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Is there proof fully algebraic that shows that $1$ is the only number that when put in the function $f(x) = x^n$, for $n = $ any number, the output is always $x$?

This is pretty obvious, but you seem to prove it using the trivial knowledge .

For example, $1^4 = 1, 1^{46373} = 1$, but $2^4 \neq 2$

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I'll provide an easy way: We need the following to be true for all $n \in \mathbb N$ $$x^n=1$$

First let $x=1$, and we could check that $1^n=1$ for all $n \in \mathbb N$, so $x=1$ is one solution.

Do we have any other solutions? NO, to see this, let $n=1$, the above equation needs to be true, and thus we get $x=1$.

So we are done.

EDIT

So the question now is changed to $x^n =x$, which makes things a bit different.

So now we want $$x^n=x$$ Still, easy to check $x=1,0$ are solutions for all $n > 0$, but notice that $0$ is not a solution when $n=0$, since $0^0$ is usually not defined.

Then let $n=2$, $x^2=x$, thus we have to have $x = 1,0$

Thus the only solution is $1$.

EDIT

I'll take comment of @rachwieb, and try not to discuss $0^0$ here. But we let $x=0,1$ if we take all $n \in \mathbb N \setminus \{0\}$; and $x= 1$ if we take $n \in \mathbb Z$, because $0^{-1}$ is not defined.

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  • $\begingroup$ Moreover, if $0^0$ is defined, then it is usually defined as $0^0=1$, thus even in that case the equation $x^x=x$ will be violated for $x=0$. $\endgroup$ – celtschk May 16 '17 at 20:03
  • $\begingroup$ @celtschk Yes, $1$ is a more common choice if you really want to define it. $\endgroup$ – Yujie Zha May 16 '17 at 20:07
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    $\begingroup$ This whole $0^0$ business is just a contextual handgrenade that should never have been part of the conversation. The assumption $n$ is "any number" is an unnecessarily floppy assumption to begin with, but it definitely eliminates $0$ on the grounds that "$0^{-1}=0$" does not hold. Cutting down to nonegative integers is a noble attempt, but runs into this $0^0$ quibble. Cutting down to positive integers seems like the most natural statement to me (because positive integral exponents do not need any magic sauce to work, they work everywhere.) $\endgroup$ – rschwieb May 17 '17 at 13:31
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    $\begingroup$ @rschwieb Tuanks for the comment. Actually I have the same feeling if we just have $n>0$, everything is going to be neat. And good point that you mention $0^{-1}$!. I'll update my answer to just for $n>0$ $\endgroup$ – Yujie Zha May 17 '17 at 13:40
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Once we rule out $x=0$ (since we typically define $0^0=1$):

Assume $x^n = x$ for any $n$

then in particular $x^2 = x$

Divide both sides by $x$ (which is ok, since $x \not = 0$), and we get:

$x=1$

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  • $\begingroup$ Another way to rule out $0$ (using the question as posted) that doesn't depend on adopting a convention on $0^0$ is to say that $0^{-1}$ isn't defined, so the relation "$0^{-1}=0$" does not hold. $\endgroup$ – rschwieb May 17 '17 at 13:24
  • $\begingroup$ @rschwieb Yes, that works ... when I posted my answer the $n$ was still a natural number but apparently the OP changed the question since. So using a negative number would take care of the $0$ case immediately without quibbling over $0^0$. $\endgroup$ – Bram28 May 17 '17 at 16:05
  • $\begingroup$ Ugh, oh really, it was natural number and they edited it to "any"? blegh, I did not see that coming :) $\endgroup$ – rschwieb May 17 '17 at 16:43
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Be $x^n=x\quad \forall n\in \mathbb{N}$ then:

Depending on how you define $0^0$ and if you include $0$ in $\mathbb{N}$ a solution might be $0$, as $0^n=0 \quad\forall n\in \mathbb{N}$

And if $x\neq0$, you can divide x from both sides of the equation $x=x^2$ and get that $x=1$

thus $0$ and $1$ are the only solutions.

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  • $\begingroup$ Typically, $0^0$ is defined as $1$ $\endgroup$ – Bram28 May 16 '17 at 18:10
  • $\begingroup$ I don't know what is more common - either way you get rid of the 0 $\endgroup$ – Felix B. May 16 '17 at 18:11
  • $\begingroup$ I do not think $0^0$ is even defined. $\endgroup$ – Yujie Zha May 16 '17 at 18:14
  • $\begingroup$ You define it sometimes as 0 and sometimes as 1, depending on the setting. You usually define it in such a way that you don't have to deal with it writing your equations. $\endgroup$ – Felix B. May 16 '17 at 18:17
  • $\begingroup$ @Bram28 fixed it, I also realized that you need to include the definition of the natural numbers, as a lot of them don't even include 0 $\endgroup$ – Felix B. May 16 '17 at 18:18
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Going off of Dr. Graubner's hint, we have:

$$f(x) - x = 0$$ $$x^n - x = 0$$ $$x(x^{n-1}-1) = 0$$.

Therefore, either $x$ or $x^{n-1}-1$ is equal to $0$.

$$x^{n-1}=1$$ $$(n-1)\ln(x) = 0$$

Now, either $(n-1)$ or $\ln(x)$ is $0$. Because $n$ is any number, we can't define it. Therefore:

$$\ln (x) = 0$$ $$x = 1$$

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For positive integer $n$, write $$f(x)=x^n=x^n-1+1=(1+x+\cdots+x^{n-1})(x-1)+1.$$

If $x > 1$, then $(1+x+\cdots +x^{n-1})(x-1) > n(x-1) $ so $f(x) > n(x+1)+1 > 1$.

If $0<x < 1$, then $(1+x+\cdots +x^{n-1})(x-1) < \frac{1}{1-x}(x-1) = -1$. Hence $f(x) < -1+1 = 0$.

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I know this question was probably written without any thought of leaving $\mathbb R$ or $\mathbb C$, but I think it's a good opportunity to sharpen the question and show the big picture. The question apparently is, or could be:

How can I algebraically prove that if $x^n=x$ for all positive integers $n$, then $x=1$ or $x=0$?

(IMO the idea of discarding $0$ as a solution -- on the grounds of a particular chosen value of $0^0$ or otherwise -- is far less natural than just restricting the powers to be positive integers.)

There's actually no problem with immediately generalizing this to:

If $R$ is a ring with identity $1$ and no nonzero zero divisors, $x^2=x$ implies $x=1$ or $x=0$.

The proof has pretty much already been given above: if $x^2-x=0$, then $x(x-1)=0$, and by assumption either $x=0$ or $(x-1)=0$. This is just assuming the proposed problem for $n=2$, so of course requiring it for $n>2$ does no add anything new. (And assuming "$n$ is any number is a bit overkill too.)

But that is not quite the end of the story. You see, it was important here that the ring did not have nonzero zero divisors. In fact, there are very natural rings that don't have that property, and the proposed problem fails piteously. Namely:

There exists an infinite Boolean ring $R$, that is, a ring in which $x^2=x$ for all $x\in R$. In such a ring, $x^n=x$ for every element of the ring, for every positive integer $n$.

I chose that one to maximize the number of things being badly behaved, but of course it has nothing to do with cardinality. A Boolean ring with $4$ elements serves equally well to show that we could potentially have more than just $0$ and $1$ doing this.

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