Primitive prime powers
edit
For
n
>
12
{\displaystyle n>12}
, each Fibonacci number
F
n
{\displaystyle F_{n}}
will have at least one primitive prime divisor, by Carmichael's theorem. By the Wall-Sun-Sun prime conjecture,
F
n
{\displaystyle F_{n}}
could have at least one primitive prime power. Let
ϕ
α
(
p
)
=
(
p
1
e
1
p
2
e
2
⋯
p
k
e
k
)
{\displaystyle \phi _{\alpha (p)}=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})}
denote the full product of primitive prime powers (one or more) that divide
F
α
(
p
)
{\displaystyle F_{\alpha (p)}}
. By definition, this product of primitive prime powers always has an equal entry point to the whole Fibonacci number itself, ie
α
(
ϕ
n
)
=
α
(
F
n
)
=
n
{\displaystyle \alpha (\phi _{n})=\alpha (F_{n})=n}
.
Lowest common multiple
edit
For any positive integers a and b, let [a,b] denote the least common multiple of a and b.
α
(
[
a
,
b
]
)
=
[
α
(
a
)
,
α
(
b
)
]
{\displaystyle \alpha ([a,b])=[\alpha (a),\alpha (b)]}
- D. W. Robinson April 1963
If
a
{\displaystyle a}
through
z
{\displaystyle z}
are relatively prime then
F
a
{\displaystyle F_{a}}
through
F
z
{\displaystyle F_{z}}
are also relatively prime.
α
(
a
b
⋯
z
)
=
α
(
[
a
,
b
,
…
,
z
]
)
=
[
α
(
a
)
,
α
(
b
)
,
…
,
α
(
z
)
]
{\displaystyle \alpha (ab\cdots z)=\alpha ([a,b,\ldots ,z])=[\alpha (a),\alpha (b),\ldots ,\alpha (z)]}
If
F
a
{\displaystyle F_{a}}
through
F
z
{\displaystyle F_{z}}
are relatively prime then we have the following.
Type A:
2
∤
n
{\displaystyle 2\not \mid n}
or else
2
∦
n
{\displaystyle 2\not \parallel n}
Type B:
2
∥
n
{\displaystyle 2\parallel n}
and also
4
∦
n
{\displaystyle 4\not \parallel n}
Twice the odd numbers, also called singly even numbers.
Type A:
α
(
F
a
F
b
⋯
F
z
)
=
α
(
[
F
a
,
F
b
,
…
,
F
z
]
)
=
[
α
(
F
a
)
,
α
(
F
b
)
,
…
,
α
(
F
z
)
]
=
[
a
,
b
,
…
,
z
]
=
(
a
⋅
b
⋯
z
)
=
a
b
⋯
z
{\displaystyle \alpha (F_{a}F_{b}\cdots F_{z})=\alpha ([F_{a},F_{b},\ldots ,F_{z}])=[\alpha (F_{a}),\alpha (F_{b}),\ldots ,\alpha (F_{z})]=[a,b,\ldots ,z]=(a\cdot b\cdots z)=ab\cdots z}
Type B:
α
(
F
2
F
b
⋯
F
z
)
=
α
(
[
F
2
,
F
b
,
…
,
F
z
]
)
=
[
α
(
F
2
)
,
α
(
F
b
)
,
…
,
α
(
F
z
)
]
=
[
1
,
b
,
…
,
z
]
=
(
1
⋅
b
⋯
z
)
=
b
⋯
z
{\displaystyle \alpha (F_{2}F_{b}\cdots F_{z})=\alpha ([F_{2},F_{b},\ldots ,F_{z}])=[\alpha (F_{2}),\alpha (F_{b}),\ldots ,\alpha (F_{z})]=[1,b,\ldots ,z]=(1\cdot b\cdots z)=b\cdots z}
The fundamental theorem of arithmetic is bi-conditional with prime powered Fibonacci numbers.
