5
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Warning: this question requires knowledge of complex numbers.


An Euler's identity is an identity, in which each of the following appears once and only once:

  • the constant $0$: neutral element for addition;
  • the constant $1$: neutral element for multiplication;
  • the constant $e$: base of natural logarithm;
  • the constant $i$: square root of $-1$;
  • the constant $\pi$: ratio of a circle's circumference to its diameter;
  • the operation $+$: addition;
  • the operation $\times$: multiplication;
  • the operation $^\wedge$: exponentiation;
  • the equal sign $=$: symbol for equality.

How many different Euler's identities exist?


Note:

  • The use of parentheses $()$ is unlimited.
  • Other than the above mentioned symbols (including parentheses), no other symbol is allowed.
  • Because of commutativity, $a + b$ and $b + a$ are considered the same. Same with $a \times b$ and $b \times a$.
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Besides the famous identity, I've also found

12 more that are of a similar flavor, using the fact that any number to the 0th power, including complex numbers, is equal to 1: $$ e^{0 \times (i + \pi)} = 1 \\ i^{0 \times (\pi + e)} = 1 \\ \pi^{0 \times (e + i)} = 1 \\ (e + i)^{0 \times \pi} = 1 \\ (i + \pi)^{0 \times e} = 1 \\ (\pi + e)^{0 \times i} = 1 \\ (e + \pi \times i)^0 = 1 \\ (\pi + i \times e)^0 = 1 \\ (i + e \times \pi)^0 = 1 \\ ((e + \pi) \times i)^0 = 1 \\ ((\pi + i) \times e)^0 = 1 \\ ((i + e) \times \pi)^0 = 1 $$

I'll keep looking for more, but a proof that all of the possible identities are found will possibly require a computer search, which I don't know how to do.

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  • $\begingroup$ Oops I just miss those patterns... Nice work! :P $\endgroup$ – Conifers Nov 13 at 2:24
  • $\begingroup$ Nice findings (: Now you just need to find the total number of these (which should be doable by hand). A small hint is that you are only asked to give the number, and don't have to list them all. This might save a small piece of time. $\endgroup$ – WhatsUp Nov 13 at 10:44
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I've only found the famous one:

$e^{i\pi}+1=0$

And possibly it's the only one due to:

The $e,\pi, i$ could only used once for each, and we must eliminate out those irrational numbers $e,\pi$ and complex number $i$ in the equation. So the possibility is to use the constants $0, 1$ with proper operations to eliminate them into integers.

Seems it's quite impossible that only 2 constants could be used but there are 3 irrational/complex numbers. So need to put $e,\pi, i$ together and only the famous one $e^{i\pi}+1=0$ found currently.

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  • $\begingroup$ You know, I'm actually quite happy to see this answer ^_^ The source of this puzzle dates back to our undergraduate time, when the professor talked about "the" beautiful identity that relates the five important constants and the three important operations... and we suddenly have many more "beautiful identities"! $\endgroup$ – WhatsUp Nov 13 at 10:49

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