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Here is a funny exercise $$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$ (If you prove it don't publish it here please). Do you have similar examples?

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closed as primarily opinion-based by Najib Idrissi, user98602, user147263, apnorton, user2345215 Jan 31 '15 at 18:16

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 7
    $\begingroup$ This is related. $\endgroup$ – J. M. is a poor mathematician Nov 3 '10 at 22:35
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    $\begingroup$ I noticed that your expression can be also written as $\sin(x - y) \sin(x + y) = (\cos y + \cos x) (\cos y - \cos x) $ $\endgroup$ – Quixotic Nov 4 '10 at 11:09
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    $\begingroup$ I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil... $\endgroup$ – user641 Dec 8 '12 at 1:23
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    $\begingroup$ @SteveD If only we could find an odd example... $\endgroup$ – peoplepower Jan 13 '13 at 0:31
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    $\begingroup$ Almost an identity: $$\sqrt{123456790}\approx 11111.11111\,.$$ $\endgroup$ – Jakob Werner Jul 12 '14 at 18:47

63 Answers 63

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$$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$

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    $\begingroup$ Sophomore's Dream? $\endgroup$ – rotskoff Jun 19 '12 at 20:50
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    $\begingroup$ $$\int_0^1 {x^x}\mathrm{d}x=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^k}$$ $\endgroup$ – user85798 Nov 2 '13 at 18:29
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$$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$$

The two on the left is not a typo.

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    $\begingroup$ If you're physics-minded, the 2 and 3 are not a surprise: $n$ and $k$ must have the same dimension, so the right hand side has this dimension to the 4. So the only possible exponent on the left is 2. $\endgroup$ – Joce Nov 16 '16 at 15:01
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$$ \infty! = \sqrt{2 \pi} $$

It comes from the zeta function.

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    $\begingroup$ Can you help me understand $\infty!$? I don't know what to make of it. $\endgroup$ – futurebird Nov 4 '10 at 0:55
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    $\begingroup$ @don: it's $\exp(-\zeta^{\prime}(0))$, where $\zeta^{\prime}(z)$ is formally $-\sum_{k=1}^\infty \frac{\ln\;k}{k^z}$ $\endgroup$ – J. M. is a poor mathematician Nov 4 '10 at 5:01
  • $\begingroup$ @a little don, You can read about it here katlas.math.toronto.edu/drorbn/MathBlog/2008-11/one/… $\endgroup$ – anon Nov 4 '10 at 17:29
  • $\begingroup$ Neat! I wonder whether "solving" this identity for $\infty$ also yields $-\frac12$ edit Hm, since $(-\frac12)!=\sqrt\pi$ not :/ $\endgroup$ – Tobias Kienzler Dec 19 '14 at 19:41
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Ah, this is one identity which comes into use for proving the Euler's Partition Theorem. The identity is as follows: $$ (1+x)(1+x^{2})(1+x^{3}) \cdots = \frac{1}{(1-x)(1-x^{3})(1-x^{5}) \cdots}$$

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Machin's Formula: \begin{eqnarray} \frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}. \end{eqnarray}

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$$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$$

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  • 3
    $\begingroup$ I thought this was going to be hard to prove...It just took three lines! $\endgroup$ – chubakueno Feb 1 '14 at 21:14
69
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The Frobenius automorphism

$$(x + y)^p = x^p + y^p$$

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    $\begingroup$ Only in a field of prime characteristic $p$ (much to the chagrin of my calculus students). $\endgroup$ – Austin Mohr Jun 19 '12 at 2:06
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    $\begingroup$ @AustinMohr: Not just in a field of prime characteristic $p$, it holds in any commutative ring of characteristic $p$. $\endgroup$ – Marc van Leeuwen Sep 30 '13 at 6:32
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\begin{eqnarray} 1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2} \end{eqnarray} More generally, let $D_{k} = ${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\mathsf{d}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then \begin{eqnarray} \sum_{d \in D} \mathsf{d}^{*}(d)^{3} = \left( \sum_{d \in D} \mathsf{d}^{*}(d) \right)^{2} \end{eqnarray}

Note that $\mathsf{d}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$.

