User:Primedivine/sandbox

test
$$F_{n}= \prod_{d \mid n} \phi_{d} \text{, for each divisor } d \text{ of } n \text{ with a squarefree quotient } {\mu(n/d) \neq 0}  \text{?}$$ - SF 2019 $$F_{\phi_{\alpha(p)}}= (\prod_{p_{i}^{\lambda_{i}} \mid \phi_{\alpha(p)}}F_{p_{i}^{\lambda_{i}}}) \cdot (\prod_{d_{i} \mid \phi_{\alpha(p)}} \phi_{d_{i}})$$

=Summary= Let $$\mu(n)$$ denote the Möbius function. $$\phi_{n}= \prod_{d \mid n} F_{d}^{\mu(n/d)} \text{, for all divisors } d \text{ of } n \text{.}$$ - T.M. Apostol 1976 $$\beta_{n}= \prod_{d \mid n} F_{d}^{-\mu(n/d)} \text{, for proper divisors } d \text{ of } n \text{.} $$ $$F_{n}=\beta_{n} \cdot \phi_{n}$$ Let $$F_{\alpha(d)}$$ denote the least positive Fibonacci number divisible by $$d$$, such that $$d|F_{n}$$. Suppose $$e \ge 2$$. If $$\phi_{\alpha(p)}=p^e$$ then $$\beta_{\phi_{\alpha(p)}}=(F_{p^{e -1}}) \neq (F_{p^{e}}) $$. Remark: The only proper divisor that leaves behind a square free quotient is $$p^{e-1}$$. Clearly, we have a contradiction with Wall-Sun-Sun primes of this form. This shortcut to the pertinent divisors is very useful for understanding the real problem with the composition of primitive prime powers with a multiplicity greater than 1. Here is an example of two or more prime powers. $$\alpha(F_{p^2} \cdot F_{r^2} \cdot \phi_{p^{2}r} \cdot \phi_{r^{2}p} \cdot \phi_{pr} \cdot \phi_{\phi_{\alpha(p)}}) = [p^2,r^2, p^{2}r, r^{2}p, pr, \phi_{\alpha(p)}]=\alpha(\beta_{\phi_{\alpha(p)}})=\phi_{\alpha(p)}$$ $$\alpha(F_{p^2} \cdot F_{r^2} \cdot \phi_{p^{2}r} \cdot \phi_{r^{2}p} \cdot \phi_{pr} \cdot \phi_{\phi_{\alpha(p^{2}r^{2})}}) = [p^2,r^2, p^{2}r, r^{2}p, pr, \phi_{\alpha(p^{2}r^{2})}]=\alpha(\beta_{\phi_{\alpha(p^{2}r^{2})}})=\phi_{\alpha(p^{2}r^{2})}$$ $$F_{n}= \prod_{d \mid n} \phi_{d} \text{, for each divisor } d \text{ of } n \text{ with a squarefree quotient } {\mu(n/d) \neq 0}  \text{?}$$ - SF 2019
 * All positive divisors $$d$$ divide some Fibonacci number.
 * By Carmichael's theorem, for $$n>12$$ each Fibonacci number $$F_{n}$$ will have at least one primitive prime divisor, or at least one primitive prime power in the case of Wall Sun Sun primes.
 * For any positive integers a and b, let [a,b] denote the least common multiple of a and b. $$\alpha([a,b])=[\alpha(a), \alpha(b)]$$ - D. W. Robinson April 1963
 * For $$n\neq 2$$, the entry point of a positive Fibonacci number $$F_{n}$$ is simply the subscript $$n$$, ie $$\alpha(F_{n})=n$$.
 * The primitive part of any Fibonacci number has an equal entry point to the whole Fibonacci number itself, ie $$\alpha(\phi_{n})=\alpha(F_{n})=n$$
 * $$\gcd(F_{m},F_{n})=F_{\gcd(m,n)}$$ - Lucas
 * For $$n<p$$, $$\gcd(F_{n},F_{p})=F_{\gcd(n,p)}=F_{1}=1$$, if $$p$$ is prime. - SF 2000
 * If $$d|n$$ then $$F_{d}|\beta_{n}$$, and consequently $$\phi_{d}|\beta_{n}$$.
 * If $$p^{y \le \lambda}|\phi_{\alpha(p)}$$ then $$\phi_{p^{y \le \lambda}}|\beta_{\phi_{\alpha(p)}}$$, for $$\lambda \ge 2$$, where $$p^{\lambda+1} \nmid \phi_{\alpha(p)}$$.

