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 3 of 6: Case 1 and Case 2: integer points, and near-block points

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In plain words

At the $2016$ integers exactly one side vanishes; just inside a block, exactly one factor on the left goes negative while every factor on the right stays positive.

x=1,\dots,2016:\ \text{one side }0,\ \text{other side }\ne0 \qquad 4k{+}1{<}x{<}4k{+}2 \text{ or } 4k{+}3{<}x{<}4k{+}4:\ \text{LHS}<0<\text{RHS}

Detailed analysis

Case 1: if $x\in\{1,\dots,2016\}$ , then $x$ makes one of the surviving factors on one side vanish (since every integer in range appears exactly once among the survivors, on one side or the other), while no factor on the other side vanishes; so the two sides differ. Case 2: if $4k+1<x<4k+2$ or $4k+3<x<4k+4$ for some $k\in\{0,\dots,503\}$ , then for the block $j=k$ the product $(x-4k-1)(x-4k-4)$ is negative (one factor positive, one negative), while for every other block $j\ne k$ the product $(x-4j-1)(x-4j-4)$ is positive; so the left-hand side is negative. Meanwhile every factor $(x-4j-2)(x-4j-3)$ on the right-hand side is a product of two same-signed numbers (checked directly for both sub-ranges), hence positive; so the right-hand side is positive. A negative number cannot equal a positive one.