Problem 5

The equation

(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots(x-2016)

is written on the 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

linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?

Step 1 of 6: Necessity: at least

2016

factors must go

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In plain words

A factor $(x-i)$ surviving unchanged on both sides makes $x=i$ an obvious real solution.

k\ge2016

Detailed analysis

There are $2016$ common linear factors $(x-1),\dots,(x-2016)$ on both sides. If any one of them, say $(x-i)$ , is left untouched on both sides, then $x=i$ solves the resulting equation. So at least one copy of every $(x-i)$ must be erased, meaning at least $2016$ factors must be erased in total.