Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $$\mathbb{R}^*=\mathbb{R}-\{0\}$$ and

$$N=\{0,...,n\}$$ and $$\mathcal{M}=\{ A\subseteq \mathbb{R}^3\times\mathbb{R}^* \mid (\forall\mathbb{x}\in N^3:\mathbb{x}\ne 0)(\exists(\mathbb{a},d)\in A)(\mathbb{a}.\mathbb{x}=d)\}$$

(here $N^3=N\times N\times N$ and $\mathbb{a}.\mathbb{x}$ is inner product)

what is $$m:=\min\{|A|:A\in \mathcal{M}\}$$?

Do you know any theorem that simplifies finding $m$?

share|cite|improve this question
You might be able to solve it for a few small values of $n$ and then look up the results in the Online Encyclopedia of Integer Sequences. Can you solve the corresponding problem one dimension down? – Gerry Myerson Feb 8 '13 at 4:37
However a 2d dimension solution is much simpler and may be generalizable to 3d. – user59671 Feb 8 '13 at 4:40
Yes, that was my point. – Gerry Myerson Feb 8 '13 at 5:51

At least you can say that $m\le 3n$ by taking a single value $a=(1,1,1)$ and taking as $d$ all possible sums of triplets in $N^3-\{(0,0,0)\}$.

On the other hand you know that $m \ge n-1$ by taking $x=(k,0,0)$ and noticing that two of these points cannot stay on the same plane.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.