8

Let $F$ be an integral domain with characteristic $2$. Let $a\in F[t]$ and $a \notin F$. Let $\alpha (a)$ be a root of the equation $x^2+ax+1=0$. We define two sequences $X_m(a), Y_m(a) \in F[t], m \in \mathbb{Z}$ by
$$X_m(a)+\alpha (a)Y_m(a)=(\alpha (a))^m=(a+\alpha (a))^{-m} \tag 1$$

Lemma.

Let $F$ be an integral domain with characteristic $p=2$. Let $a \in F[t], a \notin F$. For all $m, n \in \mathbb{Z}$ we have :

  1. $X_m(a)$ (resp. $Y_m(a)$) is equal to the polynomial obtained by substituting $a$ for $t$ in $X_m(t)$ (resp. $Y_m(t)$).
    The degree of the polynomial $X_m(t)$ is $m-2$, if $m \geq 2$.
    The degree of the polynomial $Y_m(t)$ is $m-1$, if $m \geq 2$.
    $X_{-m}=X_m(a)+aY_m(a)$
    $Y_{-m}(a)=Y_m(a)$

  2. All solutions $X, Y \in F[t]$ of the equation $$X^2+aXY+Y^2=1\tag 2$$ are given by $X_m(a), Y_m(a)$, with $m \in \mathbb{Z}$.

I want to prove this lemma but I am facing some difficulties at $2$.

First I showed that $(X_m(a), Y_m(a))$ is a solution of the equation $(2)$. But how could we show that all the solutions of the equation $(2)$ are given by $X_m(a), Y_m(a)$.

I have no idea how to do that... Could you give me some hints?

$$$$

EDIT:

At the paper that I am looking I found the corresponding lemma for the case that the characteristic is not $2$ and its proof:

enter image description here enter image description here

PROOF.

enter image description here enter image description here

$$$$

I want to try to do the same for the case $\text{char}=2$ but first I have to clarify some points at the proof above.

Why do we consider the field $K=R(t)(a)$?

Is $S$ the set of points at which the functions of $K$ are not defined?

Mary Star
  • 13,956
  • You need to solve this equation? It is reduced to the Pell equation. – individ Sep 28 '15 at 10:46
  • Could you explain it further to me? @individ – Mary Star Sep 28 '15 at 10:51
  • Standard procedure. To Express this equation using the equation Pell. For example such formulas. http://www.artofproblemsolving.com/community/c3046h1048219 – individ Sep 28 '15 at 10:56
  • do you know what a place is ? – mercio Sep 28 '15 at 11:49
  • Not really... @mercio – Mary Star Sep 28 '15 at 12:19
  • Do we have to prove which the solutions of the equation Pell are? Or is this known? @individ – Mary Star Sep 28 '15 at 12:21
  • Write a specific equation. Let us examine it. – individ Sep 28 '15 at 12:26
  • What do you mean? A specific Pell equation? I got stuck right now... @individ – Mary Star Sep 28 '15 at 12:40
  • No. The specific equation which must be solved. – individ Sep 28 '15 at 12:47
  • We want to solve the equation $X^2+aXY+Y^2=1$. @individ – Mary Star Sep 28 '15 at 12:48
  • It's obvious. $$X^2+aXY+Y^2=1$$ $$x^2+2axy+4y^2=1$$ $$x^2+2axy+a^2y^2-(a^2-4)y^2=1$$ $$(x+ay)^2-(a^2-4)y^2=1$$ – individ Sep 28 '15 at 12:59
  • I see... And the last equatlity represent an equation Pell, right? @individ – Mary Star Sep 28 '15 at 13:03
  • Yes. Of course. His decisions will determine the desired solution. – individ Sep 28 '15 at 13:06
  • Do we have to find now the solutions of such an equation Pell? @individ – Mary Star Sep 28 '15 at 13:08
  • Searching for solutions of the equation Pell. $p^2-ks^2=1$ Standard procedure. Solutions are found by the decomposition of $\sqrt{k}$ in the continued fraction. – individ Sep 28 '15 at 13:11
  • So we have that the trivial solutions of the equation $p^2-ks^2=1$ are $(p, s)=(\pm 1, 0)$. To find the non-trivial solutions we do the following: $$p^2-ks^2=1 \Rightarrow (p-\sqrt{k}s)(p+\sqrt{k}s)=1$$ We are working at the field $F[t]$, where $\text{char}F=2$, right? How do we continue? @individ – Mary Star Sep 28 '15 at 13:45
  • Since we are working in a field of characterisitc $2$ does it stand that $k=a^2-4=a^2$ ? @individ – Mary Star Sep 28 '15 at 13:54
  • Yes. Usually do. To facilitate the calculation. http://www.artofproblemsolving.com/community/c3046h1090811_the_general_pell_equation http://math.stackexchange.com/questions/128930/whenever-pells-equation-proof-is-solvable-it-has-infinitely-many-solutions/831363#831363 – individ Sep 28 '15 at 13:55
  • Why can we not do it as follows? Since $k=a^2-4=a^2$ we have $$p^2-ks^2=1 \Rightarrow p^2-a^2s^2=1 \Rightarrow (p-as)(p+as)=1$$ Since $p-as$ and $p+as$ are elements of $F[t]$, so they are polynomials we conclude that it cannot be $p-as=(p+as)^{-1}$ or $p+as=(p-as)^{-1}$, so it must be $p-as=1$ and $p+as=1$. So the only solutions are the trivial ones. @individ – Mary Star Sep 28 '15 at 14:21
  • Trivial solutions when the coefficient of the square. But he was not. – individ Sep 28 '15 at 14:38
  • What do you mean? @individ – Mary Star Sep 28 '15 at 14:39
  • The coefficient in equation Pell is not a square. – individ Sep 28 '15 at 14:40
  • So doesn't it stand that $k=a^2-4=a^2$ because the characterisitc of the field is $2$ ? @individ – Mary Star Sep 28 '15 at 14:41
  • There is a difference. Why should it be a square? Solution of Pell equation is always when the coefficient not a square. – individ Sep 28 '15 at 14:44
  • I got stuck right now... First of all, why doesn't it stand that $$k=a^2-4=a^2$$ ? @individ – Mary Star Sep 28 '15 at 14:50
  • Brought to the equation $x^2-ky^2=x^2-(a^2-4)y^2=1$ Why $k$ needs to be a square? He will not be square. – individ Sep 28 '15 at 14:56
  • At the links about the solutions of an equation Pell, do we need to know two solutions to find all the solutions of the equation? @individ – Mary Star Sep 28 '15 at 22:55
  • We need to know first the solution of the equation Pell. On it and using the formula. You can find all the solutions. Find a sequence of all solutions. – individ Sep 29 '15 at 04:14
  • Can we use the trivial solution to find all the solutions using the formula? Or do we have to use a non-trivial solution? @individ – Mary Star Sep 29 '15 at 08:06
  • Only not trivial. The first non-trivial solution. – individ Sep 29 '15 at 12:30
  • And which are the first non-trivial solution in this case? How can we find it? @individ – Mary Star Sep 29 '15 at 12:52
  • I don't think that my ideas and approach will help you. Better read a book on number theory. – individ Sep 29 '15 at 13:05

