Let $P(x) \in \mathbb {Z}[X]$ be a polynomial of degree $2016$ with no rational roots. Prove that there exists a polynomial $T(x) \in \mathbb {Z}[X]$ of degree $1395$ such that for all distinct (not necessarily real) roots of $P(x)$ like $(\alpha ,\beta):$ $$T(\alpha)-T(\beta) \not \in \mathbb {Q}$$ Note: $\mathbb {Q}$ is the set of rational numbers.
Problem
Source: Iran MO 3rd round 2016 finals - Algebra P1
Tags: algebra, polynomial
K.N
01.09.2016 20:13
Sound hard
bgn
01.09.2016 20:50
K.N wrote: Sound hard
For all positive integers $k$, we define: $T_{k}(x)= (x+k)^{1395}$
We'll prove that for some $k, T_{k}(x)$ satisfies the desired condition.
Assume that for all $k,T_{k}(x)$ doesn't satisfy the problem, this means for all integers $k$, there exists $(u,v)$ such that $P(v)=P(u)=0$ and $T_{k}(u)-T_{k}(v) \in \mathbb {Q}$
Since $T_{k}(x)$ is defined for all positive integers and $P$ has only $2016$ roots, there exists $(\alpha , \beta)$ such that for infinitely many integers $k: T_{k}(\alpha)-T_{k}(\beta) \in \mathbb {Q}$
Let $S(x)=(\alpha + x)^{1395} - (\beta +x)^{1395}$
Since $S(n) \in \mathbb {Q}$ for infinitely many $n \in \mathbb {Z}$, by Lagrange interpolation we can easily see that $S(x) \in \mathbb {Q}[X]$
Now checking the coefficients :
$[x^{1394}]=1395(\alpha - \beta) \in \mathbb{Q} \Rightarrow \alpha -\beta \in \mathbb{Q}$
$[x^{1393}]=\binom{1395}{1393}(\alpha ^{2}-\beta ^{2})\in \mathbb{Q}\Rightarrow \alpha ^{2}-\beta ^{2}\in \mathbb{Q}$
Therefore $\alpha +\beta =\frac{\alpha ^{2}-\beta ^{2}}{\alpha -\beta }\in \mathbb{Q}$ . So we get: $2\alpha =(\alpha +\beta )+(\alpha -\beta )\in \mathbb{Q}\Rightarrow \alpha \in \mathbb{Q}$ Contradiction.
(I think the hard part was that the only information we needed, was there's just 2016 non-rational (not necessarily real) numbers.)
K.N
01.09.2016 20:53
bgn wrote: K.N wrote: Sound hard
For all positive integers $k$, we define: $T_{k}(x)= (x+k)^{1395}$
We'll prove that for some $k, T_{k}(x)$ satisfies the desired condition.
Assume that for all $k,T_{k}(x)$ doesn't satisfy the problem, this means for all integers $k$, there exists $(u,v)$ such that $P(v)=P(u)=0$ and $T_{k}(u)-T_{k}(v) \in \mathbb {Q}$
Since $T_{k}(x)$ is defined for all positive integers, there exist $(\alpha , \beta)$ such that for infinitely many integers $k: T_{k}(\alpha)-T_{k}(\beta) \in \mathbb {Q}$
Let $S(x)=(\alpha + x)^{1395} - (\beta +x)^{1395}$
Since $S(n) \in \mathbb {Q}$ for infinitely $n \in \mathbb {Z}$, by Lagrange interpolation we can easily see that $S(x) \in \mathbb {Q}[X]$
Now checking the coefficients :
$[x^{1394}]=1395(\alpha - \beta) \in \mathbb{Q} \Rightarrow \alpha -\beta \in \mathbb{Q}$
$[x^{1393}]=\binom{1395}{1393}(\alpha ^{2}-\beta ^{2})\in \mathbb{Q}\Rightarrow \alpha ^{2}-\beta ^{2}\in \mathbb{Q}$
Therefore $\alpha +\beta =\frac{\alpha ^{2}-\beta ^{2}}{\alpha -\beta }\in \mathbb{Q}$ . So we get: $2\alpha =(\alpha +\beta )+(\alpha -\beta )\in \mathbb{Q}\Rightarrow \alpha \in \mathbb{Q}$ Contradiction.
(I think the hard part was that the only information we needed, was there's just 2016 irrational (not necessarily real) numbers.)
Nice solution What was your motivation?
lebathanh
23.11.2016 15:49
hehe 1395/2016 approxietly= 0.696
IMD2
08.11.2021 12:26
What an unnecesarily worded problem. We don't need $P$ being a polynomial, and we only need degree 2 for $T$. New problem: Let $\alpha_1,...,\alpha_{2016}$ be irrationals, then there exists $T\in{}\mathbb{Z}[x]$ of degree 2 such that $T(\alpha_i)-T(\alpha_j)\notin{}\mathbb{Q}$ for $i\neq{}j$. Proceed by contradiction. For a quadratic $R(x)\in{}\mathbb{Z}[x]$, let $f(R(x))$ be the number of pairs $(i,j)$ s.t $R(\alpha_i)-R(\alpha_j)\in{}\mathbb{Q}$. Let $T$ be such that $f(T)$ is minimal. As $f(T)>0$, there exist $i,j$ such that $T(\alpha_i)-T(\alpha_j)\in{}\mathbb{Q}$. It can't be true that $\alpha_i-\alpha_j\in{}\mathbb{Q}$, and $\alpha_i^2-\alpha_j^2\in{}\mathbb{Q}$, assume wlog that $\alpha_i^2-\alpha_j^2\notin{}\mathbb{Q}$, and let $T(x)=ax^2+bx+c$. For every pair $k,l$ of indices with $\alpha_k^2-\alpha_l^2\notin{}\mathbb{Q}$, there exists at most one integer $m_{k,l}$ such that if $T*(x)=m_{k,l}x^2+bx+c$, then $T*(\alpha_k)-T*(\alpha_l)\in{}\mathbb{Q}$. So choose $m$ such that this doesn't happen for any pair that satisfies that, and let $T_2=mx^2+bx+c$, we will have $f(T_2)<f(T)$, contradiction.
mojyla222
20.03.2024 01:06
Proposed by Ali Behrooz.