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 5 of 6: Case 4: strictly between the two middle quarters of a block

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In plain words

Isolating the two end factors and rewriting the rest as terms each strictly greater than $1$ shows the whole right-hand expression must exceed $1$ .

\frac{x-1}{x-2}\cdot\frac{x-2016}{x-2015}\cdot\prod_{j=1}^{503}\left(1+\frac{2}{(x-4j+1)(x-4j-2)}\right)=1

Detailed analysis

Case 4: if $4k+2<x<4k+3$ for some $k\in\{0,\dots,503\}$ , rewrite the target equation (again dividing by the right-hand side, but this time isolating the two boundary blocks $j=0$ and $j=503$ before pairing the rest) as $\dfrac{x-1}{x-2}\cdot\dfrac{x-2016}{x-2015}\cdot\prod_{j=1}^{503}\dfrac{(x-4j)(x-4j-1)}{(x-4j+1)(x-4j-2)}=1$ , and note $(x-4j)(x-4j-1)=(x-4j+1)(x-4j-2)+2$ , so each factor of the product equals $1+\dfrac{2}{(x-4j+1)(x-4j-2)}$ . In this range, $\dfrac{x-1}{x-2}>1$ and $\dfrac{x-2016}{x-2015}>1$ (both ratios of two negative-or-positive numbers of the same sign with the numerator further from zero), and each term of the product is also $>1$ by a direct sign check. A product of numbers each strictly greater than $1$ is itself strictly greater than $1$ , contradicting that it should equal $1$ .