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Calculate all real solutions $x\in\mathbb{R}$ of the equation $$ \tag1(x^2-2x+2)^2-2(x^2-2x+2)+2 = x $$

My Attempt:

I used the concept of a composite function. Let $f(x) = x^2-2x+2$. Then equation $(1)$ converts into $f(f(x)) = x$. Both $f(x) = x$ and $f(x) = -x$ satisfy the given composite function.

Case 1: If $f(x) = x$, then

$$ x^2-2x+2=x\\ x^2-3x+2=0\\ x\in\{1,2\} $$

Note that $1,2\in\mathbb{R}$.

Case 2: If $f(x)=-x$, then

$$ x^2-2x+2=-x\\ x^2-x+2=0\\ x=\frac{1\pm \sqrt{1-8}}{2}\notin \mathbb{R} $$

So only $x\in\{1,2\}$ are the real solutions of above equation.

Is my process correct? Is there is any other method by which we can solve the above question?

Fabrosi
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juantheron
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2 Answers2

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Setting $T=x-1$ you have $(T^2+1)^2-2(T^2+1)+1-T=0$ or $T^4-T=0$ that is $T(T-1)(T^2+T+1)=0$ so for real solutions you have $T=0$ or $T=1$, that is $x=1$ or $x=2$.

Omran Kouba
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To the question

Is my process right?

Unfortunately, no, it is not. There are two problems with your logic:

First, while it is true that the functions $f(x)=x$ and $f(x)=-x$ both (separately) obey the identity $f(f(x))=x$, you appear to be trying to use the converse of this, which is false: Given that $f(f(x))=x$, it is not necessarily true that either $f(x)=x$ or $f(x)=-x$. Other possibilities include $f(x)=A-x$ (for any real number $A$) and $f(x)=C/x$ (for any real number $C$, and assuming $x \ne 0$). Nor is this an exhaustive list of the possibilities; you are essentially trying to describe all involutions (functions that are their own inverse function); a complete characterization of such functions can be found at this question.

But the second problem is perhaps more serious: You seem to be confusing the statement that $f(f(x))=x$ for a particular value of x with the statement that $f(f(x))=x$ for all x. Suppose (just for example) that you have a function for which $f(3)=-3$. What does that tell you about $f(f(3))$? You know that $f(f(3))=f(-3)$, but you can't conclude that this equals $3$, because we don't know anything about $f(-3)$.

Edited to add: On the other hand, there is something right about your solution. If you are fortunate enough to find a value of $x$ for which $f(x)=x$, then it will certainly by true that $f(f(x))=x$ as well, and therefore solving the equation $f(x)=x$ can find some of the solutions to a problem like this. But it's not guaranteed to find all of them. The original equation is a 4th degree polynomial, and so in principle one can expect there to be up to 4 real solutions, but this heuristic can only find at most 2 solutions.

In this problem it happens to be the case that this heuristic actually does find all solutions, which seems to me to be a surprising coincidence. I am not sure off the top of my head how typical this is. Anybody want to take a shot at classifying those polynomials $f$ for which $f(f(x))=x$ if and only if $f(x)=x$?

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mweiss
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