Suppose $f$ is a uni-variate polynomial of degree at most $2k-1$ for some integer $k\geq1$. Let $f^{(m)}$ denote the $m$-th derivative of $f$. If $f$ and $f^{(m)}$ have $k$ distinct common roots then, Is it true that $f$ has to be a zero polynomial? Here $m<k$ is a positive integer. This statement is true for $m=1$ but is it true for larger $m$ also?
[Math] Common roots of polynomial and its derivative
differential-calculuspolynomials
Related Solutions
First, a counterexample to your conjecture. Let $\Pi = x^4+x^3+4x^2+4x = x(x+1)(x^2+4)$, so $P = 4x^3+3x^2+8x+4$. The critical values of $\omega$ are $1.06638, 3.89455 + 2.87687i, 3.89455 - 2.87687i$, and by inspection (using Mathematica) we see that for each of these values of $\omega$, $\mbox{Conv}(\Pi_\omega)$ contains a neighborhood of $0$.
Now for a calculus on convex sets. Every convex set is the intersection of a set of halfplanes. Call a halfplane in this collection essential if removing all of the halfplanes in an open set of halfplanes (in the halfplane topology) containing it from our set of halfplanes makes the intersection of the halfplanes in our set bigger. We wish to find a characterization of the essential halfplanes of $\mbox{Hull}(P)$.
First of all, I claim that any essential halfplane of $\mbox{Hull}(P)$ occurs as an essential halfplane of $\mbox{Conv}(\Pi_\omega)$ for some $\omega$. This follows from continuity - for any open set around our essential halfplane there is some $\omega$, take the limit of a subsequence of these $\omega$s...
Now, suppose that the halfplane $\mbox{Re}(x) \le 0$ occurs as an essential halfplane of some $\mbox{Conv}(\Pi_\omega)$, i.e. there are at least two roots of $\Pi_\omega$ with real part $0$, and the rest of the roots have negative real part. If the number of roots on the line $\mbox{Re}(x) = 0$ (counted with multiplicity) is two, then by holomorphicity we can always find a direction to move $\omega$ so that either both roots move to the left, or both roots stay on the line $\mbox{Re}(x) = 0$ and move towards eachother. If we can ever make both roots move to the left, then clearly the halfplane $\mbox{Re}(x) \le 0$ is not an essential halfplane of $\mbox{Hull}(P)$, otherwise we keep pushing the roots towards eachother until either they run into eachother or until a third root hits the line $\mbox{Re}(x) = 0$. In any case, we see that if a halfplane is essential for $\mbox{Hull}(P)$, then there is some $\omega$ such that the halfplane is essential for $\mbox{Conv}(\Pi_\omega)$ and such that at least three roots (counted with multiplicity) of $\Pi_\omega$ are on the boundary of the halfplane, or two of the roots are equal and $\Pi_\omega$ has no other roots.
So if we let $L$ be the set of $\omega$s such that three of the roots of $\Pi_\omega$ lie on a line, we get that $\mbox{Hull}(P) = \cap_{\omega \in L} \mbox{Conv}(\Pi_\omega)$ if $\deg P \ge 2$.
Edit: Actually, I think there is a problem with this. It's conceivable that two roots are on the line $\mbox{Re}(x) = 0$ and have derivatives (with respect to $\omega$) pointed in opposite directions, such that we can't simply push them towards eachother. For instance, the map from one root to the other root could, locally, look like the fractional linear transform sending the left halfplane to a circle contained in the right halfplane and tangent to the line $\mbox{Re}(x) = 0$ at the other root. So, we may need to enlarge the set $L$ to contain also those $\omega$s for which the ratio of the derivatives of two of the roots (with respect to $\omega$) is a negative real number.