1 Answers1

3

Suppose $(X,Y)$ is a solution where $X,Y \neq 0$.

Then, by looking at the degrees in the equation, we either have $\deg X = \deg a + \deg Y$ or $\deg Y = \deg a + \deg X$.
Suppose we are in the first case and let $X' = X + aY$. $(X',Y)$ is again a solution of the equation, and looking at the first coefficients of $X,Y,a$, we see that the coefficient of $T^{\deg X}$ in $X'$ vanishes (here we use that $F$ is an integral domain), so $\deg X' < \deg X$, and so we get a "smaller" solution.

We can repeat this process until we reach a solution where $X=0$ or $Y=0$. In that case it is easy to see that the only solutions are $(0,1)$ and $(1,0)$.

So every solution can be obtained by starting from one of those two and applying the operations $(X,Y) \mapsto (X,Y+aX)$ and $(X,Y) \mapsto (X+aY,Y)$, and those are exactly the $(X_m(a), Y_m(a))$ for $m \in \Bbb Z$

mercio
  • 50,180
  • How do we conclude that $\text{deg}X=\text{deg}a+\text{deg}Y$ or $\text{deg}Y=\text{deg}a+\text{deg}X$?$$\text{deg}(X^2+aXY+Y^2)=\text{deg}1=0\ \Rightarrow\max{\text{deg} (X^2),\text{deg} (aXY),\text{deg} (Y^2)=0\ \Rightarrow\max{2\text{deg}X,\text{deg}a+\text{deg}X+\text{deg} Y,2\text{deg} Y}=0$$ If $\max =2\text{deg}X$ or $\max =2\text{deg}Y$ then $\text{deg}X=0$ or $\text{deg}Y=0$.But then $\text{deg}a+\text{deg}X+\text{deg}Y=\text{deg}a \geq 0$.So it must be $\max =\text{deg}a+\text{deg}X+\text{deg}Y$,i.e.,$\text{deg}a+\text{deg}X+\text{deg}Y=0$. But this isn't the desired result.Or? – Mary Star Sep 26 '15 at 23:12
  • the two terms of highest degree have to cancel each other so that the sum is $1$. If $\deg X = \deg Y$ then the middle term has highest degree and can'tbe cancelled by anyone, so that's impossible. So for example $\deg X > \deg Y$ and then the first two terms have to have the same degree hence $\deg X = \deg Y + \deg a$ – mercio Sep 27 '15 at 01:27
  • Ok... I see... Why do we take then $X'=X+aY$ ? – Mary Star Sep 27 '15 at 21:06
  • because when you fix $Y$, the equation is a quadratic in $X$ so it has two solutions $(X_1,Y)$ and $(X_2,Y)$. Then it's natural to investigate the involution that permutes $X_1$ with $X_2$.. Also there are two obvious ways to put a commutative group structure on the set of solutions, and (modulo switching $X$ with $Y$), this operation is just adding or substracting the fundamental solution such that we get closer to the identity element of the group. – mercio Sep 27 '15 at 21:21
  • I have to think about it... At the paper that I am looking I found the corresponding lemma for the case that the characteristic is not $2$ and its proof, I added it in my initial post above. Could you take a look at it? – Mary Star Sep 28 '15 at 02:38