March 12 Discussion

First of all, I am not sure why Gaussian Integers exist. I mean I am having a hard time to understand what the 'point' is for Gaussian Integers. I found a lot of websites and briefly looked some of those (including Wikipedia); understand what Gaussian Integer is but still do not know why we need it.
Secondly, I am not so sure why we talked about Gaussian Integers during last class on last Thursday (and going to talk about it in today's class). I just feel like this topic is 'out of no where'; I know there are connections between this topic and the previous topics, but I just felt a bit weird after the last class.
I also did some research and I found this website, http://www.jimloy.com/algebra/gprimes.htm , there is a list of the Gaussian Primes that are less than 100 (with some of the 'steps', kind like what we did in class).

Lin

Here are instructions on how to build a 17-gon (heptadecagon) and here's an animation of the construction. I remember seeing someone who memorized the construction and could duplicate it on a blackboard, but I couldn't find it. When I do, I'll post it.

===Matt===

It was interesting to see how to construct a regular 15-gon with a circle. I looked it up in the internet to find how to construct beyond 15-gon and the easiest way I found was given below:
to first construct a regular pentagon; draw the circle in which this pentagon is inscribed, and inscribe five equilateral triangles inside that circle, each sharing one vertex with the penta- gon. The fifteen vertices of these triangles will form the vertices of a regular 15-gon. To construct a regular 2n-gon from a regular n-gon, simply bisect each of the n central angles of the n-gon. Thus, starting from the triangle, square, and pentagon it is possible to construct regular polygons with 6, 12, 24, etc. sides, with 4, 8, 16 etc. sides, with 5, 10, 20 etc. sides, and even with 15, 30, 60 etc. sides.

-Rachel

A Fermat prime is a prime of the form $p={ 2 }^{ { 2 }^{ n } }+1$. The interesting thing is that there are only 5 of them! There are:

(1)
$${ 2 }^{ { 2 }^{ 0 } }+1=2+1=3$$
(2)
$${ 2 }^{ { 2 }^{ 1 } }+1=4+1=5$$
(3)
$${ 2 }^{ { 2 }^{ 2 } }+1=16+1=17$$
(4)
$${ 2 }^{ { 2 }^{ 3 } }+1=256+1=257$$
(5)
$${ 2 }^{ { 2 }^{ 4 } }+1=65536+1=65537$$

The next one in the progression would be ${ 2 }^{ { 2 }^{ 5 } }+1=2^{ 32 }+1$, which it turns out is not prime. It has a factor of 641. In fact, there are no more Fermat primes that have been discovered after that. To see the factorization of the first (and perhaps only) Fermat primes, go to http://en.wikipedia.org/wiki/Fermat_number.

Interestingly, each of the Fermat primes are the product of all the previous Fermat primes plus 2. In other words:

(6)
\begin{align} { F }_{ 1 }+2={ F }_{ 2 }\\ \left( { F }_{ 1 }\cdot { F }_{ 2 } \right) +2={ F }_{ 3 }\\ \left( { F }_{ 1 }\cdot { F }_{ 2 }{ \cdot F }_{ 3 } \right) +2={ F }_{ 4 }\\ \left( { F }_{ 1 }\cdot { F }_{ 2 }{ \cdot F }_{ 3 }\cdot { F }_{ 4 } \right) +2={ F }_{ 5 } \end{align}

This pattern would give the already disproven $2^{ 32 }+1$ as its next one, but the pattern is sound up through the known Fermat primes. (Powell)

From internet search and my MATH 361 class, we also have:

The field consisting of all constructable points using only compass
and straight-edge is the quadratic closure of $\mathbb{Q}$.

Use this we can conclude
It is impossible to (1) square the circle (2) double the cube (3) trisect an angle

-Mingzhi Liu