I read in my class notes that there are only two polynomials functions which can satisfy the relation $$ f(x) \cdot f(\frac{1}{x}) = f(x) + f(\frac{1}{x}) $$ , which are $1 \pm x^n$ . I am not able to understand why only these 2 polynomial functions are able to satisfy the given relation , can't there be any other polynomial function which will satisfy the given relation ? Please help me to solve this query .
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The $0$ polynomial function and the constant $2$ polynomial function satisfy it as well. – JVHD2334 Mar 25 '20 at 10:04
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@JVHD2334 These correspond to the quoted solution with $n=0$. – Servaes Mar 25 '20 at 10:12
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Of course, you are right. – JVHD2334 Mar 25 '20 at 11:57
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Hint: The relation is equivalent to $$\Big(f(x)-1\Big)\Big(f(\tfrac1x)-1\Big)=1.$$
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Let $d=\deg f$ so that $g(x)=x^df(\tfrac1x)$ is a polynomial. Then the above shows that $$x^d=\Big(f(x)-1\Big)\Big(g(x)-x^d\Big).$$ It follows that $f(x)-1=\pm x^n$ for some integer $n$ with $0\leq n\leq d$, and so $f(x)=1\pm x^n$. Conversely it is easily verified that for every integer $n\geq0$ you have $$(1\pm x^n)(1\pm x^{-n})=(1\pm x^n)+(1\pm x^{-n}),$$ when all the signs agree.
Servaes
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