# Wilsons Theorem

Wilsons Theorem says, that p is prime if and only if (p-1)!+1 is a multiple of p.

To prove this, we must show, that if (p-1)!+1 is a multiple of p, that then p is prime and that if p is prime, that then (p-1)!+1 is a multiple of p.

First we will show, that p must be prime so that  (p-1)!+1 is a multiple of p. To do this we first assume, that p is not prime and that (p-1)!+1 is a multible of p. Then p can be written as x*y with x and y < p. It follows, that (p-1)! is a multible of x and y. Now we can say, that (p-1)!+1 is not a multible of x or y. It follows, that (p-1)!+1 is not a multiple of p. We also need to consider the case that x=y. It does not work for x=y=2. For all cases wit x bigger than 2, 2x will also be a factor in (p-1)!. It follows, that (p-1)!+1 is not a multiple of p. This is a contradiction which tells us, that p has to be prime.

Now we have to show the other part of the theorem. It says, that (p-1)!+1 is a multible of p for every prime p . Considering that p is prime, for every remainder modulo p there exists a unique inverse remainder, so that the product of the two remainders is 1 modulo p. The only two remainders that have them selves as a inverse are 1 and -1. The factors of (p-1)! are all possible remainders of p. It follows, that for every factor other than 1 and p-1 there exists a second factor so that when multiplied the product is congruent to 1 modulo p. The two numbers that are left multiplied together equal -1 modulo p. It follows, that (p-1)! is congruent to -1 modulo p and therefore (p-1)!+1 a multiple of p.

This proves Wilsons theorem.

# Paul Erdős’ favorite Problem

It is said, that the following problem was Paul Erdős’ favorite problem.
The Problem:
Let n be any positiv whole number. Let n+1 numbers randomly be taken from the numbers 1 to 2n. Show that there will always exist two numbers x and y which are in the n+1 chosen numbers, so that x*z=y for some positive whole number z.

Solution:
It is possible to write every number in the form q*2^h for an uneven number q. If two numbers have the same q, one can always be devided by the other. It follows, that no two of the n+1 chosen numbers are allowed to have the same q. We know that there only exist n uneven numbers in the numbers from 1 to 2n. Since we need to choose n+1 numbers from this set, two numbers must have the same q when written in the form q*2^h.
It follows that there will always be 2 numbers x and y with x*z=y for some positive whole z.

# Napoleon’s Theorem

First I want to thank an anonymous Person who Formulated this Post for me.

Napoleon’s Theorem:

Let ABC be a triangle. Let equilateral triangles be constructed outward at each side of the triangle ABC. Then the midpoints of these equilateral triangles are the vertices of another equilateral triangle.

Proof:

We use vector calculus. Let A, B, C denote the vectors pointing to the vertices.  By translating the tirangle so that the barycenter of it becomes the origin, we may assume that

(1)       A+B+C=0

Let R be a the linear map which turns each vector by 60 degrees to the right. Note that R^3 = -1and for every vector X we have the identity

(2)     R^2 X  = RX – X

The tip of the equilateral triangle above the side from A to B is

B+R(A-B)

The center of this equilateral triangle is then

(A+B+(B+R(A-B)))/3

Using (1), we may write this as

(3)    ((B-C)+ R(A-B))/3

By symmetry the centers of the other equilateral triangles are

(4)   ((C-A)+R(B-C))/3

(5)   ((A-B)+R(C-A))/3

We need to prove that these three points are permuted when we rotate by 120 degrees aound the origin, that is R^2. For simplicity we may multiply the three vectors by 3. Applying R^2 to 3 times expression (3) gives

R^2(B-C)+R^3(A-B)

Applying (2) and the fact R^3=-1 turns this into

R(B-C)-(B-C)-(A-B)=(C-A)+R(B-C)

This evidently is three times (4). By symmetry, the three points (3),(4),(5) are permuted when rotated by R^2, which proves that they form an equilateral triangle.

# Huygens Problem

A few months ago, I stumbled a pond a new problem solving idea. It is called the telescope method. I did know about it but I just used it subconsciously. I want to present a rather easy problem that can be solved with this method. In 1672 Huygens asked Leibnitz to solve this problem, which he did using the same method.

The problem is this: What is the sum off all reciprocals off the triangle numbers.

First we find a formula to express a triangle number. it is well known that this formula is:        (n*(n+1))/2 . Let x be the solution to the problem. Let dn denote the nth triangle number.

Now there is a nice trick that you can use:

This has a great benefit for infinite sums. We add this into the equation above and get:

Lets se what happens to the first few n. 1/1-1/2+1/2-1/3+1/3-14+1/4-1/5. As you can see, everything except the 1/1 cancels itself out. 1/1+(-1/2+1/2)+(-1/3+1/3)+(-14+1/4)+(-1/5…  =1/1+0+0+0+0….

We have 1=1/2x. Simple algebra gives us x=2, which is the solution to the problem.

# Nesbitt’s Inequality

The Nesbitt’s inequality goes like this:

I want to prove this in this post:

We define x to be a+b, y to be b+c and z to be a+c.

Now we have to prove, that x/y+y/x is bigger or equal to 2. We multiply the equation with x times y times 2. We get: x^2+y^2 is bigger or equal to 2xy. Now we subtract 2xy on both sides and we get (x-y)^2 is bigger or equal to 0. This is true. Therefor we proved that x/y+y/x is bigger or equal to 2.

Now we have:

This completes the proof.

# The Farmer Problem

The problem goes like this:

There are 2 points on a plane, which are not the same. One represents a farmer and the other point represents a pig. Then there is a straight line on the plane, which does not go through either of the points. The line represents a river. The farmer wants to give the pig water. To do his, he has to first go to the river and then to the pig. How can you find the fastest way for him to do so.

The answer is astonishingly easy. You have to reflect the (pig) point at the river as in the picture (a=b) :

It is known, that the shortest path between two points on a plane is a straight line. Now we connect the farmer with the reflected point is a straight line. Then we call the point where the line cuts the river c. Now we just have to connect the farmer with c and the pig with c. This is the shortest path, since the left side of the river was just reflected to the right side and we had the shortest path before we reflected back..

# Proof of irrationality of the Square root of 2

We will do a proof by contradiction:

Lets assume, that the square root of 2 is rational. Then you can write the square root of 2 as a fraction a over b. Lets assume that a and b are hole numbers and that you can not reduce a over b. Then you can do this:

It follows, that a is even.

Now we prove that b is even. For a we can write 2k.

It follows that b is even. That means, that a/2 and b/2 are hole numbers. It follows, that you can reduce a over b with 2. This is a contradiction, since we assumed that a over b can not be reduced.