Do there exist real polynomials $P(x)$ and $Q(x)$ with nonnegative coefficients, and $n>20$ a natural number such that
$$\left(\sum\limits_{k=0}^{n} x^k\right)^2=(x-3)^2P(x)+(x+4)^2 Q(x)?$$
I have asked this question here (*), but no answer.
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$\begingroup$ Fix $n$ and by non-negativity of the coefficients it's necessary that $\deg(P) \le 2n-2$, $\deg(Q) \le 2n-2$. You get a linear programming problem, which you can just throw at a standard solution procedure. How far have you attempted this numerically? $\endgroup$– Peter TaylorCommented Jun 5 at 11:43
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$\begingroup$ My goal is to grow this list : mathoverflow.net/questions/472089/… $\endgroup$– DattierCommented Jun 5 at 11:48
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1$\begingroup$ Does the right hand side, i.e. the factors $(x-3)^2$ and $(x+4)^2$, have a specific or motivated reason, or is this just some "random" question without special meaning? $\endgroup$– Peter MuellerCommented Jun 5 at 16:36
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$\begingroup$ I came across this question during my research. The goal of my research is to prove that even for problems that are considered difficult, there are probably short and very clever answers that no one has yet thought of. $\endgroup$– DattierCommented Jun 5 at 17:03
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1 Answer
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There is no solution for any $n$: For the values $x=2$ and $x=4$ we get \begin{align} P(2)+36Q(2) &= (2^{n+1}-1)^2\\ P(4)+64Q(4) &= ((4^{n+1}-1)/3)^2. \end{align} Note that $P$ and $Q$ have degrees $\le 2n-2$. Set $\alpha=\frac{64}{36}2^{2n-2}$. As $4^i-2^i\alpha \le0$ and $64\cdot4^i-36\cdot2^i\alpha\le0$ for $0\le i\le 2n-2$, we see that if we subtract $\alpha$ times the first equation from the second one, then the left handside is nonpositive. However, an easy calculation shows that $((4^{n+1}-1)/3)^2-(2^{n+1}-1)^2\alpha>0$, that is the right handside becomes positive: Setting $m=2^n$, hence $\alpha=(2m/3)^2$, we need to see that $(4m^2-1)/3>(2m-1)2m/3$, which clearly holds for $m>1/2$.
answered Jun 6 at 15:45