To what is X_{-m}+α(a)Y_{-m} equal?

[mathmari]

Hey!

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 the two sequences $X_m(a), Y_m(a) \in F[t], m \in \mathbb{Z}$ as followed:
$$X_m(a)+\alpha (a)Y_m(a)=(\alpha (a))^m=(a+\alpha (a))^{-m}$$

Lemma.

Let $F$ be an integral domain with characteristic $p=2$. Let $a \in F[t], a \notin F$. $X_m(s)$ (resp. $Y_m(a)$) is equal to the polynomial that we obtain if we substitute $t$ with $a$ at $X_m(t)$ (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 $m \geq 2$.
• $X_{-m}=X_m(a)+aY_m(a)$
• $Y_{-m}(a)=Y_m(a)$

To prove this lemma I have done the following:

For the first two sentences about the degree I used induction on $m$. Is this correct?

As for the last two relations:

$$X_{-m}+\alpha (a)Y_{-m}=(a+\alpha (a))^m=((a+\alpha (a))^{-m})^{-1}=(X_m(a)+\alpha (a)Y_m(a))^{-1}=\frac{1}{X_m(a)+\alpha (a)Y_m(a)}$$

How could we continue?

 

[mathmari]

Could we maybe do it as follows:

$$X_{-m}(a)+\alpha (a)Y_{-m}(a)=(a+\alpha (a))^m=(a+\alpha (a))^{2m}(a+\alpha (a))^{-m}=[(a+\alpha (a))^2]^m(X_m(a)+\alpha (a)Y_m(a))=[a^2+(\alpha (a))^2]^m(X_m(a)+\alpha (a)Y_m(a))=[a^2+a\alpha (a)+1]^m(X_m(a)+\alpha (a)Y_m(a))$$

But how could we continue?

Or is this the wrong way?

 

[mathmari]

I have also an other idea:

$$X_{-m}(a)+\alpha (a)Y_{-m}(a)=(a+\alpha (a))^m=X_m(a)+(a+\alpha (a))Y_m(a)\\ =X_m(a)+aY_m(a)+\alpha (a)Y_m(a) \\ \Rightarrow X_{-m}(a)=X_m(a)+aY_m(a)\ \ , \ \ Y_{-m}(a)=Y_m(a)$$

Is this correct?

 

[Fallen Angel]

Hi,

I don't have the solution, in fact, I can not even try to do it right now, but just as a comment, you can't take inverses (rise to -1), because you are working in an integral domain, not in a field.

And using induction is OK if you haven't made any mistakes. :)

I will try to solve it if I can have some rest this weekend =P

 

[mathmari]

you can't take inverses (rise to -1), because you are working in an integral domain, not in a field.

At my last idea, at post #3, I don't inverse the elemnts to $-1$. Is this correct to show it in that way?

And using induction is OK if you haven't made any mistakes. :)

I did it as follows:

Base case: For $m=2$ we have $X_2(a)+\alpha (a)Y_2(a)=(\alpha (a))^2=a\alpha (a)+1$. So $X_2(a)=1, Y_2(a)=a$. That means that $\text{deg}(X_2(t))=0=2-2$ and $\text{deg}(Y_2(t))=1=2-1$.

Inductive hypothesis: We suppose that it holds for $m=k$, i.e., $\text{deg}(X_k(t))=k-2$ and $\text{deg}(Y_k(t))=k-1$.

Inductive step: We will show that it holds for $n=k+1$, i.e., $\text{deg}(X_{k+1}(t))=k-1$ and $\text{deg}(Y_{k+1}(t))=k$.
$$X_{k+1}(a)+\alpha (a)Y_{k+1}(a)=(a+\alpha (a))^{-(k+1)}=(a+\alpha (a))^{-k}(a+\alpha (a))^{-1}=(X_k(a)+\alpha (a)Y_k(a))(a+\alpha (a))^{-1}=(X_k(a)+\alpha (a)Y_k(a))\alpha (a)=\alpha (a)X_k(a)+\alpha (a)^2Y_k(a)=\alpha (a)X_k(a)+(a\alpha (a)+1)Y_k(a)=Y_k(a)+\alpha (a)[X_k(a)+aY_k(a)] \\ \Rightarrow X_{k+1}(a)=Y_k(a), Y_{k+1}(a)=X_k(a)+aY_k(a) \\ \Rightarrow \text{deg}(X_{k+1}(a))=\text{deg}(Y_k(a))=k-1 \ \ , \ \ \text{deg}(Y_{k+1}(a))=\max \{\text{deg}(X_k(a)), \text{deg}(aY_k(a))\}=\max \{k-2, 1+k-1\}=k$$
I am a little confused about "a" and "t".