Proving midpoint is an integer

Question

the questions below seem very simple and evident but I'm having a hard time answering the second part. I would really appreciate your help!

1) Let $n_1, n_2$, and $n_3$ be integers, show that there is $k$ and $d$ such that $1 \leq k < d \leq 3$ and the average of both is an integer.

This is my answer: each of $n_k$ and $n_d$ can be either $n_1, n_2$, or $n_3$ so they're both integers. There are 4 cases:

• They're all even: the sum of two of them is even so it can be divided by 2.

• They're all odd: the sum of two is even so it can be divided by 2.

• One is even and two are is odd: an even integer can be obtained by summing up the two odds, so it exists.

• One is odd and two are even: an even integer can be obtained by summing up the two evens, so the integer exists.

Now comes the trickier part: A point in space is integer if all its coordinates have integer values. Let $a_1$ ... $a_9$ be distinct integer points in space. We need to show that there exists $k$ and $d$ such that $1 \leq k < d \leq 9$ and the midpoint between $a_k$ and $a_d$ is an integer point.

I feel like there are so many possibilities for this one, is there a more "direct" way of proving it rather than listing down all cases?

Thank you very much!

Answer 1

A simpler way of proving the 1D case is to use the pigeonhole principle. Two of $n_1,n_2,n_3$ must have the same parity. The average of those two is an integer. For your 3D case, there are $2^3=8$ patterns of even and odd coordinates. As you have nine points, two of them have the same pattern. The midpoint of those two has all integer coordinates.

Answer 2

Think of odd and even as minus and plus. Then in $R^3$ there will be 8 octants. Given $9$ points, if $8$ of them are in different octants, then the $9$-th must be in one of them and it must pair up with the other one resulting in their mid-point integer. In general, in $R^n$, given $2^n+1$ points, there can be found two points, whose mid-point is also integer.