Let
n
=
p
1
e
1
p
2
e
2
⋯
p
k
e
k
{\displaystyle n=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}}}
.
Type A:
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
e
k
)
=
α
(
[
F
p
1
e
1
,
F
p
2
e
2
,
…
,
F
p
k
e
k
]
)
=
[
α
(
F
p
1
e
1
)
,
α
(
F
p
2
e
2
)
,
…
,
α
(
F
p
k
e
k
)
]
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
e
k
]
=
(
p
1
e
1
p
2
e
2
⋯
p
k
e
k
)
=
n
=
α
(
F
n
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,F_{p_{k}^{e_{k}}}])=[\alpha (F_{p_{1}^{e_{1}}}),\alpha (F_{p_{2}^{e_{2}}}),\ldots ,\alpha (F_{p_{k}^{e_{k}}})]=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{e_{k}}]=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=n=\alpha (F_{n})}
.
Type B:
2
α
(
F
2
F
p
2
e
2
⋯
F
p
k
e
k
)
=
2
α
(
[
F
2
,
F
p
2
e
2
,
…
,
F
p
k
e
k
]
)
=
2
[
α
(
F
2
)
,
α
(
F
p
2
e
2
)
,
…
,
α
(
F
p
k
e
k
)
]
=
2
[
1
,
p
2
e
2
,
…
,
p
k
e
k
]
=
2
(
1
⋅
p
2
e
2
⋯
p
k
e
k
)
=
n
=
α
(
F
n
)
{\displaystyle 2\alpha (F_{2}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}})=2\alpha ([F_{2},F_{p_{2}^{e_{2}}},\ldots ,F_{p_{k}^{e_{k}}}])=2[\alpha (F_{2}),\alpha (F_{p_{2}^{e_{2}}}),\ldots ,\alpha (F_{p_{k}^{e_{k}}})]=2[1,p_{2}^{e_{2}},\ldots ,p_{k}^{e_{k}}]=2(1\cdot p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=n=\alpha (F_{n})}
.
α
(
F
F
3
F
p
2
e
2
⋯
F
p
k
e
k
)
=
α
(
[
F
F
3
,
F
p
2
e
2
,
…
,
F
p
k
e
k
]
)
=
[
α
(
F
F
3
)
,
α
(
F
p
2
e
2
)
,
…
,
α
(
F
p
k
e
k
)
]
=
[
2
,
p
2
e
2
,
…
,
p
k
e
k
]
=
(
2
⋅
p
2
e
2
⋯
p
k
e
k
)
=
n
=
α
(
F
n
)
{\displaystyle \alpha (F_{F_{3}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}})=\alpha ([F_{F_{3}},F_{p_{2}^{e_{2}}},\ldots ,F_{p_{k}^{e_{k}}}])=[\alpha (F_{F_{3}}),\alpha (F_{p_{2}^{e_{2}}}),\ldots ,\alpha (F_{p_{k}^{e_{k}}})]=[2,p_{2}^{e_{2}},\ldots ,p_{k}^{e_{k}}]=(2\cdot p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=n=\alpha (F_{n})}
.
α
(
F
F
F
4
F
p
2
e
2
⋯
F
p
k
e
k
)
=
α
(
[
F
F
F
4
,
F
p
2
e
2
,
…
,
F
p
k
e
k
]
)
=
[
α
(
F
F
F
4
)
,
α
(
F
p
2
e
2
)
,
…
,
α
(
F
p
k
e
k
)
]
=
[
2
,
p
2
e
2
,
…
,
p
k
e
k
]
=
(
2
⋅
p
2
e
2
⋯
p
k
e
k
)
=
n
=
α
(
F
n
)
{\displaystyle \alpha (F_{F_{F_{4}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}})=\alpha ([F_{F_{F_{4}}},F_{p_{2}^{e_{2}}},\ldots ,F_{p_{k}^{e_{k}}}])=[\alpha (F_{F_{F_{4}}}),\alpha (F_{p_{2}^{e_{2}}}),\ldots ,\alpha (F_{p_{k}^{e_{k}}})]=[2,p_{2}^{e_{2}},\ldots ,p_{k}^{e_{k}}]=(2\cdot p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=n=\alpha (F_{n})}
.