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  • $\begingroup$ There is no reason to restrict this to unitary divisors: Liouville's result still works if you replace "unitary divisor" with "divisor" throughout, which affords a much richer variety of sets $D_k$. The present formulation does not even generalize the standard sum $\sum n^3$ and describes a vanishingly small collection of subsets. $\endgroup$ – Erick Wong Mar 23 '17 at 23:25
  • $\begingroup$ Sure, but then you could still greatly simplify the description by replacing "unitary divisors" with "divisors" and restricting $k$ to be squarefree, there's no need to introduce two new notations. But at this point it feels like trying to dissect a proverbial joke :). $\endgroup$ – Erick Wong Apr 3 '18 at 19:15
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$$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$

and

$$\large{111,111,111 \times 111,111,111 = 12,345,678,987,654,321}$$

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  • 17
    $\begingroup$ is there any way to generalise $$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$? $\endgroup$ – pipi Nov 16 '12 at 7:32
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    $\begingroup$ Do a search for Armstrong numbers and/or narcissistic numbers. Or type 1741725 into the Online Encyclopedia of Integer Sequences. $\endgroup$ – Gerry Myerson Sep 26 '13 at 13:22
  • $\begingroup$ @ThomasWeller The same trick works with $11\times11$, $111\times 111$, $1111\times 1111$, etc., if the given version is too large to fit. (The given example is the largest of its type, though, because otherwise the digits overflow into adjacent locations and it doesn't look as nice.) $\endgroup$ – Mario Carneiro Jul 12 '16 at 2:21
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$$\sec^2(x)+\csc^2(x)=\sec^2(x)\csc^2(x)$$

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  • $\begingroup$ Is the natural logarithm function, or the exponential function related to this? $\endgroup$ – Doug Spoonwood Feb 13 '12 at 3:24
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    $\begingroup$ @Doug Spoonwood: If you multiply both sides by $\sin^2(x)\cos^2(x)$ you get the Pythagorean identity. Whether that's related to logarithm/exponential, I don't know. Just a test question I gave my students that I thought looked neat. $\endgroup$ – Joe Johnson 126 Feb 13 '12 at 14:15
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    $\begingroup$ And because of this identity we have $\frac{\,d}{\,dx} \left[e^{\tan{x}} \cdot e^{-\cot{x}}\right] = \frac{\,d}{\,dx} \left[e^{\tan{x}}\right] \cdot \frac{\,d}{\,dx} \left[e^{-\cot{x}}\right]$. $\endgroup$ – Ant Jun 10 '17 at 3:56
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\[\sqrt{n^{\log n}}=n^{\log \sqrt{n}}\]

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    $\begingroup$ $a^{\log b} = b^{\log a}$ for a and b at least 1. $\endgroup$ – Wok Nov 30 '10 at 10:03
  • $\begingroup$ Yeah, I'm sure there's a zillion related identities and generalisations (and I should probably be more careful about the domain of n). But this one in particular came up in my research and I thought it was funny -- I couldn't decide whether or not to write $\sqrt{n}^{\log n}$ or $n^{\log \sqrt{n}}$. $\endgroup$ – Douglas S. Stones Nov 30 '10 at 10:22
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    $\begingroup$ Isn't this kind of trivial? $\endgroup$ – Chantry Cargill Feb 26 '14 at 4:28
  • $\begingroup$ @ChantryCargill Not when you consider the fact that not many people know (surprisingly) that $\sqrt{x} \equiv x^{\frac{1}{2}}$ $\endgroup$ – Cole Johnson Jul 12 '14 at 18:31
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$\displaystyle\big(a^2+b^2\big)\cdot\big(c^2+d^2\big)=\big(ac \mp bd\big)^2+\big(ad \pm bc\big)^2$

