We call polynomials $A(x) = a_n x^n +. . .+a_1 x+a_0$ and $B(x) = b_m x^m +. . .+b_1 x+b_0$ ($a_n b_m \neq 0$) similar if the following conditions hold: $(i)$ $n = m$; $(ii)$ There is a permutation $\pi$ of the set $\{ 0, 1, . . . , n\} $ such that $b_i = a_{\pi (i)}$ for each $i \in {0, 1, . . . , n}$. Let $P(x)$ and $Q(x)$ be similar polynomials with integer coefficients. Given that $P(16) = 3^{2012}$, find the smallest possible value of $|Q(3^{2012})|$. Proposed by Milos Milosavljevic
Problem
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Tags: number theory
jowanar
28.05.2015 16:06
Solution?
Tom_Hulla
18.03.2019 22:43
Clearly $3^{2012}\equiv 1 \pmod{5}$, hence $Q(3^{2012})\equiv Q(1)=P(1) \equiv P(16) \equiv 1 \pmod{5},$ hence $|Q(3^{2012})| \ge 1.$ Now, let $P(x)=ax^2+bx+c$ and $Q(x)=cx^2+ax+b$. We are asked to solve the system \begin{align*} a*16^2+b*16+c &= 3^{2012} \\ c*(3^{2012})^2+a*3^{2012}+b &= 1 \end{align*} Substituting for $c$ in the second equation yields \begin{align*} (3^{2012})^2(3^{2012}-a*16^2-b*16)+a*3^{2012}+b &=1 \\ \end{align*}I will finish later
shinhue
18.09.2022 13:12
any solution to this problem?
pavel kozlov
19.09.2022 22:19
shinhue wrote: any solution to this problem? Just expand the last expression in Tom_Hullas's solution, you get: $$(256\cdot 3^{4024} - 1)a+(16\cdot 3^{4024}-1)b=3^{6036}-1.$$ As $(256\cdot 3^{4024} - 1,16\cdot 3^{4024}-1)=5|3^{6036}-1$ one can find such integers $a,b$.