TITLE: IMO 2016 Problem 5
QUESTION [5 upvotes]: The equation
$$
(x-1)(x-2)(x-3)\dots(x-2016)=(x-1)(x-2)(x-3)\dots(x-2016)
$$
is written on a board, with $2016$ linear factors on each side. What is the least possible value of $k$ for which it is possible to erase exactly $k$ of these $4032$ factors so that at least one factor remains on each side and the resulting equation has no real solutions?

REPLY [4 votes]: An interpretation of the solution from here. Sorry my Chinese.

$k\ge 2016$ (see @McFry comment above).
For $t=1,2,\ldots,504$, remove the following (exactly $2016$ many) factors
$$
\begin{align*}
& (x-(4t-2)),\quad (x-(4t-1)) &\qquad \text{ from the LHS},\\
& (x-(4t-3)),\quad (x-4t) &\qquad \text{ from the RHS}. 
\end{align*}
$$
We are to prove that what's left has no real solutions. It would mean that the smallest $k$ is $2016$. We are left with the equation
$$
\begin{align*}
& (x-1)(x-4)(x-5)(x-8)\ldots(x-2013)(x-2016)=\\
= & (x-2)(x-3)(x-6)(x-7)\ldots (x-2014)(x-2015). 
\end{align*}\tag{1}
$$
Let's prove that for all $x\in\mathbb{R}$ it holds
$$
\begin{align*}
(x-1)(x-4)&<(x-2)(x-3),\\
(x-5)(x-8)&<(x-6)(x-7)\qquad \text{etc.}\tag{2}
\end{align*}
$$
In other words, we prove that for $t=1,\ldots,504$
$$
(x-(4t-3))(x-4t)<(x-(4t-2))(x-(4t-1)).\tag{3}
$$
Denote $y=x-4t$. Then "RHS minus LHS" in (3) is
$$
(y+2)(y+1)-(y+3)y=y^2+3y+2-y^2-3y=2>0.
$$
Hence (2) is proven for all $x\in\mathbb{R}$.
Obviously, $x\in \{1,2,\ldots,2016\}$ is not a solution
If $x<1$, $x>2016$ or $\exists m\colon 1\le m\le 503$ such that $4m<x<4m+1$ then all sides of the inequalities in (2) are positive, therefore, (1) has no real roots.
If $\exists n\colon 1\le n\le 504$ such that $4n-3<x<4n-2$ or $4n-1<x<4n$ then one inequality among first $n$ has negative LHS and positive RHS and the rest 503 inequalities have both sides positive. Then equality in (1) is again impossible.
What's left is to prove that $4n-2<x<4n-1$ for some $n$, $1\le n\le 504$, cannot be a solution to (1). In this case the following $503$ inequalities holds true (similar prove as above) with both sides being positive
$$
\begin{align*}
(x-4)(x-5)&>(x-3)(x-6),\\
(x-8)(x-9)&>(x-7)(x-10),\\
&\vdots\\
(x-2012)(x-2013)&>(x-2011)(x-2014).
\end{align*}
$$
Moreover, for $1\le n\le 504$ we have $2\le 4n-2<x<4n-1\le 2015$, hence
$$
\begin{align*}
(x-1)&>(x-2)&>0,\\
-(x-2016)&>-(x-2015)&>0,
\end{align*}
$$
which implies
$$
\begin{align*}
& -(x-1)(x-4)(x-5)(x-8)\ldots(x-2013)(x-2016)>\\
> & -(x-2)(x-3)(x-6)(x-7)\ldots (x-2014)(x-2015),
\end{align*}
$$
so again (1) is impossible.