$, where $^{b}a$ is tetration. Feb 13, 2016 at 23:21