Edit 2: It turns out that this isn't an issue. Call the two roots on the line $\mbox{Re}(x) = 0$ $r_1$ and $r_2$. Suppose that locally, $r_1(\epsilon) = \epsilon$, $r_2(\epsilon) = i - m\epsilon + a\epsilon^k + O(\epsilon^{k+1})$, $a \ne 0$, $m > 0$. Note that if we had $r_2(\epsilon) = i-m\epsilon$, then the intersection of the halfplanes corresponding to $r_1(\epsilon), r_2(\epsilon)$ and $r_1(-\epsilon), r_2(-\epsilon)$ would be contained in the halfplane $\mbox{Re}(x) \le 0$, and the intersection of their boundaries would be located at $i/(m+1)$. Now if $k$ is even, then the correction term shifts the intersection of the boundaries by $a\epsilon^k/(m+1) + O(\epsilon^{k+1})$, so if we choose $\epsilon$ small such that $a\epsilon^k$ is real and negative, then we see that the halfplane $\mbox{Re}(x) \le 0$ is not essential. If $k$ is odd, then if we choose $\epsilon$ small such that $\mbox{Re}(\epsilon) < 0$ and $a\epsilon^k$ is a positive real times $i$, then $r_2(\epsilon)$ is shifted up and $r_2(-\epsilon)$ is shifted down, so the intersection of the boundaries will be shifted to the left (draw a picture), so again the halfplane $\mbox{Re}(x) \le 0$ is not essential.
Suppose one wishes to solve a system $P_1(n_1,\dots,n_k) = \dots = P_m(n_1,\dots,n_k)=0$ of diophantine equations involving polynomials $P_1,\dots,P_m$ of various degrees. We have the following probabilistic heuristic (discussed for instance in this blog post of mine): if we pick $n_1,\dots,n_k$ to be random integers of size $N$ (here we make the assumption that the dominant scenario is one in which the $n_1,\dots,n_k$ all have comparable magnitude), then each equation $P_j(n_1,\dots,n_k)=0$ heuristically has a probability $\sim N^{-\mathrm{deg} P_j}$ to hold, so (assuming no unusual correlations) the whole system should be satisfied with probability $\sim N^{-\mathrm{deg} P_1 - \dots - \mathrm{deg} P_m}$. On the other hand, we have $\sim N^k$ choices of candidates $(n_1,\dots,n_k)$ here. Thus, if we define the degree surplus (or degree-of-freedom deficit) of this system to be $\mathrm{deg} P_1 + \dots + \mathrm{deg} P_m - k$, we arrive at the following naive predictions:
- For a negative surplus, solutions should be quite abundant (unless there are some obvious local obstructions, or maybe some less obvious obstructions such as Brauer-Manin obstructions), and locatable by brute force search.
- For a positive surplus, one would typically only expect a small number of sporadic or degenerate solutions, or no solutions at all, unless there is some algebraic miracle that allows one to lower the surplus, for instance through some birational embedding. If the surplus is close to zero, e.g., if it is equal to $1$, then it might not be too much to ask for such a miracle, but when the surplus is large then this begins to look unreasonable.
- For zero surplus, the situation is delicate and could go either way.
Very roughly speaking, this surplus is a crude measure of how many "algebraic miracles" would be needed to be present in this problem in order to produce a large number of non-trivial integer solutions.
This heuristic has to be taken with several grains of salt - for instance, one needs the polynomials $P_1,\dots,P_m$ to be "independent" in some sense for this prediction to be accurate, one is implicitly assuming that the $n_1,\dots,n_k$ are probably of comparable magnitude, and this notion of surplus is not a birational invariant, in contrast to more intrinsic notions such as the genus of an algebraic curve - but it can still serve as a starting point for making initial guesses. Some examples:
- The Fermat equation $a^n + b^n = c^n$ has surplus $n-3$ and so one expects a lot of integer solutions for $n=2$, almost no non-trivial solutions for $n>3$, and no definitive conclusion for $n=3$. (Of course, as it turns out, there are no non-trivial solutions for $n=3$, but the story for say $a^3 + b^3 = c^3 + h$ for other constants $h$, e.g., $h=33$, can be different; see https://en.wikipedia.org/wiki/Sums_of_three_cubes .)
- Rational points on elliptic curves $y^2 = x^3 + ax + b$ in Weierstrass form, after clearing out denominators, becomes a cubic equation in three integer variables with a surplus of $3-3=0$, which is consistent with the fact that some elliptic curves have positive rank and thus infinitely many rational points, and others have rank zero and only finitely many rational points (or none at all).
