18th Turkish Mathematical Olympiad

We will show that there exists $Q(x)=c{x}^2+dx$ for $P(x)=a{x}^2+bx$ if and only if $2^{8}1005^9\mid a$ and $(2010,b)=1$. Then it follows that the answer is $$2\cdot 2010^9\cdot 2010^{18}\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{67}\right)=2^5\cdot 3\cdot 11\cdot 2010^{26}.$$ Assume that $Q(P(n))\equiv n(\mathrm{mod}2010^{18})$ for all $n$. Then $n\mapsto P(n)$ is one-to-one modulo $2010^{18}$ and using the Chinese Remainder Theorem we deduce that $n\mapsto P(n)$ is one-to-one modulo $p^{18}$ for each $p\in \{2,3,5,67\}$. Let $p\in \{2,3,5,67\}$. If $p\mid b$, then $P(p^{17})\equiv P(0)(\mathrm{mod}{p}^{18})$ gives a contradiction. Hence $p\nmid b$. If $p\nmid a$, then $P(-a^{-1}b)\equiv P(0)(\mathrm{mod}{p}^{18})$ gives a contradiction. Hence $p\nmid a$. Therefore $2010\nmid a$ and $(2010,b)=1$. In particular, $(b(a^2-b^2){)}^{-1}$

(mod $2010^{18}$) exists. Since $$\begin{array}{l}Q(P(1))\equiv 1⟹c(a+b{)}^2+d(a+b)\equiv 1\\ Q(P(-1))\equiv -1⟹c(a-b{)}^2+d(a-b)\equiv -1\\ ⟹\{\begin{array}{l}2b(a^2-b^2)c\equiv 2a(\mathrm{mod}2010^{18})\\ 2b(a^2-b^2)d\equiv -2(a^2+b^2)(\mathrm{mod}2010^{18})\end{array}\end{array}$$ we have $$c\equiv (b(a^2-b^2){)}^{-1}a+\epsilon \frac{2010^{18}}{2}(\mathrm{mod}2010^{18})$$ $$d\equiv -(b(a^2-b^2){)}^{-1}(a^2+b^2)+\epsilon \frac{2010^{18}}{2}(\mathrm{mod}2010^{18})$$ where $\epsilon$ is 0 or 1. Therefore $$\begin{gathered} Q(P(x))-x\equiv -(b(a^2-b^2){)}^{-1}{a}^{2}x(x-1)(x+1)(ax+2b)+\epsilon \frac{2010^{18}}{2}x(x-1)(\mathrm{mod}2010^{18}) \\ \equiv 0 \end{gathered}$$ Now if we let $x=2$, we obtain $2010^{18}\mid 2^2\cdot 3\cdot a^2$, and hence $2^{8}1005^9\mid a$. Conversely, if $2^{8}1005^9\mid a$ and $(2010\cdot b)=1$, then we can define $c$ and $d$ as above. Since $2\mid n(n-1)$ and $2\mid an+2b$ for all $n$, $Q(P(n))\equiv n(\mathrm{mod}2010^{18})$ follows.