Problem 6

Let

n

be a positive integer. Consider

S=\{(x,y,z) : x,y,z\in\{0,1,\ldots,n\},\ x+y+z>0\}

as a set of

(n+1)^3-1

points in three-dimensional space. Determine the smallest possible number of planes, the union of which contains

S

but does not include

(0,0,0)

Step 3 of 5: Peel off the slice x = n_1 by polynomial division

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In plain words

Dividing $P$ by $x - n_1$ leaves a remainder $R(y,z)$ that vanishes on the entire top layer $x = n_1$ , so $P = (x - n_1)Q$ on all remaining grid points where $x < n_1$ .

P(x,y,z) = (x - n_1) Q(x,y,z) + R(y,z), \qquad R(y,z) = 0 \text{ on } S_2 \times S_3

Detailed analysis

When $n_1+n_2+n_3 > 0$ , assume WLOG that $n_1 > 0$ . Dividing $P$ by $(x - n_1)$ in $\mathbb{R}[y,z][x]$ gives $P(x,y,z) = (x - n_1)Q(x,y,z) + R(y,z)$ . For every $(y,z) \in S_2 \times S_3$ , the point $(n_1,y,z)$ has $n_1+y+z \ge n_1 > 0$ , so $P(n_1,y,z) = 0$ , giving $R(y,z) = 0$ on all of $S_2 \times S_3$ . Hence for every point $(x,y,z)$ in the smaller grid $\{0,\ldots,n_1-1\}\times S_2 \times S_3$ , we have $P(x,y,z) = (x - n_1)Q(x,y,z)$ with $x - n_1 \ne 0$ .