I'm very interested in studying number theory and I post an other conjecture between the golden ratio and Fibonacci numbers. N.B: it is well known that is a link between the golden ratio and Fibonacci numbers. I just want to know if my proof already exists in the literature.
Conjecture:
I'm looking for a formula between Fibonacci numbers and the golden ratio $\varphi$. The formula is:
$$\varphi(n) = n \cdot \prod_{p | n} \left(1 - \frac{1}{p}\right).$$
where the product is taken over all distinct prime factors $p$ of $n$.
Proof
If we let $n$ be a power of $2$, say $n = 2^k$ ($k$ is an integer $> 1$), then we can simplify this formula further using the fact that $1 - \frac{1}{2} = 1/2$.
Applying this identity repeatedly, we get:
$$\varphi(2^k) = 2^k \cdot \prod_{i=0}^k (1 - \frac{1}{2}) = 2^{k-1},$$
where $\prod_{i=0}^k$ denotes the product of $(k+1)$ terms, and we have used the fact that there are $k+1$ distinct prime factors of $2^k$, namely $2^0$, $2^1$, ..., $2^k$.
Now, we can use the fact that the limit of the ratio of consecutive Fibonacci numbers approaches the golden ratio as the sequence progresses. Specifically, if we let $F_n$ denote the $n^{th}$ Fibonacci number, then we have:
$$\lim_{n \to +\infty} {\frac{F_{n+1}}{F_n}} = \varphi$$
Finally, we can combine these results to obtain a formula relating Fibonacci numbers to the golden ratio in terms of Fibonacci numbers:
$$\varphi(2^k) = 2^{k-1}= \left(\frac{F_k}{\sqrt{5}}\right)^{\frac{1}{\varphi}}$$
This formula expresses the value of $\varphi(2^k)$ in terms of both the totient function and the golden ratio, with the Fibonacci numbers serving as an intermediate link between the two.
Thanks.
I'm not sure what all the downvote are all about. You have a well-formed conjecture with a nice proof.
Your conjecture is indeed correct. To answer your question: this is already well-known and can be found on the wiki page for the Euler Totient function.
Here is another question with the given proof.
I'm not aware & couldn't find any "well-known" proof using the technique you give.