Let $P$ be a polynom of degree $n \geq 5$ with integer coefficients given by $P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_0 \quad$ with $a_i \in \mathbb{Z}$, $a_n \neq 0$. Suppose that $P$ has $n$ different integer roots (elements of $\mathbb{Z}$) : $0,\alpha_2,\ldots,\alpha_n$. Find all integers $k \in \mathbb{Z}$ such that $P(P(k))=0$.
Problem
Source: French TST 2005 pb 6.
Tags: inequalities, triangle inequality, algebra unsolved, algebra
Valentin Vornicu
27.05.2005 19:01
Nice problem! Let me post my solution of this: Let $ P(x) = a_nx(x - \alpha_2)\cdots (x - \alpha_n)$. We shall prove that the equation $ P(x) = \alpha_2$ has no integer solution, so WLOG, none of the equations $ P(x) = \alpha_k$ has integer solutions. Thus all the integer solutions of $ P(P(x)) = 0$ are exactly the roots of $ P(x)$. Since all the $ \alpha_k$-s are distinct, and $ n > 4$ it follows that if $ |P(k)| = |a_nk(k - \alpha_2)\cdots (k - \alpha_n)| \geq 4 = 1 \cdot 1 \cdot 2 \cdot | - 2| \cdot | - 1|$ for all integers $ 0\neq k \neq \alpha_i$, $ i = \overline{2,n}$. Therefore if $ k$ is an integer solution of $ P(x) = \alpha_2$ then $ |\alpha_2|\geq 4$. Now we have $ (|k|\cdot |k - \alpha_2|) \mid \alpha_2$, and as such $ |k| \neq |\alpha_2|$ and also $ |k - \alpha_2| \neq |\alpha_2|$ (for obvious reasons), therefore we have $ 2 |k| \leq |\alpha_2|$ and also $ 2 |k - \alpha_2 | \leq |\alpha_2|$. Adding these last two relations up side by side we obtain \[ |\alpha_2| \leq |k| + |k - \alpha_2| \leq |\alpha_2| \] by the triangle inequality, so we must have $ |\alpha_2 | = 2|k|$ and also $ |\alpha_2| = 2|k - \alpha_2|$ but then $ |k|\cdot |k - \alpha_2| = \frac {|\alpha_2|^2}4 \geq |\alpha_2|$ which means that $ 2\leq |a_n (k - \alpha_3)\cdots (k - \alpha_n)| \leq 1$ which is a contradiction. The problem is thus finished
pbornsztein
27.05.2005 20:11
Try the others Valentin, maybe you will qualify for french team Pierre.
Valentin Vornicu
27.05.2005 20:38
pbornsztein wrote: Try the others Valentin, maybe you will qualify for french team Pierre. The ones I solved were already posted I was too late in posting this year
pbornsztein
27.05.2005 20:47
Bad luck..., maybe will you qualify next year. Pierre.
bellahsayn
03.06.2005 01:03
hi i have the same probleme but generally find the solutions of$P(P(x))=0$ if $deg(p(x))=n$and$p(x)$has n root
indybar
31.07.2005 11:33
Valentin Vornicu wrote: \[ |\alpha_2| \leq |k| + |k-\alpha_2| \leq |\alpha_2| \] by the triangle inequality, so we must have $|\alpha_2 |=2|k|$ and also $|\alpha_2|=2|k-\alpha_2|$ Why is that? I don't understand.