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    $\begingroup$ $(a^2+b^2)\cdot(c^2+d^2)$ is obviously $c^4+c^2d^2$ $\endgroup$ – Mateen Ulhaq Apr 18 '11 at 2:30
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    $\begingroup$ $c^2(c^2+d^2)$??... what do you mean? $\endgroup$ – Neves Apr 18 '11 at 13:43
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    $\begingroup$ $a^2+b^2=c^2$ $\endgroup$ – Mateen Ulhaq Apr 18 '11 at 16:06
  • $\begingroup$ There are also the related Lagrange's identity and the corresponding one for eight squares. $\endgroup$ – Yuval Filmus Nov 16 '11 at 13:28
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    $\begingroup$ The Brahmagupta-Fibonacci identity. $\endgroup$ – The Chaz 2.0 May 1 '12 at 5:17
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Facts about $\pi$ are always fun!

\begin{equation} \frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\ldots\\ \end{equation} \begin{equation} \frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots\\ \end{equation} \begin{equation} \frac{\pi^2}{6} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\ldots\\ \end{equation} \begin{equation} \frac{\pi^3}{32} = 1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}-\ldots\\ \end{equation} \begin{equation} \frac{\pi^4}{90} = 1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots\\ \end{equation} \begin{equation} \frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots\\ \end{equation} \begin{equation} \pi = \cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ldots}}}}}\\ \end{equation}

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Well, i don't know whether to classify this as funny or surprising, but ok it's worth posting.

  • Let $(X,\tau)$ be a topological space and let $A \subset X$ . By iteratively applying operations of closure and complemention, one can produce at most 14 distinct sets. It's called as the Kuratowski's Closure complement problem.
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    $\begingroup$ An example achieving the is $[0,1] \cup (2,3) \cup \{(4,5) \cap \mathbb{Q}\} \cup \{(6,8) - \{7\}\} \cup \{9\}$. See section 9 of austinmohr.com/Work_files/730.pdf for details. $\endgroup$ – Austin Mohr Jun 19 '12 at 2:09
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    $\begingroup$ I think you mean $[0, 1]\cup (2, 3)\cup((4, 5)\cap\mathbb{Q})\cup(6, 7)\cup(7, 8)\cup\{9\}$. The set you wrote isn't a subset of $\mathbb{R}$ as it contains $(4, 5)\cap\mathbb{Q}$ as an element. $\endgroup$ – Michael Albanese Jan 13 '13 at 15:08
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    $\begingroup$ @MichaelAlbanese: Anyone able to understand the contents of this thread instantly recognizes those braces are there to give the intersection operator a higher precedence than the neighboring union operators. One must travel far out of one's mathematical way to arrive at your alternative interpretation... $\endgroup$ – mathematrucker Jun 3 '17 at 15:42
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The following number is prime

$p = 785963102379428822376694789446897396207498568951$

and $p$ in base 16 is

$89ABCDEF012345672718281831415926141424F7$

which includes counting in hexadecimal, and digits of $e$, $\pi$, and $\sqrt{2}$.

Do you think this's surprising or not?

$$11 \times 11 = 121$$ $$111 \times 111 = 12321$$ $$1111 \times 1111 = 1234321$$ $$11111 \times 11111 = 123454321$$ $$\vdots$$

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    $\begingroup$ The prime is unsurprising -- the final F7 doesn't seem to mean anything, and about one in 111 numbers of that size is prime. So it's not very remarkable that there's a prime among the 256 40-hex-digit numbers that start with those particular 38 chosen digits. $\endgroup$ – Henning Makholm Nov 20 '13 at 18:04
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    $\begingroup$ I remember that last from reading "The number devil"! And it works for other bases too; for a base $b$, until $\left(\sum_{n=0}^{b-1}\left(b^n\right)\right)^2=123...\ \text{digit } b-1\ ...321$. $\endgroup$ – JMCF125 Nov 24 '13 at 11:06
  • $\begingroup$ I find (1....1)^n interesting, it's also nearly impossible to caluclate by hand without messing it up. $\endgroup$ – HopefullyHelpful Mar 4 '16 at 23:19
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$$ \frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots $$ [Nick Pippenger, Amer. Math. Monthly, 87 (1980)]. Set all the exponents to 1 and you get the Wallis formula for $\pi/2$.

