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 2 of 6: A candidate erasure of exactly

2016

factors

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In plain words

Splitting the indices $1,\dots,2016$ into $504$ blocks of $4$ and keeping opposite residues on each side reduces the problem to showing one explicit equation has no real root.

\prod_{j=0}^{503}(x-4j-1)(x-4j-4)=\prod_{j=0}^{503}(x-4j-2)(x-4j-3)

Detailed analysis

Erase every factor $(x-k)$ with $k\equiv2,3\pmod4$ from the left-hand side, and every factor $(x-m)$ with $m\equiv0,1\pmod4$ from the right-hand side. This erases exactly $2\cdot504+2\cdot504=2016$ factors. Writing the $2016$ indices as $504$ blocks $\{4j+1,4j+2,4j+3,4j+4\}$ for $j=0,\dots,503$ , the surviving equation is exactly $\prod_{j=0}^{503}(x-4j-1)(x-4j-4)=\prod_{j=0}^{503}(x-4j-2)(x-4j-3)$ . It remains to show this has no real solution.