Let
ϕ
α
(
p
)
=
(
p
1
e
1
p
2
e
2
⋯
p
k
e
k
)
{\displaystyle \phi _{\alpha (p)}=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})}
.
Suppose
α
(
p
2
)
=
α
(
p
)
≠
α
(
p
3
)
{\displaystyle \alpha (p^{2})=\alpha (p)\neq \alpha (p^{3})}
.
Suppose
p
i
=
p
{\displaystyle p_{i}=p}
, for one or more Wall-Sun-Sun primes. In this particular instance, take
p
k
=
p
{\displaystyle p_{k}=p}
for the sake of notation below.
Suppose
p
2
∥
F
α
(
p
)
{\displaystyle p^{2}\parallel F_{\alpha (p)}}
and also
p
3
∦
F
α
(
p
)
{\displaystyle p^{3}\not \parallel F_{\alpha (p)}}
, Type A .
If
p
2
|
F
α
(
p
)
{\displaystyle p^{2}|F_{\alpha (p)}}
then
F
p
2
|
F
F
α
(
p
)
{\displaystyle F_{p^{2}}|F_{F_{\alpha (p)}}}
.
If
p
y
≤
λ
|
ϕ
α
(
p
)
{\displaystyle p^{y\leq \lambda }|\phi _{\alpha (p)}}
then
ϕ
p
y
≤
λ
|
F
ϕ
α
(
p
)
{\displaystyle \phi _{p^{y\leq \lambda }}|F_{\phi _{\alpha (p)}}}
, for
λ
≥
2
{\displaystyle \lambda \geq 2}
, where
p
λ
+
1
∤
ϕ
α
(
p
)
{\displaystyle p^{\lambda +1}\nmid \phi _{\alpha (p)}}
.
Claim 1 (Right side b)
edit
If
F
p
2
|
F
F
α
(
p
)
{\displaystyle F_{p^{2}}|F_{F_{\alpha (p)}}}
then
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
2
⋅
b
)
=
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
⋅
b
)
=
F
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot b)=\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}}\cdot b)=F_{\alpha (p)}}
.
Proof 1 (Right side b)
edit
α
(
ϕ
F
α
(
p
)
)
=
F
α
(
p
)
{\displaystyle \alpha (\phi _{F_{\alpha (p)}})=F_{\alpha (p)}}
. Solve for the products with the Robinson equality.
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
2
⋅
ϕ
F
α
(
p
)
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
2
,
F
α
(
p
)
]
=
F
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot \phi _{F_{\alpha (p)}})=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},F_{\alpha (p)}]=F_{\alpha (p)}}
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
⋅
ϕ
F
α
(
p
)
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
,
F
α
(
p
)
]
=
F
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}}\cdot \phi _{F_{\alpha (p)}})=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{},F_{\alpha (p)}]=F_{\alpha (p)}}
If
ϕ
F
α
(
p
)
|
b
{\displaystyle \phi _{F_{\alpha (p)}}|b}
, then
α
(
b
)
=
α
(
ϕ
F
α
(
p
)
)
{\displaystyle \alpha (b)=\alpha (\phi _{F_{\alpha (p)}})}
, for divisors
b
i
{\displaystyle b_{i}}
of
F
F
α
(
p
)
{\displaystyle F_{F_{\alpha (p)}}}
.
