<s>[INST] Find all continuous functions satisfying $\int_0^xf=(f(x))^2+C$ for some constant $C \neq 0$. Find all continuous functions $f$ satisfying 
$$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.
I have a question about the solution to this problem. It says that 
Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and 
$$f(x)=2f(x)f'(x),$$
so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$ 
so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.
I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?
Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?
Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?
I would appreciate it if anyone answers my questions.
 [/INST] putting $x = 0,$ gives $C = -f^2(0)$  differentiating gives you $f = 2ff'$ which has solution $f = 0$ which implies $C = 0$ or $f = \frac{x}{2} + f(0).$  
$\int_0^x(f(0) + t/2) \, dt = xf(0) + \frac{x^2}{4}$  and $f^2(x) + C = \frac{x^2}{4} + xf(0) + f^2(0) - f^2(0) = xf(0) + \frac{x^2}{4}$ it verifies that 
$$f(x) = \frac{x}{2} + f(0), f(0) \neq 0, C = -f^2(0).$$
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