An algebra problem by João Vitor

In this solution, we will apply modular arithmetics properties for polynomials. You can recall them here

• Solution

Using $p(x) \equiv 0\quad mod\quad p(x)$ we have:

${ x }^{ 2 }-x+1\equiv 0\quad mod\quad p(x)\Longleftrightarrow (x-1)\equiv { x }^{ 2 }\quad mod\quad p(x)\Longrightarrow { (x-1) }^{ 2016 }{ \equiv x }^{ 4032 }\quad mod\quad p(x)$

But

${ x }^{ 3 }+1=0\Longleftrightarrow (x+1)({ x }^{ 2 }-x+1)=0\quad \therefore \quad { x }^{ 3 }+1\equiv 0\quad mod\quad p(x)\Longleftrightarrow { x }^{ 3 }\equiv -1\quad mod\quad p(x)\Longrightarrow { ({ x }^{ 3 } })^{ 1344 }\equiv 1\ mod\quad p(x)$

Hence

${ (x-1) }^{ 2016 }\equiv 1\quad mod\quad p(x)\quad \Longleftrightarrow \quad { (x-1) }^{ 2017 }\equiv (x-1)\quad mod\quad p(x)\quad \therefore \quad { P }^{ 1 }(x)=x-1$

Because $deg({ P }^{ 1 }(x))<deg(d(x))$

Doing the first cases, we can conjecture that $P^{ k }\left( x \right) =x-k$

Which is clearly true for $k=1$

Suppose it is true for a natural number $k=n$

Then $P^{ n }\left( x \right) =x-n$ which yields

${ P }^{ n+1 }(x)={ P }^{ 1 }({ P }^{ n }(x))\quad \Longleftrightarrow \quad { P }^{ n+1 }(x)=(x-n)-1\quad \Longleftrightarrow \quad { P }^{ n+1 }(x)=x-(n+1)$

$\therefore \quad P^{ k }\left( x \right) =x-k\quad \blacksquare$

Therefore, ${ P }^{ 2017 }(x)=x-2017\quad \therefore \quad \boxed { { P }^{ 2017 }(2016)=-1 }$

Solve $d(x)=0$ first (using whatever means) to obtain the two roots $u=\text{cis}\frac{\pi}{3}$ and $v=\text{cis}\frac{-\pi}{3}$ . Note that $p(u)=(u-1)^{2017}=u-1=-v$ and $p(u)=(u-1)^{2017}=u-1=-v$ . If we let $P^1(x)=ax+b$ for some constants $a$ and $b$ , then we can write $p(x)=Q(x)d(x)+ax+b$ . Substituting $x=u$ and $x=v$ into this result gives $-v=au+b$ and $-u=av+b$ . Subtracting the last two equations give $u-v=a(u-v)\Rightarrow a=1\Rightarrow b=-u-v=-1$ . Hence $P^1(1)=x-1$ . Applying this polynomial $2017$ times to the number $x=2016$ gives $P^{2017}(2016)=\boxed{-1}$ .

p(x)=(x-1)²⁰¹⁷, d(x)=x²-x+1. Let x-1=t Thus, p(x)=t²⁰¹⁷, d(x)=t²+t+1. Therefore p(x)=t²⁰¹⁷-t+t=t(t²⁰¹⁶-1)+t=t(t²⁰¹³+t²⁰¹⁰+....1)(t-1)d(x)+t. Therefore P1(x)=t=x-1. Pn(x)=x-n