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Does there exist a function $f$ such that $\forall x \in \mathbb{R}:f^3(x) + f^2(x) \cdot x^2 = 1$?

I haven't studied functional equations so I have no idea how to solve this problem. I think I proved it's impossible if $f$ is polynomial (it would have to be $f(x) = 1 - x^2$, but that doesn't work). But what I really want to do is the opposite, I want to find $f$ with this property, as it would work as a counter-example to some phenomena I'm investigating about derivatives.

Is it possible?

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    Does $f^2(x)$ mean $f(x)^2=f(x)\cdot f(x)$ or $f(f(x))$? – mr_e_man Oct 16 '20 at 01:54
  • Assuming $f(x)^n$ stands for $(f(x))^n$, here is an argument: If we write $$F(x,y)=y^3+x^2y^2-1,$$ then $F(x,0)=0$ and $y\mapsto F(x,y)$ is strictly increasing without bound for $y>0$. So, for each $x$, there exists a unique point $y=f(x)\in(0,\infty)$ such that $F(x,f(x))=0$. Also, the implicit function theorem tells that $f(x)$ defined in this way is smooth. Finally, although we can also write down a closed-form of $f(x)$ as an algebraic function (since the equation is cubic), that formula is not so interesting in analytical sense because it involves combination of complex square roots. – Sangchul Lee Oct 16 '20 at 02:00

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Assuming that we face the problem of $$y^3+x^2y^2-1=0 \qquad \text{with} \qquad y=f(x)$$ following the steps given here, we have $$\Delta=4 x^6-27 \qquad p=-\frac{x^4}{3}\qquad q=\frac{2 x^6}{27}-1$$

If $\Delta <0$, using the hyperbolic method for only one real root $$y=\frac{1}{3} x^2 \left(2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{27}{2 x^6}-1\right)\right)-1\right)$$

If $\Delta > 0 $, the equation has then three real solutions given by

$$y_k=\frac{1}{3} x^2 \left(2 \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{27}{2 x^6}-1\right)\right)\right)-1\right)$$ with $k=0,1,2$.

Claude Leibovici
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