- An elliptic equation in quartic form $y^2 = a_4 x^4 + \dots + a_0$ has a surplus of $4-3=1$ so one would naively expect very few rational points here; but (assuming the existence of a rational point) one can use a (rational) Mobius transformation (moving the rational point to infinity in a suitable fashion) to convert the equation into Weierstrass form, thus lowering the degree surplus down to $0$.
The question here is that of finding integer solutions $a_1,\dots,a_d,b_1,\dots,b_{d-1}$ to the equation $$ \frac{d}{dx} (x-a_1) \dots (x-a_d) = d (x-b_1) \dots (x-b_{d-1}).$$ Comparing coefficients, the naive surplus here is $$ 1 + 2 + \dots + d-1 - (d + d-1) = \frac{d^2-5d+2}{2}.$$ As noted in comments, this is negative for $d=1,2,3,4$, suggesting plentiful solutions in those cases. For $d=5$ the surplus is $1$, suggesting a minor miracle is needed to find solutions. For $d>5$ the surplus is quite large and one would not expect many solutions. However, with symmetry reductions one can do a bit better. For $d$ even one can look at the symmetrised equation $$ \frac{d}{dx} (x^2-a^2_1) \dots (x^2-a^2_{d/2}) = d x(x^2-b^2_1) \dots (x^2-b^2_{d/2-1})$$ and now the surplus is $$ 2 + 4 + \dots + (d-2) - (\frac{d}{2} + \frac{d}{2}-1) = \frac{d^2-6d+4}{4}.$$ In particular for $d=6$ the surplus is reduced from $4$ to $1$, so one is only "one miracle away" from finding many solutions in some sense. But it gets rapidly worse, for instance for $d=8$ this symmetrised problem has surplus $5$ (though this still improves substantially from the non-symmetric surplus of $13$).
In the paper of Choudry linked by the OP, some ad hoc substitutions and birational transformations are used in the non-symmetric $d=5$ case and the symmetric $d=6$ case (both of which have a surplus of $1$, as noted before) to extract a subfamily of solutions parameterized by a specific elliptic curve in quartic form; the surplus is still $1$, but as discussed previously one can then use a Mobius transform to send the surplus to $0$, giving hope that solutions exist. (One still has to be lucky in that the rank of the elliptic curve is positive, but this happens to be the case for both of the elliptic curves Choudry ends up at, so that paper in fact succeeds in obtaining an infinite family of solutions in both cases.)
Thus I would be doubtful of any non-degenerate solutions (beyond perhaps some isolated sporadic ones) for $d \geq 7$, even if one imposes symmetry; one would need quite a long sequence of miraculous surplus-lowering transformations to be available, and absent some remarkable algebraic structure in this problem (beyond the affine and permutation symmetries) I don't see why one should expect such a sequence to be present. Establishing this rigorously (or even conditionally on conjectures such as the Bombieri-Lang conjecture) looks quite hard though. Being homogeneous, the problem is one of finding rational points to a certain projective variety (avoiding certain subvarieties corresponding to degenerate solutions); if one could somehow show this variety to be of general type, Bombieri-Lang would then say that one should only expect solutions that are trapped inside lower dimensional subvarieties, but a complete classification of such subvarieties looks extremely tedious at best given the relatively high dimension and degree. (And in view of the undecidability of Hilbert's tenth problem, which already shows up at relatively low choices of degree and dimension, there is little chance of a systematic way to resolve these questions even with the aid of conjectures such as Bombieri-Lang.) Still I would very much doubt that there are any non-trivial infinite families of solutions for $d>6$.
Best Answer
Assume $a,b,c \in \mathbb{R}$ solve $$2(a^3+b^3+c^3)-3(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2)+12abc=0,$$ e.g. $(a,b,c)=(-1,1,3)$. Then $$ \begin{eqnarray} f(x)&:=&(x-a)(x-b)(x-c)(3x^2-2(a+b+c)x+3(ab+bc+ca)-2(a^2+b^2+c^2))\\ &=&3x^5-5(a+b+c)x^4+10(ab+bc+ca)x^3\\ &&+(2(a^3+b^3+c^3)-3(a^2b+ab^2+\dots)-18abc)x^2+\dots \end{eqnarray} $$ satisfies $$f^{(2)}(x)=60(x-a)(x-b)(x-c)$$ and is therefore a counterexample for $m=2$ and $k=3$.