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
2
⋅
ϕ
F
α
(
p
)
)
=
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
⋅
ϕ
F
α
(
p
)
)
=
F
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot \phi _{F_{\alpha (p)}})=\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}}\cdot \phi _{F_{\alpha (p)}})=F_{\alpha (p)}}
Claim 2 (Left side a)
edit
If
F
p
2
|
F
F
α
(
p
)
{\displaystyle F_{p^{2}}|F_{F_{\alpha (p)}}}
then
α
(
a
⋅
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
2
)
=
α
(
a
⋅
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
)
=
F
α
(
p
)
{\displaystyle \alpha (a\cdot F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}})=\alpha (a\cdot F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}})=F_{\alpha (p)}}
.
Proof 2 (Left side a)
edit
If
ϕ
α
(
p
)
|
F
α
(
p
)
{\displaystyle \phi _{\alpha (p)}|F_{\alpha (p)}}
then
F
ϕ
α
(
p
)
|
F
F
α
(
p
)
{\displaystyle F_{\phi _{\alpha (p)}}|F_{F_{\alpha (p)}}}
.
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
e
k
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
F
ϕ
α
(
p
)
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}}
.
Establish the hypothetical equality conjectured by Wall-Sun-Sun.
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
2
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
F
p
1
e
1
F
p
2
e
2
⋅
F
p
k
⋅
ϕ
ϕ
α
(
p
)
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdot F_{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})}
?
Solve for the products with the Robinson formula to prove that hypothetically a Wall Sun Sun prime would cause this equality to be true.
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
2
⋅
ϕ
ϕ
α
(
p
)
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
2
,
ϕ
α
(
p
)
]
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot \phi _{\phi _{\alpha (p)}})=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},\phi _{\alpha (p)}]=\phi _{\alpha (p)}}
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
⋅
ϕ
ϕ
α
(
p
)
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
,
ϕ
α
(
p
)
]
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{},\phi _{\alpha (p)}]=\phi _{\alpha (p)}}
Claim 3 (Invalidate the conditional of Claim 2)
edit
α
(
ϕ
ϕ
α
(
p
)
⋅
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
2
)
≠
α
(
ϕ
ϕ
α
(
p
)
⋅
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
)
{\displaystyle \alpha (\phi _{\phi _{\alpha (p)}}\cdot F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}})\neq \alpha (\phi _{\phi _{\alpha (p)}}\cdot F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}})}
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
2
)
≠
α
(
F
p
1
e
1
F
p
2
e
2
⋯
F
p
k
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}})\neq \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}})}
Proof 3 (by contradiction)
edit
By the greatest common divisor, we have
F
p
=
ϕ
p
,
F
p
2
=
ϕ
p
ϕ
p
2
{\displaystyle F_{p}=\phi _{p},F_{p^{2}}=\phi _{p}\phi _{p^{2}}}
.
By Wall's hypothesis,
α
(
F
p
1
F
p
2
⋯
F
p
k
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
[
F
p
1
,
F
p
2
,
…
,
F
p
k
,
ϕ
ϕ
α
(
p
)
]
)
=
[
p
1
,
p
2
,
…
,
p
k
,
ϕ
α
(
p
)
]
=
α
(
F
ϕ
α
(
p
)
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{}}F_{p_{2}^{}}\cdots F_{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([F_{p_{1}^{}},F_{p_{2}^{}},\ldots ,F_{p_{k}^{}},\phi _{\phi _{\alpha (p)}}])=[p_{1}^{},p_{2}^{},\ldots ,p_{k}^{},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}}
By the Wall-Sun-Sun prime conjecture,
α
(
F
p
1
e
1
F
p
2
e
2
⋯
ϕ
p
k
2
⋅
ϕ
p
k
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
[
F
p
1
e
1
,
F
p
