Prove that there does not exist $f : \mathbb{R}\to\mathbb{R}$ so that $f'(x)=k(f(x))^{2n} + x^{2m-1}\,\forall x\in\mathbb{R}$ 
Let $a,b$ be positive integers and let $k$ be a nonzero real number. Prove that there does not exist a differentiable function $f : \mathbb{R}\to\mathbb{R}$ so that $f'(x)=k(f(x))^{2a} + x^{2b-1}\,\forall x\in\mathbb{R}$.

This can most likely be solved using a proof by contradiction. A hint I got was that the key is that the derivative of arctan can't be always greater than a constant. I should consider $\arctan(f(x)).$ Suppose for the sake of contradiction that such a function $f$ exists. Consider $g(x) = \arctan (f(x)).$ $g'(x) = \dfrac{f'(x) }{1+f(x)^2} = \dfrac{k(f(x))^{2a} +x^{2b-1}}{ 1+ f(x)^2 }$. Is there some way to get a useful lower bound for this derivative?
 A: Let's start with the case $k>0$. Assume by contradiction that such a function exists. Note that for $x\geq x_0>0$ we have (by the fundamental theorem of calculus)
$$f'(x) = k (f(x))^{2n} + x^{2m-1} \geq x^{2m-1} \geq x_0^{2m-1}. $$
In particular, we have
$$ f(x) = f(x_0)+\int_{x_0}^{x} f'(s) ds \geq f(x_0) + (x-x_0) x_0^{2m-1}. $$
Hence, there exists $c>0$ such that for $x\geq c$, we get that $f(x)\geq 1$.
On the other hand, we have (again by the fundamental theorem of calculus)
$$ \pi/2 \geq \arctan(f(x)) = \arctan(f(c)) + \int_c^x \frac{f'(s)}{1+f(s)^2} ds
= \arctan(f(c)) + \int_c^x \frac{k(f(s))^{2n}+s^{2m-1}}{1+f(s)^2} ds
\geq \arctan(f(c)) + \int_c^x \frac{k(f(s))^{2n}}{1+f(s)^2} ds
\geq \arctan(f(c)) + \frac{1}{2}\int_c^x k f(s)^{2n} ds
\geq \arctan(f(c)) + \frac{k}{2}(x-c), $$
where we used in the last and second last line that $f(s) \geq 1$ for $x\geq c$. However, for $x\rightarrow \infty$ the RHS of the inequality blows up, while the LHS is bounded, which yields a contradiction.
Similarly we can deal with the case $k<0$. Let $x\leq -1$, then we have by the fundamental theorem of calculus
$$ \pi/2 \geq \arctan(f(x)) = \arctan(f(-1)) + \int_x^{-1} \left((-k) \frac{f(s)^{2n}}{1+f(s)^2} - \frac{s^{2m-1}}{1+f(s)^2} \right) ds. $$
Note that for $\vert f(s) \vert \geq 1$ we have
$$ \frac{f(s)^{2n}}{1+f(s)^2} = \frac{f(s)^{2n}+1-1}{1+f(s)^2} \geq 1 - \frac{1}{1+f(s)^2} \geq \frac{1}{2}. $$
On the other hand, we have for $\vert f(s) \vert \leq 1$ and $s\leq -1$
$$-\frac{s^{2m-1}}{1+f(s)^2} \geq -\frac{s^{2m-1}}{2} \geq \frac{1}{2}.$$
Thus, we get for $x\leq -1$
$$ \pi/2 \geq \arctan(f(-1)) + \int_x^{-1} \frac{(-k)+1}{2} ds = \arctan(f(-1)) + (-1-x)\frac{\min\{(-k), 1\}}{2}.$$
The RHS blows up for $x\rightarrow -\infty$ and we have again a contradiction.
