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Let $ K $ be a field and $f:K-\left \{ 0 \right \}\rightarrow K-\left \{ 0 \right \}$ a function with $f(f(x))=x^{-1},\forall x\in K-\left \{ 0 \right \}$ and $ f(1)\neq 1.$
Knowing that $ f^{2}(x)-f(x)+1=0 $ has an unique solution in $K-\left \{ 0 \right \} $, determine $ f(2)$.

I haven't found something useful yet.

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    so $f^2(x)$ is $f(x)^2$ in this context right? – 2017-01-28
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    Yes, it's $ f(x)^2 $. – 2017-01-28
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    $f$ is bijective. – 2017-01-28
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    I think that I proved that any field in which $x^2-x+1$ has only one root must have characteristic $2$, so $2=0$. – 2017-01-28
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    nevermind, the characteristic was $3$. – 2017-01-28
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    for an example consider $\mathbb Z_3$ and the function $f(x)=-x$ – 2017-01-28

1 Answers 1

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Suppose that $a$ is a root of the polynomial $x^2-x+1$.

Then $(x-a)$ divides the polynomial $x^2-x+1$.

This means that $(x-a)(x-a)=x^2-x+1$ (because the root is unique).

it follows that $x^2-2a+1=x^2-x+1$ and so $2a=1$ and $a^2=1$.

we conclude that $1=1^2=(2a)^2=4a^2=4$.

So the field must have characteristic $3$, and so $2=-1$.

Let $z=f(1)$. Notice $z^{-1}=f(f(z))=f(f(f(1))=f(1)=z$. It follows that $z^2=1$. The only two roots of the polynomial $x^2-1$ are $1$ and $-1$. We conclude $z=-1$.

So we have $f(1)=-1$, and since $1=f(f(1))=f(-1)$ we have $1=f(-1)=f(2)$.

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    doesn't it follow $x^2-x+1=(x-a)(x-c)$? – 2017-01-28
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    I think 1 = 4 means that char $K$ is $3$ not $2$. – 2017-01-28
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    oh yeah, good point. – 2017-01-28
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    How does that help finding $f(2)$? – 2017-01-28
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    @SSepehr thanks for the help, that did the trick! – 2017-01-28
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    @EugenCovaci I think the OP means a field as in mathematics, these ones are always commutative. Maybe you are thinking of the fields where horses run around and stuff? – 2017-01-28
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    If $ 2a=1 $ and $ a^{2}=1 $ multiplying the first relation with $ a$, we get that $ 2a^{2}=a\Leftrightarrow 2=a $, so $f(2)=0$. It's correct, isn't it? – 2017-01-28
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    @ztefelina I don't see how you conclude $f(2)=0$ from $2=a$. (although the other steps look good) – 2017-01-28
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    That's because $ a$ is a root of $f^{2}(x)-f(x)+1$.(or $f(a)$ is a root of $x^2-x+1$). – 2017-01-28
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    Well, something is wrong with your reasoning. I proved $f(2)=1$. And there is an example of a función that works in the comments. – 2017-01-28
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    Its not clear how you finish to show $f(-1)=1$. – 2017-01-28
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    @ReneSchipperus let $z=f(1)$, we have $f(f(z))=f(f(f(1))=f(1)=z$. We conclude $z=z^{-1}$ and so $z^2=1$. The only other element satisfying this is $-1$. – 2017-01-28
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    Oh, I made a mistake in my "proof"..You're right. :) – 2017-01-28
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    Yeah I just got that, so $f(1)=-1$ and then $f(-1)=1$. – 2017-01-28
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    yeah, I got a bit confused, but hopefully it's good now. Thanks for the improvements. – 2017-01-28