Problem for week eight

Problem Eight (Jan. 9th, 2011): Prove theorem 1 using theorem 2.

Theorem 1. Let $n > 1$ . Then there exists an integer $S(n)$ such that for all primes $p > S(n)$ the congruence $x^n + y^n \equiv z^n \pmod p$ has a solution in the integers, such that $p$ does not divide $xyz$ .

Theorem 2. (Schur’s theorem)Let $r> 1$ . Then there is a natural number $S(r)$ , such as if $N > S(r)$ and if the numbers $\{1, 2, . . . ,N\}$ are colored with $r$ colors, then there are three of them $x, y, z$ of the same color satisfying the equation: $x + y = z$ .

Solutions to last problem: Suppose a abstract simplicial $K$ have $m+1$ vertices $a^0,a^1,\cdots,a^m$ . We say $a^0,\cdots,a^n$ are geometric independent when the vectors $a^1-a^0,a^2-a^0,\cdots,a^n-a^0$ are linearly independent. We first show the fact that we can find in $R^{2n+1}$ $m+1$ points such that any $q\leq 2n+2$ of them are geometric independent.

We induct on the number $k$ and suppose that we have already found points $A^0,A^1,\cdots, A^k$ such that any $2n+2$ (or less) vertices are geometric independent. Since any $2n+1$ (or less) vertices are geometric independent, they have to lie in a space of at most $2n$ dimension. These $2n$ dimensional subspaces cannot make full of the whole $R^{2n+1}$ . We then choose a point $A^{k+1}$ which doesn’t lie in any of these $2n$ dimensional subspaces.Thus we complete our induction by choosing an additional $A^{k+1}$ making any $2n+2$ (or less) vertices choosing from $A^0,A^1,\cdots, A^k$ are geometric independentm, completing the induction step.

Then we can choose points $A^{0},A^{1},\cdots,A^{k+1}$ to corespond the abstract vertices $a^0,a^1,\cdots,a^{k+1}$ . We now verify they are the geomeric realization of the abstract simplicial $K$ . We only need to check the the intersection of any two simplices $\sigma_1, \sigma_2\in K$ is a face of both $\sigma_1$ and $\sigma_2$ . This is easily seen because there are totally no more than $2n+2$ points involved and thus can form a simplicial $\sigma'$ . This means $\sigma_1\cap \sigma_2$ is either empty set or a common face of $\sigma_1$ and $\sigma_2$ .

Like this:

Be the first to like this post.

Explore posts in the same categories: One Problem A Week

This entry was posted on 2011/01/09 at 9:22 pm and is filed under One Problem A Week. You can subscribe via RSS 2.0 feed to this post's comments. You can comment below, or link to this permanent URL from your own site.

One Comment on “Problem for week eight”

Problem For Week Nine « Xiaochuan Liu's Weblog Says:

2011/01/20 at 1:19 am
[...] Xiaochuan Liu's Weblog mathematics and other aspects of life « Problem for week eight [...]

Reply

Xiaochuan Liu's Weblog