(IMO 2016 P5) The equation

$$
(x-1) \times \ldots \times(x-2016)=(x-1) \times \ldots \times(x-2016)
$$

is written on the board. We want to erase some of the linear factors so that the remaining equation has no real solutions. Determine the smallest number of linear factors to erase to achieve this goal.
Answer: 2016

First, if we remove fewer than 2016 factors, by the pigeonhole principle, there exists a factor $x-k$ that is not removed on either side, so the equation has the real solution $x=k$.

Now, let's show that we can remove 2016 factors so that the equation has no solution.

For this, we remove from the left the factors of the form $x-4k+3, x-4k$ (for $1 \leqslant k \leqslant 504$) and from the right the factors of the form $x-4k+2$ and $x-4k+1$. Let's show that the equation

$$
(x-1)(x-4) \ldots(x-2013)(x-2016)=(x-2)(x-3) \ldots(x-2014)(x-2015)
$$

has no real solution.

- : The result is true if $x \in\{1, \ldots, 2016\}$ since one side of the equation is then zero and not the other.

-: If $x > 2016$, then $(x-(4k+2))(x-(4k+3)) > (x-(4k+1))(x-(4k+4))$ for all $k \in\{0, \ldots, 503\}$, so the right side is greater than the left side.

-: If $4k < x < 4k+1$, then $(x-2)(x-2015) > (x-1)(x-2016)$ and $(x-4j)(x-4j-1) > (x-4j+1)(x-4j-2)$ for all $j$, so the left side is strictly greater than the right side.