2
e
2
,
…
,
ϕ
p
k
2
,
ϕ
p
k
,
ϕ
ϕ
α
(
p
)
]
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
2
,
p
k
,
ϕ
α
(
p
)
]
=
α
(
F
ϕ
α
(
p
)
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}}\cdot \phi _{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{2}},\phi _{p_{k}^{}},\phi _{\phi _{\alpha (p)}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},p_{k}^{},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}}
α
(
F
p
1
e
1
F
p
2
e
2
⋯
ϕ
p
k
2
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
[
F
p
1
e
1
,
F
p
2
e
2
,
…
,
ϕ
p
k
2
,
ϕ
ϕ
α
(
p
)
]
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
2
,
ϕ
α
(
p
)
]
=
α
(
F
ϕ
α
(
p
)
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{2}},\phi _{\phi _{\alpha (p)}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}}
α
(
F
p
1
e
1
F
p
2
e
2
⋯
ϕ
p
k
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
[
F
p
1
e
1
,
F
p
2
e
2
,
…
,
ϕ
p
k
,
ϕ
ϕ
α
(
p
)
]
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
,
ϕ
α
(
p
)
]
=
α
(
F
ϕ
α
(
p
)
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{}},\phi _{\phi _{\alpha (p)}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}}
α
(
F
p
1
e
1
F
p
2
e
2
⋯
ϕ
p
k
2
)
=
α
(
[
F
p
1
e
1
,
F
p
2
e
2
,
…
,
ϕ
p
k
2
]
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
2
]
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{2}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2}]=\phi _{\alpha (p)}}
α
(
F
p
1
e
1
F
p
2
e
2
⋯
ϕ
p
k
2
⋅
ϕ
p
k
)
=
α
(
[
F
p
1
e
1
,
F
p
2
e
2
,
…
,
ϕ
p
k
2
,
ϕ
p
k
]
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
2
,
p
k
]
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}}\cdot \phi _{p_{k}^{}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{2}},\phi _{p_{k}^{}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},p_{k}^{}]=\phi _{\alpha (p)}}
α
(
F
p
1
e
1
F
p
2
e
2
⋯
ϕ
p
k
2
⋅
ϕ
p
k
)
=
α
(
F
p
1
e
1
F
p
2
e
2
⋯
ϕ
p
k
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}}\cdot \phi _{p_{k}^{}})=\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{}})=\phi _{\alpha (p)}}
However, we can measure that equality to verify that it is false.
α
(
F
p
1
e
1
F
p
2
e
2
⋯
ϕ
p
k
)
=
α
(
[
F
p
1
e
1
,
F
p
2
e
2
,
…
,
ϕ
p
k
]
)
=
[
p
1
e
1
,
p
2
e
2
,
…
,
p
k
]
=
ϕ
α
(
p
)
/
p
{\displaystyle \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{}]=\phi _{\alpha (p)}/p}
By Carmichael's theorem, for
n
>
12
,
F
n
{\displaystyle n>12,F_{n}}
will have at least one primitive prime divisor that has not appeared as a divisor of an earlier Fibonacci number. By the Wall-Sun-Sun prime conjecture, let
ϕ
α
(
p
)
=
(
p
1
e
1
p
2
e
2
⋯
p
k
e
k
)
{\displaystyle \phi _{\alpha (p)}=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})}
denote the full product of primitive prime powers (one or more) that divide
F
α
(
p
)
{\displaystyle F_{\alpha (p)}}
.
For proper divisors
d
i
{\displaystyle d_{i}}
of
ϕ
α
(
p
)
{\displaystyle \phi _{\alpha (p)}}
,
α
(
ϕ
d
1
ϕ
d
2
⋯
ϕ
d
j
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
[
ϕ
d
1
,
ϕ
d
2
,
…
,
ϕ
d
j
,
ϕ
ϕ
α
(
p
)
]
)
=
[
α
(
ϕ
d
1
)
,
α
(
ϕ
d
2
)
,
…
,
α
(
ϕ
d
j
)
,
α
(
ϕ
ϕ
α
(
p
)
)
]
=
[
d
1
,
d
2
,
…
,
d
j
,
ϕ
α
(
p
)
]
=
α
(
F
ϕ
α
(
p
)
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (\phi _{d_{1}}\phi _{d_{2}}\cdots \phi _{d_{j}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([\phi _{d_{1}},\phi _{d_{2}},\ldots ,\phi _{d_{j}},\phi _{\phi _{\alpha (p)}}])=[\alpha (\phi _{d_{1}}),\alpha (\phi _{d_{2}}),\ldots ,\alpha (\phi _{d_{j}}),\alpha (\phi _{\phi _{\alpha (p)}})]=[d_{1},d_{2},\ldots ,d_{j},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}}
.
