I don't know if we can use convolution to prove your statement, and it's the first time i see it so I don't know any reference , But I have a proof by induction.
The first terms $a_1,a_2$ coincide with Fibonacci numbers.
Let us suppose that:$a_1,a_2,\cdots,a_{x},a_{x+1}$ are the first Fibonacci numbers (this assumption is used in the proposition) and prove that $a_{x+2}$ is the next Fibonacci number.
we compute $a_{x+2}-a_{x+1}$:
$$a_{x+2}-a_{x+1}=\sum_{n:\alpha(n)=x} \varphi(n)+\sum_{n:\alpha(n)<x-1}\varphi(n) \left (\left \lfloor{ \frac{x}{\alpha(n)} }\right \rfloor - \left \lfloor{ \frac{x-1}{\alpha(n)} }\right \rfloor\right )$$
And we know that $ \lfloor x/k \rfloor - \lfloor (x-1)/k \rfloor = 1 $ when $k|x$ (for $k\geq1$) and $0$ otherwise, so
$$a_{x+2}-a_{x+1}=\sum_{n:\alpha(n)|x} \varphi(n)\,\,\,\, (1)$$
but we have
Proposition for every positive integer $n$ $$\alpha(n)|x \Leftrightarrow n|a_x$$
proof
first it's clear that if $\alpha(n)=k$ with $k|x$ then $n|a_k$, and because $a_k$ and $a_x$ are Fibonacci numbers then $a_k|a_x$ finally $n|a_x$.
Second, given a divisor $n$ of $a_x$,let $\alpha(n)=k\leq x$ then $n|a_k$ so $n|gcd(a_k,a_x)=a_{gcd(x,k)}$ ( because $a_x$ and $a_k$ are Fibonacci numbers) using the definition of $\alpha(n)$ we have $k\leq gcd(x,k)$ hence $k|x$.
Using the proposition, the sum $(1)$ becomes:
$$a_{x+2}-a_{x+1}=\sum_{n:n|a_x} \varphi(n)$$
using Euler's identity $\sum_{d:d|n} \varphi(d)=n$ :
$$a_{x+2}-a_{x+1}=a_x $$
Finally $a_{x+2}$ is the next Fibonacci number
Try formulating the induction step like this:
$$ \begin{align}\Phi(n) = & \text{$f(3n)$ is even ${\bf and}$}\\
& \text{$f(3n + 1)$ is odd ${\bf and}$}\\
& \text{$f(3n+2)$ is odd. }
\end{align}$$
Then use induction to prove that $\Phi(n)$ is true for all $n$. The base case $\Phi(0)$ is as easy as usual; it's just $\text{“$0$ is even and $1$ is odd and $1$ is odd”}$.
Best Answer
HINT: Induction is much easier, but if you really must use Binet’s formula, note that your summations are geometric, with ratios $\alpha^2$ and $\beta^2$, respectively; you can use the familiar formula for the sum of a finite geometric series. You also know that $\alpha$ and $\beta$ satisfy the equation $x^2-x-1=0$, so $\alpha^2-1=\alpha$ and $\beta^2-1=\beta$; this information will help you simplify the sums of the geometric progressions.