Assume natural number n≥2. Amin and Ali take turns playing the following game: In each step, the player whose turn has come chooses index i from the set {0,1,⋯,n}, such that none of the two players had chosen this index in the previous turns; also this player in this turn chooses nonzero rational number ai too. Ali performs the first turn. The game ends when all the indices i∈{0,1,⋯,n} were chosen. In the end, from the chosen numbers the following polynomial is built: P(x)=anxn+⋯+a1x+a0Ali's goal is that the preceding polynomial has a rational root and Amin's goal is that to prevent this matter. Find all n≥2 such that Ali can play in a way to be sure independent of how Amin plays achieves his goal.
Problem
Source: Iran MO Third Round N1
Tags: polynomial, number theory, Rational roots
08.09.2022 12:16
Firstly , for each even n we'll show that Ali has the winning strategy. Consider that the coefficient ai stays undetermined till the last move ( which Ali will make it. ). Then Ali can choose an arbitrary rational number r and one can see that if we put ai like below , then we have P(r)=0 and we're done. ai=∑j≠iajrjri∈QNow if n is an odd number , then since n≥3 Amin can play somehow a0 be determined before his last move and if he wants to determine coefficient ai at last , we'll show that there exist such a rational number , such that P(x) doesn't have a rational root. Suppose that for a ai , number r=mn′ is a rational root for P(x) such that gcd(m,n′)=1 , then one can see that : n'^nP(r) \equiv n'^na_0 \equiv 0 \pmod m \implies m|a_0So all possible rational roots are numbers in the form \frac{a_0}{j} for a j \in \mathbb{Z} which are all in the interval [-a_0 , a_0]. So if we procced rational function f(x) such below , it's enough to show that this function , doesn't procced all rational values for x=\frac{a_0}{j}. f(x)=\frac{\sum_{j \neq i}{a_jr^j}}{x^i}Which is true since this function values are limited in interval [-a_0 , a_0] because the limit of this function doesn't tend to infinity while x tends to 0. \lim_{x \to 0}{\frac{\sum_{j \neq i}{a_jr^j}}{x^i}}=\lim_{x \to 0}{\frac{a_nx^n}{x^i}}=\lim_{x \to 0} a_nx^{n-i}=a_n \text{or} 0
17.09.2022 09:23
18.09.2022 12:29
Mehrshad wrote:
Yes , actually there is the step that the second player should make all the coefficients integer and then determine a_i , which corrects the m|a_{j}m^{k} part and I forgot to wrote. And while all possible roots are going to be in the form \frac{a_0}{j} , where is that function part problem ? :// Anyway , this is not totally my sol in exam , I changed the last part for posting that here )
19.09.2022 16:53
Shayan-TayefehIR wrote: And while all possible roots are going to be in the form \frac{a_0}{j} , where is that function part problem ? :// Okay, I didn't say the roots aren't in the form \frac{a_0}{j}. But the limit that you wrote was wrong. When x approaches 0, the a_nx^n is not so effective but a_0 is the most important nominal and \frac{a_0}{x^i} approaches infinity.
28.03.2023 14:28
Mehrshad wrote: Shayan-TayefehIR wrote: And while all possible roots are going to be in the form \frac{a_0}{j} , where is that function part problem ? :// Okay, I didn't say the roots aren't in the form \frac{a_0}{j}. But the limit that you wrote was wrong. When x approaches 0, the a_nx^n is not so effective but a_0 is the most important nominal and \frac{a_0}{x^i} approaches infinity. how can you find the official solution
20.06.2023 18:00
ylt_chn wrote: Mehrshad wrote: Shayan-TayefehIR wrote: And while all possible roots are going to be in the form \frac{a_0}{j} , where is that function part problem ? :// Okay, I didn't say the roots aren't in the form \frac{a_0}{j}. But the limit that you wrote was wrong. When x approaches 0, the a_nx^n is not so effective but a_0 is the most important nominal and \frac{a_0}{x^i} approaches infinity. how can you find the official solution There is a channel in the Bale messenger named دوره تابستانی المپیاد ریاضی 1401 here which in addition to solutions also has marking schemes.
19.09.2024 17:30
Not the most interesting polynomial game (as everything could be made up to depend only on the last move), but certainly a nice number theory thing to try out!