I've been working on a proof involving Dickson polynomials, and I'm having trouble finding any mistakes and completing this proof.
We have the equality \begin{equation} D_n(x, a)=\sum_{i=0}^{\lfloor \frac{n}{2}\rfloor}d_{n,i}x^{n-2i} \ \text{, where} \ d_{n,i}=\frac{n}{n-i}{n-i\choose i}(-a)^i. \end{equation} It simplifies to the conditions that $$D_n(x,a)=x^n-nax^{n-2}+ \cdots + \frac{n}{2}^2(-a)^{(n/2)-1}x^2+2(-a)^{n/2},$$ when $n$ is an even number, and $$D_n(x,a)=x^n-nax^{n-2}+ \cdots + \frac{n}{2}^2(-a)^{(n/2)-1}x,$$ for odd values of $n$.
We will prove this expression using mathematical induction. Let's check if the equality holds for $n = 1$ and $n = 2$. For $n = 1$: $$ D_1(x, a) = x. $$ According to the equality: $$ \sum_{i=0}^{\lfloor \frac{1}{2}\rfloor}d_{1,i}x^{1-2i} = d_{1,0}x^1 = x, $$ which is true. For $n = 2$: $$ D_2(x, a) = x^2 - 2ax. $$ According to the equality: $$ \sum_{i=0}^{\lfloor \frac{2}{2}\rfloor}d_{2,i}x^{2-2i} = d_{2,0}x^2 + d_{2,1}x^0 = x^2 - 2ax, $$ which is also true. Assume that the equality holds for $n = k$ and $n = k - 1$. Now, we will show that the equality also holds for $n = k + 1$. Dickson polynomials satisfy the recursive relation: $$ D_{k+1}(x, a) = xD_k(x, a) - aD_{k-1}(x, a). $$ Substituting the expressions for $D_k(x, a)$ and $D_{k-1}(x, a)$, we have: $$ D_{k+1}(x, a) = x\sum_{i=0}^{\lfloor \frac{k}{2}\rfloor}d_{k,i}x^{k-2i} - a\sum_{i=0}^{\lfloor \frac{k-1}{2}\rfloor}d_{k-1,i}x^{k-1-2i}. $$ Now, transform the sums so that they can be added together, taking into account the index shifts. $$ D_{k+1}(x, a) = \sum_{i=0}^{\lfloor \frac{k}{2}\rfloor}d_{k,i} x^{k+1-2i} - a\sum_{i=0}^{\lfloor \frac{k-1}{2}\rfloor} d_{k-1,i} x^{k-1-2i+2}. $$ Next, we obtain: $$ D_{k+1}(x, a) = d_{k,0}x^{k+1} + \sum_{i=1}^{\lfloor \frac{k+1}{2}\rfloor} [d_{k,i} x^{k+1-2i} - a d_{k-1,i-1} x^{k+1-2i}]. $$ To complete the proof, we need to show that $d_{k+1,i} = d_{k,i} - a d_{k-1,i-1}$. Using the definition of $d_{n,i}$: $$d_{n,i} = \frac{n}{n-i} {n-i \choose i} (-a)^i$$ Substitute the values for $d_{k+1,i}$, $d_{k,i}$, and $d_{k-1,i-1}$: $$\frac{k+1}{k+1-i} {k+1-i \choose i} (-a)^i = \frac{k}{k-i} {k-i \choose i} (-a)^i - a \frac{k-1}{k-i} {k-i-1 \choose i-1} (-a)^{i-1}$$
And i don't know what next.
