@Rishabh Jain
–
wen I used this Daum eqn editor , I wasn't able to copy the equation and paste it in the Field..
It just copies the code only !! Any solution ??

Let \(x=\frac{a}{b}\) and \(x^x=\frac{c}{d}\), where \(a\), \(b\), \(c\), and \(d\) are integers such that \(a\) and \(b\) are coprime as well as \(c\) and \(d\). After some rearranging, we attain

\[(a^a)(d^a)=(b^b)(c^b)\]

Since both \(a\) and \(b\) are coprime, \(a^a=c^b\), and \(d^a=b^b\). Neither of these statements are possible, because \(a\) and \(b\) are coprime, unless the case where \(b=1\) and \(d=1\).

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestLet x = a/b, and x^x = c/d, where a, b, c, d, are integers, such that a & b and c & d are co-primes. Then we have after some re-arranging:

(a^a) (d^a) = (b^b) (c^b)

Since both a & b and c & d are co-primes, a^a = c^b and d^a = b^b, neither which is possible because a & b are co-primes, unless b = 1 and d = 1.

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Ow, nice! The only thing that I think your proof lack is LaTeX. Anyway, thank you.

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Yeah, I should learn how to us LaTeX, you're right about that.

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It's right to assume that a,b,c,d are relative prime?

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Here you go @Michael Mendrin:

Let \(x=\frac{a}{b}\) and \(x^x=\frac{c}{d}\), where \(a\), \(b\), \(c\), and \(d\) are integers such that \(a\) and \(b\) are coprime as well as \(c\) and \(d\). After some rearranging, we attain

\[(a^a)(d^a)=(b^b)(c^b)\]

Since both \(a\) and \(b\) are coprime, \(a^a=c^b\), and \(d^a=b^b\). Neither of these statements are possible, because \(a\) and \(b\) are coprime, unless the case where \(b=1\) and \(d=1\).

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