N.T.TUAN
03.02.2007 18:03
Valentin Vornicu wrote: which means that $2\leq |a_{n}(k-\alpha_{3})\cdots (k-\alpha_{n})| = 1$ . No, It is $2\leq |a_{n}(k-\alpha_{3})\cdots (k-\alpha_{n})| \leq 1$ . @indybar: If $A\leq B,C\leq D$ and $A+C=B+D$ then $A=B$ and $C=D$
Maharjun
21.03.2010 09:17
since $ 0$ is a root of $ P(x)$ $ \implies a_0 = 0$
let $ P(x_i) = \alpha_i$
$ \& {a_n}{x_i}^n + {a_{n - 1}}{x_i}^{n - 1} + {a_{n - 2}}{x_i}^{n - 2} + \ldots \ldots + {a_2}{x_i}^2 + {a_1}{x_i} = {\alpha _i} \\
\\
\& \implies {x_i}|{\alpha _i} \implies {x_i}|{\alpha _i} - {x_i}$
$ \& P\left( {{\alpha _i}} \right) = 0 \\
\& P\left( {{x_i}} \right) = {\alpha _i} \\
\& \implies {\alpha _i} - {x_i}|P\left( {{\alpha _i}} \right) - P\left( {{x_i}} \right) \implies {\alpha _i} - {x_i}| - {\alpha _i} \implies {\alpha _i} - {x_i}|{x_i}$
$ \implies {\alpha _i} - {x_i} = \pm {x_i} \implies {\alpha _i} = 0{\text{ or }}{x_i} = \frac {{{\alpha _i}}} {2}$
the case of $ \alpha_i = 0$ is trivial. let us consider the case of $ {\alpha_i \over 2} = x_i$
let $ P'\left( x \right) = {a_n}{x^{n - 1}} + {a_{n - 1}}{x^{n - 2}} + {a_{n - 2}}{x^{n - 3}} + \ldots \ldots + {a_2}x + {a_1}$
$ \& P'\left( {{\alpha _i}} \right) = 0 \\
\& P'\left( {{x_i}} \right) = \frac {{{\alpha _i}}} {{{x_i}}} = 2 \\
\\
\& P'\left( x \right) = {a_n}\left( {x - {\alpha _1}} \right)\left( {x - {\alpha _2}} \right)\left( {x - {\alpha _3}} \right) \ldots \ldots \left( {x - {\alpha _n}} \right),{\text{ }}n \ge 4 \\
\\
\& \implies \left| {P'\left( x \right)} \right| = \left| {{a_n}\left( {x - {\alpha _1}} \right)\left( {x - {\alpha _2}} \right)\left( {x - {\alpha _3}} \right) \ldots \ldots \left( {x - {\alpha _n}} \right)} \right| \\
\ge \left| {{a_n}} \right|\left| { - 1} \right|\left| 1 \right|\left| 2 \right|\left| { - 2} \right| \ge 4{\text{ }}\left[ {If{\text{ }}x \in {\Bbb Z}} \right] \\
\\
\& \implies 2 \ge 4$
Contradiction
Hence for no $ x \in \Bbb Z$ we have $ P(x) = \alpha_i$
hence the only values of $ k$ are the roots of $ P(x) = 0$
Sahil
16.01.2012 12:07
My solution: $P(x)=0$ only at those points where $x$ $\epsilon$ $\{0,\alpha_2,\alpha_3...\alpha_n\}$ Hence $P(P(k))=0$ iff $P(P(k))$ $\epsilon$ $\{0,\alpha_2,\alpha_3...\alpha_n\}$ $P(k)=0$ implies that $k$ $\epsilon$ $\{0,\alpha_2,\alpha_3...\alpha_n\}$. $P(k)=\alpha_i$ for some $i$ $\epsilon$ $\{0,\alpha_2,\alpha_3...\alpha_n\}$ implies that $a_{n}k(k-\alpha_{2})\cdots (k-\alpha_i) \cdots (k-\alpha_{n})=\alpha_i$ with $a_n,\alpha_i,k \ne 0$. Hence we have $k|\alpha_i$ implies that $k|k-\alpha_i$. Whereas we have $k-\alpha_i|\alpha_i$ implies that $k-\alpha_i|k$. Hence $|k-\alpha_i|=|k|$ meaning that $2k=\alpha_i$ as $\alpha_i$ is non-zero . Plugging it back into the equation , we have $a_{n}(k-\alpha_{2})\cdots (k-\alpha_i) \cdots (k-\alpha_{n})=2$ . As all the $\alpha_i$'s are distinct this means that all the $k-\alpha_i$'s are distinct , but they are at least four in number and must belong to the set $\{-2,-1,1,2\}$ with both $2$ and $-2$ not occuring together , which is impossible. Hence as a conclusion , $k$ can take only the values of roots of $P(x)$