## Saturday, January 31, 2009

### Another quick "2009" stuff

Let

and let

be a function satisfying

Then f has at least one fixed point, that is, there exists an element x in X such that f(x)=x.

Indeed, it is not hard to see that f is a bijection and induces a permutation of X in which all cycles are either of length 3 or 1 (those of length one correspond to the fixed points of f ). Since 2009 is not divisible by 3, there must be a cycle of length 1.

### A complicated way of characterizing the constant functions

Show that the constant functions are the only continuous functions

satisfying

for all real numbers x,y.

## Wednesday, January 28, 2009

### Squares and walks

The (quasi-) random walk represented in the picture corresponds to the cubic polynomial
f(X) = X^3-X-1 mod 2003.
Every step is a unit step. The n-th step is made to the right if f(n) is a perfect square modulo 2003, and to the left, if f(n) is a non-square.

## Sunday, January 25, 2009

### Factorials and squares

If N > 1, then N! is not a perfect square.

Indeed, let N > 1 and let p be the largest prime that is less than or equal to N. Then 2p cannot be less than or equal to N - otherwise, a prime q greater than p and less than 2p (such a q exists by Bertrand's theorem), would be a prime greater than p that is less than or equal to N. Thus p defined as above appears with an exponent of 1 in the prime decomposition of N! and therefore N! cannot be a perfect square if N > 1.

## Tuesday, January 20, 2009

### Un sir neperiodic

Sa consideram sirul binar a_n, n=0,1,2,... definit dupa cum urmeaza:

a_n=0, daca scrierea lui 2^n in baza 10 are un numar par de cifre, si a_n=1, in caz contrar.
Sa se arate ca {a_n} nu este "eventual periodic" (nu exista un intreg pozitiv k astfel incat egalitate a_{n+k}=a_n are loc pentru orice n suficient de mare). Aceasta problema (plus variatii pe aceeasi tema) a constituit o tema de "undergraduate research" pe care am propus-o unui student, acum 6 ani. Este o modalitate frumoasa de a introduce teorema de distributie uniforma in [0,1] a partilor fractionare ale multiplilor unui numar irational.

## Saturday, January 17, 2009

### On Putnam A1/2008

This is problem A1 from the 2008 Putnam Exam.
The proof is quite straightforward:

## Thursday, January 15, 2009

### "modulo n"...

`Pentru fiecare n > 1 "inelul intregilor modulo n" Z/nZ consta din 0,1,...,n-1, adica toate resturile ce pot fi obtinute la impartirea cu n. Aceste elemente nu sunt "numere intregi obisnuite", similaritatea in notatie putand fi inselatoare. Astfel, "0"-ul lui Z/nZ reprezinta toti intregii care dau restul 0 la impartirea cu n, "1"-ul lui Z/nZ reprezinta toti intregii care dau restul 1 la impartirea cu n, si asa mai departe. Z/nZ este un univers aparte, in care se definesc o "adunare modulo n" si o "inmultire modulo n". Ideea este simpla: sa ne imaginam ca adunam/inmultim "normal" si, in cele din urma, pentru a identifica rezultatul adunarii/inmultirii in Z/nZ vom scrie numai restul adunarii/inmultirii "normale" la impartirea cu n. De exemplu: in Z/3Z = {0,1,2} avem 1+2=0, 2+2=1, 2^9=2, iar (X+Y)^3=X^3+Y^3 este o identitate valida! Putem considera "19" ca element al lui Z/3Z? Sigur! Iarasi, pentru a-l identifica pe 19, vom lua restul lui 19 la impartirea cu 3 - astfel, 19=1 in Z/3Z. Asemenea identificari sunt uneori foarte utile. De exemplu care dintre elementele 0,1,...,8 ale lui Z/9Z il reprezinta pe 11^99? In sfarsit, ca aplicatie la ecuatii in numere intregi, sa se arate ca ecuatia 3*X^2-Y^2=1 nu are solutii in numere intregi. `

## Wednesday, January 14, 2009

### 2009

Since we are at the beginning of 2009, I will try to break the ice with a list of problems involving the number 2009.
1. Factor 2009.
2. Write 2009 as a sum of two squares: 2009=x^2+y^2 with integers x and y.
3. If x and y are integers and if x^2+y^2 is a multiple of 2009, is it true that both x and y must be multiples of 7?
4. How many zeros are there at the end of 2009! ?