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\begin{align} \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1} &\text{Power Rule?}&\ \text{False}\\ \frac{\mathrm d}{\mathrm dx}(x^x) &= x^{x}\ln(x) &\text{Exponential Rule?}&\ \text{False}\\ \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1}+x^{x}\ln(x) &\text{Sum of these?}&\ \text{True}\\ \end{align}

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    $\begingroup$ This is a special case of $\frac{d}{dx} h(f(x),g(x)) = \partial_1 h f' + \partial_2h g'$ $\endgroup$ – ronno Dec 20 '13 at 13:56
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    $\begingroup$ "I just add those 2 together so that I can get partial credit." $\endgroup$ – Derek 朕會功夫 Aug 11 '14 at 5:33
  • $\begingroup$ @Derek朕會功夫 God the puns xD $\endgroup$ – Simply Beautiful Art Mar 21 '17 at 1:16
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\begin{eqnarray} \sum_{i_1 = 0}^{n-k} \, \sum_{i_2 = 0}^{n-k-i_1} \cdots \sum_{i_k = 0}^{n-k-i_1 - \cdots - i_{k-1}} 1 = \binom{n}{k} \end{eqnarray}

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$$\sum\limits_{n=1}^{\infty} n = 1 + 2 + 3 + \cdots \text{ad inf.} = -\frac{1}{12}$$

You can also see many more here: The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation

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    $\begingroup$ @fmartin: it's $\zeta(-1)$, which can be shown to be expressible in terms of Bernoulli numbers. $\endgroup$ – J. M. is a poor mathematician Nov 8 '10 at 0:55
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    $\begingroup$ @J.M.: I still fail to see how an infinite summation of positive numbers can result in a negative number. $\endgroup$ – F M Nov 15 '10 at 23:26
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    $\begingroup$ @fmartin: I agree it's counterintuitive; properly explaining this mathematical joke requires a foray into complex analysis (the magic words are "analytic continuation"), which I'll leave to more eloquent users to explain. $\endgroup$ – J. M. is a poor mathematician Nov 16 '10 at 7:05
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    $\begingroup$ It's particularly a string theory joke, since this is the trick they use to regularize certain sums in their theories. That's how they arrive at 26 dimensions (in non-supersymmetric theories), because the regularization only works for that many dimensions. I suppose the argument works in the same way in supersymmetric theories, but they then get 10 dimensions. $\endgroup$ – Raskolnikov Nov 27 '10 at 11:25
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    $\begingroup$ Isn't this Ramanujan's interpretation of $\zeta(-1)$. $\endgroup$ – Pixel Aug 14 '12 at 11:43
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Two related integrals:

$$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$$

$$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=-\gamma$$

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    $\begingroup$ can you give a hint for those of us who don't see it? :) $\endgroup$ – anon Nov 3 '10 at 22:48
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    $\begingroup$ They are Abel-summable integrals; e.g. the first one is properly interpreted as $\lim_{\epsilon\to 0}\int\exp(-\epsilon x)\sin\;x\quad \mathrm{d}x$ $\endgroup$ – J. M. is a poor mathematician Nov 3 '10 at 22:53
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    $\begingroup$ Doesn't the first integral diverge? $\endgroup$ – Hawk Jun 19 '12 at 1:52
  • $\begingroup$ @jak, see my previous comment. $\endgroup$ – J. M. is a poor mathematician Jun 19 '12 at 2:05
  • $\begingroup$ @J.M. I have not thought about this, but is that unique? I don't think so, but why look at the $\exp$ "kernel"? $\endgroup$ – AD. Jul 4 '12 at 19:31
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M.V Subbarao's identity: an integer $n>22$ is a prime number iff it satisfies,

$$n\sigma(n)\equiv 2 \pmod {\phi(n)}$$

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$32768=(3-2+7)^6 / 8$

Just a funny coincidence.

