Saturday, June 8, 2013

On Beal's conjecture and related topics

Recently, Beal's conjecture award has raised to $1 million, and this was one of the big headlines in the math community in the past days. This is a very interesting statement in number theory inspired by, as Andy Beal himself says, on Fermat's last theorem

It says that if the positive integers $A,B,C,x,y,z$ satisfy
with $x,y,z >2$, then $A,B,C$ have a common prime factor.

Notice that this statement work on the basis that there is a solution for the equation, that is, it talks about a property that solutions of the equation must have. This then could be used to easily prove Fermat's last theorem by contradiction using the infinite decent method.

Analyzing these type of equations, it is not hard to find some nice properties arising from them. When I first saw the equation, I was doubtful about the statement, so I went to look for some examples. It was precisely this that led Beal to formulate the conjecture, having looked for numerical solutions with the aid of computer power for values of all the variables up to 1000. So my next inquire was to find a way to create solutions to  these equations.

My approach was to have $x,y,z$ fixed and try to cook up solutions $A,B,C$. The idea is to play around with the prime factors that appear in the game. Basically one has to balance the prime factors in the equation and make them to have the right power.

Grab any pair of positive integers $A,B$ and analyze $A^x+B^y$. The equation requires for this to be a $z$-power, that is, each prime factor appearing in $A^x+B^y$ must have a power multiple of $z$. The idea is to balance those powers by multiplying by powers of $p$.

Suppose that $p$ has a power of $\alpha x$ in $A^x$, $\beta y$ in $B^y$ and $\gamma$ in $A^x+B^y$. Hence, the idea is to multiply by $p^m$ to the expression, where $m$ is such that $\alpha x+m$ is multiple of $x$, $\beta y+m$ is multiple of $y$ and $\gamma+m$ is multiple of $z$. This is a classical problem of chinese remainder, which can be solve by finding solutions to the system of congruences

$$m\equiv 0\text{ mod lcm}(x,y)$$
$$m\equiv -\gamma\text{ mod }z$$.

In order to solve this, we must require that $\text{lcm}(x,y)$ and $z$ to be relative prime. Performing this, we can make $A'=p^{m/x}A$ and $B'=p^{m/y}B$ and then $A'^x+B'^y$ has the right powers for $p$. 

Following this idea for every prime factor $p$ appearing in $A^x+B^y$ leads to a solution of the equation 

Therefore, every pair $(A,B)$ leads to a solution of the previous equation. Here then is necessary to make the assumption that $(xy,z)=1$, which makes Fermat's last theorem to fail for this methodology ($x=y=z$).

In general, something similar can be made for $(xz,y)=1$ or $(yz,x)=1$ just by rewriting the equation as 
$$C^z-A^x=B^y$$ or $$C^z-B^y=A^x$$
the only constrain is that $(C,A)$ (respectively $(C,B)$) should be such that the left hand side must be possitive.

This is totally unrelated with Beal's conjecture, as it provides a way to construct solutions, but the method agrees with its statement.   

Friday, June 7, 2013

Solución paramétrica de una elipse

Hace unos días estaba analizando una ecuación diferencial parcial de segundo grado y me topé con la necesidad de encontrar todas las soluciones de una ecuación homogénea de segundo orden


Esta no es más que una elipse rotada, así que las soluciones $(x,y)$ pueden encontrarse de forma paramétrica por medio de las soluciones de una elipse. Un problema bastante rutinario y nada fuera de lo común. Sin embargo, estuve buscando en internet alguna solución paramétrica de una elipse, encontrando únicamente la solución para una elipse normal, no rotada. 

Desde que estaba en los primeros cursos de cálculo, recuerdo el truco de rotar los ejes para eliminar el termino cruzado $xy$ en la ecuación de una cónica, y con esto simplificar la expresión, así que me decidí a escribir la ecuación paramétrica de una elipse general


Esta es la ecuación general de una elipse, esto es, una elipse trasladada con centro $(h,k)$ y con ejes rotados por un ángulo $\theta$. 

Sustituyendo $x\mapsto x-h$ e $y\mapsto y-k$, se tiene que la ecuación se puede expresar como


$$h=\frac{be-2cd}{4ac-b^2}$$ y $$k=\frac{bd-2ae}{4ac-b^2}.$$

Ahora, resta encontrar las soluciones de una ecuación del tipo

$$\alpha x^2+\beta xy+\gamma y^2+\delta=0.$$

Esto se logra por medio de la rotación de ejes

$$x=u \cos(\theta)-v \sin(\theta)$$ e $$y=u\sin(\theta)+v\cos(\theta)$$.

Sustituyendo esto se obtiene que para anular el factor $uv$ es necesario hacer 


Luego de esto, se tiene una ecuación del tipo 


en donde se puede suponer que $Au^2=\cos(t)$ y $Bv^2=\sin(t)$, con lo que se puede obtener entonces las soluciones paramétricas



Por lo que regresando a $(x,y)$ da



Para finalizar, solo basta incorporar $h,k$ y hacer la correspondencia de $\alpha, \beta, \gamma, \delta$ en las ecuaciones,