$\forall x \in \Bbb R ,$ Find all functions that satisfy: $$\left(f(x)\right)^2 - f\left(x^2\right)=2x$$
My solution: $$f(x) \cdot f(x) - f(x \cdot x)=x+x$$
$$f(x) \cdot f(y) - f(x \cdot y)=x+y$$ $$f(x) \cdot f(0) -f(0)=x$$ $$f(x)=\frac{x}{f(0)}+1 $$ $x=0 \to f(0)=1$ $$f(x)=x+1$$
On
Partial approach. This will define a large class of answers that are not necessarily continuous.
Let $g(x)=f(x)-(x+1).$
Then $f(x)^2=g(x)^2+2(x+1)g(x)+(x+1)^2$ and $f(x^2)=g(x^2)+(x^2+1).$
So:
$$2x = f(x)^2-f(x^2)= g(x)^2+2(x+1)g(x)-g(x^2)+2x$$
Or $$g(x)^2+2(x+1)g(x)=g(x^2).\tag 1$$
So if we define $g(x)$ as a positive function on $(1/4,1/2)$ we can define inductively for $x\in [2^{-2^{k+1}},2^{-2^{k}}],$ as $g(x)=g(\sqrt x)^2+2(x+1)g(\sqrt x).$
And for negative $k$ and $x\in[2^{-2^{k+1}},2^{-2^k}):$
$$g(x)=-(x+1)+\sqrt{(x+1)^2+g(x^2)},\tag2$$ which is gotten by solving the quadratic equation $(1)$ and choosing the positive case.
We can choose either $g(0)=1$ or $g(0)=0.$ Likewise we can choose $g(1)$ from $1,2.$
This defines $g$ on $(0,1).$ We can define $g(0)$ as $0$ or $-1$ and likewise $g(1)$ as $0$ or $-3.$ But we need $g(1)=0$ since $g(-1)^2=g(1).$ Note that these definitions ensure $g(x)$ is always positive.
We can likewise define $g$ as positive on $[2,4)$ and again define inductively to get $g$ defined on all of $(1,\infty).$
Then the problem is the negative cases. But again, we can define the negative cases from $(2)$ again, except the problem is that some $g(x^2)$ might be negative.
So, we can find a large set of such functions if they don't have to be continuous.
I doubt these are ever continuous except when $g(x)$ is everywhere $0,$ but I might be wrong.
And this might not be all the possible examples, since this defines only the cases when $g$ is positive on the positive reals.
On
I can show that $f(-1)=0$ and $f(1)=2$ although I cannot use a similar approach to nail down a value for $f(0)\in\{0,1\}$:
Taking $x=1$ we have
$$f(1)^2-f(1)=2\Rightarrow f(1)\in \{-1,2\}$$
Taking $x=-1$ and both possibilities of $f(1)$ from above we have
$$f(-1)^2-(-1)=-2\Rightarrow f(-1)^2+3=0\Rightarrow \text{impossible}$$
$$f(-1)^2-(2)=-2\Rightarrow f(-1)^2=0\Rightarrow f(-1)=0$$
This also gives us that $f(1)=-1$ is impossible.
For $x = 1$ you must have $f(1)^2 - f(1) = 2$, so $f(1) = -1$ or $2$ (EDIT: but $-1$ will make a real value for $f(-1)$ impossible, as QC_QAOA points out, so it should be $2$). For $x = 0$, $f(0)^2 - f(0) = 0$, so $f(0) = 0$ or $1$. On $(\sqrt{2},2]$, and $[1/2, 1/\sqrt{2})$ let $f(x)$ be arbitrary with $f(x) \ge x+1$. On $(2,\infty)$ and $(0,1/2)$ define $f$ by iterating $f(x^2) = f(x)^2 - 2 x$, and on $[1/\sqrt{2}, 1)$ and $(1,\sqrt{2})$ by iterating $f(\sqrt{x}) = \sqrt{f(x)+2\sqrt{x}}$. Note that in each case this preserves the property $f(x) \ge x+1$. Finally, for $x < 0$ define $f(x) = \sqrt{f(x^2) + 2 x}$.