# Solve the functional equation $\left(f(x)\right)^2 - f\left(x^2\right)=2x$

$$\forall x \in \Bbb R ,$$ Find all functions that satisfy: $$\left(f(x)\right)^2 - f\left(x^2\right)=2x$$

My solution: $$f(x) \cdot f(x) - f(x \cdot x)=x+x$$

$$f(x) \cdot f(y) - f(x \cdot y)=x+y$$ $$f(x) \cdot f(0) -f(0)=x$$ $$f(x)=\frac{x}{f(0)}+1$$ $$x=0 \to f(0)=1$$ $$f(x)=x+1$$

• How do you justify going from $f(x) \cdot f(x) - f(x \cdot x) = x+x$ to $f(x) \cdot f(y) - f(x \cdot y) = x + y$? Commented Mar 6, 2023 at 16:20
• $$f(x) \cdot f(y) - f(x \cdot y)=x+y$$ If we write $x=y$ in this equation, we will get the equation given in the condition. If this solution is not correct. I will be glad, if you explain why. Commented Mar 6, 2023 at 16:29
• The fact that an equation holds when $x=y$ doesn't mean that it holds for all pairs $(x,y)$.
– lulu
Commented Mar 6, 2023 at 16:36
• Going from $$f(x)^2-f(x^2)=2x$$ to $$f(x)f(y)-f(xy)=xy$$ is not at all justified. You cannot just freely change some parts of it but not others. However it is true that if a function satisfies the second equation, then it also satisfied the first; indeed as you pointed out we can go from the second to the first by taking $y=x$. This does mean that we can find some solution(s) by solving the second equation. This does not mean, however, that we will find all, as there could (potentially) be functions satisfying the first but not the second equation. Commented Mar 6, 2023 at 16:37
• Also, it's not clear to me how you know that $f(0)=1$. Letting $y=f(0)$ we get $y^2-y=0$ which only tells you that $f(0)\in \{0,1\}$.
– lulu
Commented Mar 6, 2023 at 16:42

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 $$f(x) = \sqrt{f(x^2) + 2 x}$$.

• That's much easier than my equiva lent approach. But do we know all functions must satisfy $f(x)\geq x+1$ for $x$ positive? Commented Mar 6, 2023 at 17:20
• You definitely need $f(x)\geq -2\sqrt x$ when $x$ is positive. Commented Mar 6, 2023 at 17:24
• To define $f(-\sqrt{x})$, you need $f(x) \ge 2 \sqrt{x}$ when $x > 0$. No, I'm not sure that you need $f(x) \ge x+1$ for $x > 0$. Commented Mar 6, 2023 at 18:47

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.

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.