Find $f(x)$ that satisfies the equation $f(x^2) + (f(x))^2 = 2x^2 - 2$

Question

We know that $f(x) = x - \frac{1}{x}$ , for $x \neq 0$ , is a particular solution of the equation, but what would be the method to arrive at this solution?

Answer 1

Note that you can find values immediately for $x = 0,1$ $$ [f(0)]^2 + f(0) + 2 = 0 \implies f(0) = \frac{-1 \pm i\sqrt{3}}{2}. $$ Therefore, if you are only interested in $f: \mathbb R \to \mathbb R$ you need to restrict yourself to $x \neq 0$. For $f(1)$ we see $$ [f(1)]^2 + f(1) = 0 \implies f(1) = 0,-1. $$ Look at the action from $x \to -x$, which yields the relation $f(x) = \pm f(-x) $ (hence the function is either even or odd.

Assuming the function $f$ is differentiable, we obtain $$ xf'(x^2) + f(x)f'(x) = 2x. $$ Evaluating this at $x = 1$, $$ [1 + f(1)]f'(1) = 2 $$ and we see that if $f(1) = -1$, then the equation cannot be satisfied and therefore if our function did satisfy $f(1) = -1$, then it can't be differentiable at $x = 1$.

Lets try a series solution (noting that it must be even or odd, lets try odd first) $f(x) = \sum_n a_{2n+1} x^{2n+1}$, note that this is differentiable everywhere and so $f(1) = 0 = \sum_n a_{2n+1} = 0$ and $f'(1) = 2 = \sum_n (2n+1) a_{2n+1}$. Plugging this into the functional equation \begin{align} \sum_n a_{2n+1} x^{2(2n+1)} + \sum_n \sum_m a_{2n+1} a_{2m+1} x^{2(n+m+1)} = 2x^2 - 2 \end{align} Equating powers shows that $$ a_{2n+1} + \sum_k a_{2k+1} a_{2(2n-k)+1} = 2 \delta_{n0} \tag{1}$$ and $$ \sum_k a_{2k+1} a_{2(2n-k)-1} = -2 \delta_{n0} \tag{2}$$ for all $n$. At this point, I am not sure if there is a systematic way to proceed here, but you can show certain properties if the solution is of this form. First, note that you require at least two of the $a$'s to be nonzero. Searching for a solution of this type implies that it must be the pair $a_{2p+1},a_{-(2p+1)}$ to satisfy the second equality with $n = 0$, $a_{2p+1} a_{-(2p+1)} = -1$. With this, the first equality at $n = 0$, leads to $a_1^2 + a_1 = 2$ (the other terms drop out because the products are all 0 if $k \neq 0$). Solutions for $a_1 = 1,-2$. This also shows that $p = 0$ and so $a_{-1} = -1,1/2$.

Checking the case in equality (1) where $n \neq 0$: $$ 0 = a_{2n+1} + \sum_k a_{2k+1} a_{2(2n-k)+1} = a_{2n+1} + a_{2n+1}^2 \implies a_{2n+1} = 0, -1. $$ So all is fine (equality (2) is trivially satisfied when $n \neq 0$). Finally, the condition $f(1) = 0 = a_1 + a_{-1}$ implies that we choose the solution $a_{\pm 1} = \pm 1$ hence $f(x) = x - \frac{1}{x}$ is the only power series solution with two terms.

I am not sure if much more is possible.