old
$$\mu(n)=\delta_{\omega(n)}^{\Omega(n)}\lambda(n)$$ where $$\delta$$ is the Kronecker delta, $λ(n)$ is the Liouville function, $ω(n)$ is the number of distinct prime divisors of $n$, and $Ω(n)$ is the number of prime factors of $n$, counted with multiplicity. $$\phi_{n}= \prod_{d \mid n} F_{d}^{\mu(n/d)} \text{, for all divisors } d \text{ of } n \text{.}$$ [T.M. Apostol 1976] $$\beta_{n}= \prod_{d \mid n} F_{d}^{-\mu(n/d)} \text{, for proper divisors } d \text{ of } n \text{.} $$ $$F_{n}=\beta_{n} \cdot \phi_{n}$$ Let $$F_{\alpha(d)}$$ denote the least positive Fibonacci number divisible by $$d$$, such that $$d|F_{n}$$. For any positive integers a and b, let [a,b] denote the least common multiple of a and b.
 * $$\alpha([a,b])=[\alpha(a), \alpha(b)]$$ - D. W. Robinson April 1963
 * For $$n\neq 2$$, the entry point of a positive Fibonacci number $$F_{n}$$ is simply the subscript $$n$$, ie $$\alpha(F_{n})=n$$.
 * All positive divisors $$d$$ divide some Fibonacci number.
 * By Carmichael's theorem, every Fibonacci number $$F_{n}, n>12$$ will have at least one primitive prime divisor, or at least one primitive prime power in the case of Wall Sun Sun primes.
 * $$\gcd(F_{a},F_{b})=F_{\gcd(a,b)}$$ - Lucas
 * For $$n<p$$, $$\gcd(F_{n},F_{p})=F_{\gcd(n,p)}=F_{1}=1$$. - SF
 * If $$d|n$$ then $$F_{d}|\beta_{n}$$, and consequently $$\phi_{d}|\beta_{n}$$.
 * If $$p^{y \le \lambda}|\phi_{\alpha(p)}$$ then $$\phi_{p^{y \le \lambda}}|\beta_{\phi_{\alpha(p)}}$$, for $$\lambda \ge 2$$, where $$p^{\lambda+1} \nmid \phi_{\alpha(p)}$$.

Result
Suppose $$e \ge 2$$. If $$\phi_{\alpha(p)}=p^e$$ then $$\beta_{\phi_{\alpha(p)}}=(F_{p^{e -1}}) \neq (F_{p^{e}}) $$. Remark: The only proper divisor that leaves behind a square free quotient is $$p^{e-1}$$. Clearly, we have a contradiction with Wall-Sun-Sun primes of this form. This shortcut to the pertinent divisors is very useful for understanding the real problem with the composition of primitive prime powers with a multiplicity greater than 1. Here is an example of two or more prime powers. $$\alpha(F_{p^2} \cdot F_{r^2} \cdot \phi_{p^{2}r} \cdot \phi_{r^{2}p} \cdot \phi_{pr} \cdot \phi_{\phi_{\alpha(p)}}) = [p^2,r^2, p^{2}r, r^{2}p, pr, \phi_{\alpha(p)}]=\alpha(\beta_{\phi_{\alpha(p)}})=\phi_{\alpha(p)}$$