Prove or disprove the existence of polynomials $p$ and $q$ for which $pe^p+qe^q=1$. Prove or disprove that there exist non-constant polynomials $p$ and $q$ such that $p(z)e^{p(z)}+q(z)e^{q(z)}=1$ for all $z\in \mathbb{C}$.  This question was first asked here Prove or disprove the existence of polynomials., and I tried following the hint but could not solve the problem.
Taking the derivative of both sides yields $e^pp'(p+1)+e^qq'(q+1)=0$, and since $e^p$ and $e^q$ are never zero, we can conclude that $p'(p+1)=0$ if and only if $q'(q+1)=0$. I could not see how to finish the problem using only this information, so I must be missing something else. Thank you.
 A: We cannot have $p=q$ else $2pe^p=1$ leading to $2p=e^{-p}$ which for nonconstant $p$ cannot hold. And neither can we have $p-q=k$ for a constant $k,$ otherwise from $pe^p+(p+k)e^{p+k}=1$ follows $p+(p+k)e^k=e^{-p},$ which again cannot hold for nonconstant $p.$
Now consider the relatin $e^pp'(p+1)+e^qq'(q+1)=0$ and let $r=q-p,$ and say $a=p'(1+p),\ b=q'(1+q).$ Then we arrive at 
$a+b e^r=0,$ which cannot hold since taking enough derivatives makes the first term zero but not the second, since the polynomial $r$ is not a constant as noted above.
[Note that $b$ is not $0$ otherwise either $q' \equiv 0$ or $q \equiv 1$ and $q$ is assumed nonconstant.] 
A: I will attempt an answer based on Kelenner's comment.  Taking repeated derivatives of the the equation $pe^p+qe^q=1$ yields that $p'(p+1)$ and $q'(q+1)$ have the same distinct roots each with the same multiplicity.  Hence $q'(q+1)=\lambda p'(p+1)$ for some $\lambda \in \mathbb{C}$.
Hence $p'(p+1)[e^p+\lambda e^q]=0$ for all $z$.  Since $p'(p+1)$ has only finitely many zeros, $e^p+\lambda e^q=0$ for all $z$ by the uniqueness principle.  Thus $p=q+\log (-\lambda)$ on some branch of the logarithm on which $log(-\lambda)$ is defined. Hence $p'=q'$, so we must have that $(q+1)=\lambda (p+1)$. But this implies $p$ is constant, which it cannot be. Thus no such nonconstant polynomials $p$ and $q$ exist.
