Inspired by Yash Dev Lamba

Algebra Level 3

How many integer solutions are there to the equation $x^x = x$ ?

$0^0$ is indeterminate.

3 1 Finitely many Infinitely many 2

1 solution

Rishabh Jain
Mar 6, 2016

$x\neq 0$ as $0^0$ is indeterminate . $\large x^x = x\\\large \implies \dfrac{x^x}{x}=1 ~~(x\neq 0)\\\large \implies x^{x-1}=1$ We know now three case arises: $\large\bullet~ x=0 ,\\ \large\bullet~ x=1 ,\\\large \bullet~ x=-1 \text{ such that x-1(= -2) is even}$ $x=0$ is rejected while from 2nd and 3rd cases we get $x=\pm 1$ i.e $\boxed 2$ solutions.

Can $x$ be negative in $x^{x}$ ?

Yash Dev Lamba - 5 years, 3 months ago

Yes , it can be.

Nihar Mahajan - 5 years, 3 months ago

Yash Dev Lamba - 5 years, 3 months ago

@Yash Dev Lamba – Ya I have also searched on many graphing apps and the graph of $x^x$ is only visible for $x>0$ . But isn't $(-1)^{-1}$ defined ?

Rishabh Jain - 5 years, 3 months ago

@Yash Dev Lamba – Ya.. Right but for $x<0$ , it's imaginary part vanishes at x=-1,-2,-3,... (Negative integers!!) So I genuinely think that its defined at $x=-1$ which can also be verified by direct substitution.

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – I also think you are right.

Yash Dev Lamba - 5 years, 3 months ago

Same way! Nice solution.

Rishik Jain - 5 years, 3 months ago

Inspired by Yash Dev Lamba

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