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By excluding the first two primes, Euler's Prime Product becomes a square:

$$\prod _{n=3}^{\infty } \frac{1}{1-\frac{1}{(p_n)^{2}}}=\frac{\pi ^2}{9}$$

By using multiples of the product of the first two primes, we get the square root:

$$\prod _{n=1}^{\infty } \frac{1}{1-\frac{1}{(n p_1 p_2)^{2}}}=\frac{\pi }{3}$$

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    $\begingroup$ It doesn't make sense to speak of "perfect squares" for positive real numbers... but this is a nice identity though. $\endgroup$ – Patrick Da Silva Jun 19 '12 at 19:53
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    $\begingroup$ @PatrickDaSilva It might, if you know that the values of $L$-functions sometimes land in a special ring which is strictly between algebraic numbers and transcendental numbers. This is the ring of 'periods'. I don't believe that it is closed under taking square roots, so to say that something is the square of a period might not be completely silly. $\endgroup$ – Bruno Joyal Sep 26 '13 at 21:29
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    $\begingroup$ @BrunoJoyal If I'm not mistaken, periods include $\zeta(3)$ and many more - they are basically anything you can get with integration. If I recall correctly, it is not known whether or not $\dfrac1\pi$ is a period. $\endgroup$ – Akiva Weinberger Aug 28 '14 at 3:56
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$$ 10^2+11^2+12^2=13^2+14^2 $$

There's a funny Abstruse Goose comic about this, which I can't seem to find at the moment.

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$${\Large% \sqrt{\,\vphantom{\huge A}\color{#00f}{20}\color{#c00000}{25}\,}\, =\ \color{#00f}{20}\ +\ \color{#c00000}{25}\ =\ 45} $$

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Parallelogram

$$\left|z+z'\right|^{2}+\left|z-z'\right|^{2}=2\times\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)$$

The sum of the squares of the sides equals the sum of the squares of the diagonals.

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  • $\begingroup$ The so-called parallelogram identity. $\endgroup$ – abnry Aug 19 '13 at 19:38
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What is 42?

$$ 6 \times 9 = 42 \text{ base } 13 $$ I always knew that there is something wrong with this universe.

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The product of any four consecutive integers is one less than a perfect square.

To phrase it more like an identity:

For every integer $n$, there exists an integer $k$ such that $$n(n+1)(n+2)(n+3) = k^2 - 1.$$

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  • $\begingroup$ I wonder now if this isn't a proper identity because of the existential quantifier. Any thoughts? $\endgroup$ – Austin Mohr Jun 19 '12 at 3:15
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    $\begingroup$ You can also write that as $$n(n+1)(n+2)(n+3)=((n+1)^2+1)^2 - 1$$ $\endgroup$ – AD. Jun 19 '12 at 6:53
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I have one: In a $\Delta ABC$, $$\tan A+\tan B+\tan C=\tan A\tan B\tan C.$$

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    $\begingroup$ Also $\cot(A/2)+\cot(B/2)+ \cot(C/2)=\cot(A/2)\cot(B/2)\cot(C/2)$. $\endgroup$ – N. S. Apr 11 '13 at 22:39
  • $\begingroup$ @N.S. there's no need for them to be divided by 2 right? $\endgroup$ – InertialObserver May 26 at 22:36
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Considering the main branches

$$i^i = \exp\left(-\frac{\pi}{2}\right)$$

$$\root i \of i = \exp\left(\frac{\pi}{2}\right) $$

And $$ \frac{4}{\pi } = \displaystyle 1 + \frac{1}{{3 +\displaystyle \frac{{{2^2}}}{{5 + \displaystyle\frac{{{3^2}}}{{7 +\displaystyle \frac{{{4^2}}}{{9 +\displaystyle \frac{{{n^2}}}{{\left( {2n + 1} \right) + \cdots }}}}}}}}}} $$

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