Minimize the number of roots

Algebra Level 5

Suppose $p(x)$ and $q(x)$ are polynomials of degree $100$ with complex coefficients, having no common zeroes. Find the smallest possible total number of complex zeroes of the polynomials $p,$ $q,$ and $p-q,$ counted without multiplicity.

Details and assumptions

When the zeroes are counted without multiplicity, the polynomial $x^2(x-1)^3$ has two zeroes: $x=0$ and $x=1.$

The answer is 101.

2 solutions

Michael Tong
Aug 1, 2013

Note: This solution is somewhat incomplete or not as thorough as would be completely ideal. Unfortunately I must type this from a phone so I will give a general overview of the problem.

Clearly, the roots of p(x) and q(x) can be minimized when they have the least amount of roots themselves as possible. WLOG, say $p(x) = (x - a)^{100}$ and $q(x) = (x - b)^{100}$ , giving us roots $a$ and $b$ .

Then, we must find the minimum number of roots in $p(x) - q(x)$ . This can be done by either getting rid of leading terms to decrease the degree or by having roots a or b.

Note that $p(x) - q(x) = 100(b-a)x^{99} + (100 C 2)(a^2-b^2))x^{98} + ...$ (the degree was decreased by one because the first term of both polynomials is $x^{100}$ ). By observing Vieta's formulas, it becomes clear that all of the roots of this polynomial must be distinct from each other and from a and b. Thus, there are a minimum of $1 + 1 + 99 = 101$ roots.

Again, sorry for not explaining much, it's very hard to type a comprehensive solution on a phone (no internet here!). As soon as I can use my computer I will elaborate the solution in a reply.

Moderator note:

Minimizing the roots of $p(x)$ and $q(x)$ do not necessarily mean that we would minimize the total roots of $p, q, p-q$ . There is no evidence that the total number cannot be less than 100.

The ABC conjecture has a proof that is currently being checked, and hasn't been verified as yet. This problem is related to Mason-Stothers theorem , which is the "polynomial analogue".

Now that I read the comments, I realize that some of the assumptions I make in the problem involve knowledge of the abc conjecture. http://en.wikipedia.org/wiki/Abc_conjecture

Michael Tong - 7 years, 10 months ago

Cody Johnson
Jul 28, 2013

Note that this solution is incomplete.

Let $p(x)=(x-r_1)^{100}$ and $q(x)=(x-r_2)^{100}$ . Since the $x^{100}$ terms cancel, $p(x)-q(x)$ must be of degree 99 (begin flaw) and therefore have 99 distinct roots (end flaw). This gives $1+1+99=\boxed{101}$ roots.

Moderator note:

This only shows that there are 2 polynomials which have a total of 101 complex zeros (given your flaw).

It needs to be shown that $p, q, p-1$ must have at least 101 complex zeros.

What is the proof that $101$ indeed is the smallest possible number of zeroes? What if $p(x) - q(x)$ has some repeated roots?

Sreejato Bhattacharya - 7 years, 10 months ago

Seems like the proof that the answer is at least 101 requires some form of the abc conjecture for polynomials, which is a theorem of Mason and Stothers. Here is an elementary proof: http://topologicalmusings.wordpress.com/2008/03/03/mason-stothers-theorem-and-the-abc-conjecture/

Patrick Corn - 7 years, 10 months ago

Thanks for the link, Patrick.

There is one slight flaw in that proof -- namely, the way it invokes symmetry after having previously said "So, without any loss of generality, assume f and g are non-constant polynomials." But that's easily fixed: just note that if, say, g'=0 then f'g-fg'=f'g is nonzero as desired.

Peter Byers - 7 years, 10 months ago

Minimize the number of roots

The answer is 101.

2 solutions

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