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Find all $f(x)$ such that $$(f(x))^2-f(x^2)=constant$$

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  • $\begingroup$ I Think this proble is very nice, and Now I can't solution.Thank you everyone help $\endgroup$
    – math110
    Jul 5, 2013 at 12:26
  • $\begingroup$ What is the underlying field of constants? If it is of characteristic two, then, for example, all the functions $f(x)\in\mathbb{F}_2[x,1/x]$ are solutions ;-> $\endgroup$ Jul 5, 2013 at 13:55
  • $\begingroup$ Thank you ,what is $F_{2}[x,1/x]?$ $\endgroup$
    – math110
    Jul 5, 2013 at 14:01
  • $\begingroup$ $$\mathbb{F}_2[x,1/x]= \{\sum_{i=-n}^ma_ix^i\mid n,m\in\mathbb{Z}, a_i\in\mathbb{F}_2\,\text{for all $i$}\}.$$ $\endgroup$ Jul 5, 2013 at 14:02
  • $\begingroup$ In characteristic two we have, for example $$ (x^3+x)^2=x^6+x^2,$$ so $f(x)=x^3+x$ is a solution in that case. I do expect that you mean that the field of constants is some subfield of the complex numbers. That's why I added the evil grin emoticon. But I do have the point that the term "rational function" does not specify the ground field. $\endgroup$ Jul 5, 2013 at 14:05

2 Answers 2

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Assume that $a$ is a pole of $f$ or order $n$ then $f^2$ has $a$ as a pole of order $2n$ and therefore so does $f(x^2)$. Therefore $a^2$ is a pole of $f$ of order $2n$. Therefore $f$ has no poles different from zero.

Let $g(x):=f(1/x)$. Notice that $g(x)$ also satisfy the equation. Subtract the two equations to get $$(f(x)-g(x))(f(x)+g(x))=f(x^2)-g(x^2).$$

Therefore, for the function $h(x):=f(x)-g(x)$ we have $h(x)(f(x)+g(x))=h(x^2)$. If $a$ is a zero of $h(x)$ of order $n$, then $a^2$ is a zero of $h(x)$ too. Therefore, either $a=0$ or $h(x)=constant$.

Now we have $f(x)=P(x)/x^k$ for some polynomial $P$. Then $P(x)/x^k-P(1/x)/x^{-k}=C$.

If $n$ is the degree of $P$, comparing degrees we get that $n=2k$. Comparing coefficients we get that $P$ is a symmetric polynomial (the coefficients can be put in reverse order), i.e. $P(x)=a_{2k}x^{2k}+a_{2k-1}x^{2k-1}+\ldots+a_{2k-1}x+a_{2k}$.

So, so far we have that $f$ is a linear combination of $x^k+1/x^k$. Plugging a linear combination into $f(x)^2-f(x^2)=constant$,

... to be continued, but from there you should get the solutions ...

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  • $\begingroup$ Nice! You should also add $f(x) =$ constant under $k = 0$. $\endgroup$
    – hot_queen
    Jul 5, 2013 at 12:59
  • $\begingroup$ The constant should be $0$ or $1$. Oh, in the recapitulation forgot to mention $1$. Adding it. $\endgroup$
    – OR.
    Jul 5, 2013 at 13:04
  • $\begingroup$ Okay. You are solving $f(x^2) - (f(x))^2 = 0$. It doesn't really matter. $\endgroup$
    – hot_queen
    Jul 5, 2013 at 13:08
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    $\begingroup$ In view of Michael's comment, you probably want to fix the part that says "f can only have 0 as its pole and/or zero implies $f(x) = ax^k$" to account for solutions of form $x^k + 1/x^k$. $\endgroup$
    – hot_queen
    Jul 5, 2013 at 13:14
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    $\begingroup$ I saw that poles of $g$ are not the zeros of $f$ long ago. Typing. $\endgroup$
    – OR.
    Jul 5, 2013 at 13:18
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The earlier answers have forgotten $f(x)=x+1/x$

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  • $\begingroup$ yes, I think this problem have other $\endgroup$
    – math110
    Jul 5, 2013 at 13:10
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    $\begingroup$ @S.B. that wasn't the question. Look again. $\endgroup$
    – Empy2
    Jul 5, 2013 at 13:31

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