Category Archives: algebra

The 15-puzzle of Sam Loyd

The 15-puzzleis a sliding puzzle that consists of a frame of numbered square tiles in random order with one tile missing. Starting from the initial position of pic.1 we apply a finite number of planar reordering in such way that the bottom right square is not filled.

Diagram2

Every such reordering is a permutation of {1,2,…,15} , which is an element of S_{15}. So the set G of all the permutations is a group of order 15.

The reordering is done by the means of the altering permutations (i,j) like pic.2

15_puzzle

Which are all the possible re orderings ? The answer is that G is equal to the group of even permutations. So G\leq A_{16}\cap S_{15} = A_{15}

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New problem after ages. Elementary algebra ideas like symmetric polynomials can be connected in a beautiful way like  lightening the leaves of a tree with a torch at night slowly…

problem:

Find all solutions to the system:

a+b+c = 1

a^2+b^2+c^2 = 2

a^4+b^4+c^4 = 3

 

if you are looking for a hint see this

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Unique Factorization Domains

Lets consider the number 2. Looking for a factorization of 2 over $ latex \mathbb{Z}$ we have 2 \cdot 1. Is this factorizartion unique ? If we are looking over another domain like $latex \mathbb{Z}[\sqrt{-6}]$ , then how easy is to find all the posible factorizations of 2 ? If we find one is it indeed unique?

2=(a+b\sqrt{-6})(c+d\sqrt{-6}) ,a,b,c,d \in \mathbb{Z}  considering the norm of 2 : N(2)=N((a+b\sqrt{-6}))(c+d\sqrt{-6}). But N(2)=(2+0\sqrt{-6})(2-0\sqrt{-6})=4.Taking norms for both sides we have :4=(a^2+6b^2)(c^2+6d^2) in $latex \mathbb{Z}$. On the RHS the factors can be a) either both of them 2 ,b) one of them 4 and the other 1.

a) It’s easy to so that the equation 2=x^2+6y^2 has no integer solutions (just consider (x,y)mod3).

b) 1=x^2+6y^2 \rightarrow 1=1+06 , which gives the factorization of norms: 4=4\cdot1 which is trivial factorization on $latex \mathbb{Z\sqrt{-6}}$ (we are looking for non-trivial ones).

One can read a little more diophantine_eq and how the integer equation 2x^3=y^2+1 can be solved over a unique factorization domain (UFD) such as  \mathbb{Z[i]}.

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