For
d
i
≠
1
,
2
,
6
,
12
{\displaystyle d_{i}\neq 1,2,6,12}
,
n
=
(
p
1
e
1
p
2
e
2
⋯
p
k
e
k
)
=
[
d
1
,
d
2
,
…
,
d
j
,
n
]
=
[
α
(
ϕ
d
1
)
,
α
(
ϕ
d
2
)
,
…
,
α
(
ϕ
d
j
)
,
α
(
ϕ
n
)
]
=
α
(
[
ϕ
d
1
,
ϕ
d
2
,
…
,
ϕ
d
j
,
ϕ
n
]
)
=
α
(
ϕ
d
1
ϕ
d
2
⋯
ϕ
d
j
⋅
ϕ
n
)
{\displaystyle n=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=[d_{1},d_{2},\ldots ,d_{j},n]=[\alpha (\phi _{d_{1}}),\alpha (\phi _{d_{2}}),\ldots ,\alpha (\phi _{d_{j}}),\alpha (\phi _{n})]=\alpha ([\phi _{d_{1}},\phi _{d_{2}},\ldots ,\phi _{d_{j}},\phi _{n}])=\alpha (\phi _{d_{1}}\phi _{d_{2}}\cdots \phi _{d_{j}}\cdot \phi _{n})}
.
For example, if
ϕ
α
(
p
)
=
p
2
⋅
r
{\displaystyle \phi _{\alpha (p)}=p^{2}\cdot r}
then
α
(
ϕ
p
ϕ
p
2
ϕ
r
ϕ
p
r
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
[
ϕ
p
,
ϕ
p
2
,
ϕ
r
,
ϕ
p
r
,
ϕ
ϕ
α
(
p
)
]
)
=
[
α
(
ϕ
p
)
,
α
(
ϕ
p
2
)
,
α
(
ϕ
r
)
,
α
(
ϕ
p
r
)
,
α
(
ϕ
ϕ
α
(
p
)
)
]
=
[
p
,
p
2
,
r
,
p
r
,
ϕ
α
(
p
)
]
=
α
(
F
ϕ
α
(
p
)
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (\phi _{p}\phi _{p^{2}}\phi _{r}\phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([\phi _{p},\phi _{p^{2}},\phi _{r},\phi _{pr},\phi _{\phi _{\alpha (p)}}])=[\alpha (\phi _{p}),\alpha (\phi _{p^{2}}),\alpha (\phi _{r}),\alpha (\phi _{pr}),\alpha (\phi _{\phi _{\alpha (p)}})]=[p,p^{2},r,pr,\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}}
.
α
(
ϕ
p
ϕ
r
ϕ
p
r
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
[
ϕ
p
,
ϕ
r
,
ϕ
p
r
,
ϕ
ϕ
α
(
p
)
]
)
=
[
α
(
ϕ
p
)
,
α
(
ϕ
r
)
,
α
(
ϕ
p
r
)
,
α
(
ϕ
ϕ
α
(
p
)
)
]
=
[
p
,
r
,
p
r
,
ϕ
α
(
p
)
]
=
α
(
F
ϕ
α
(
p
)
)
=
ϕ
α
(
p
)
{\displaystyle \alpha (\phi _{p}\phi _{r}\phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([\phi _{p},\phi _{r},\phi _{pr},\phi _{\phi _{\alpha (p)}}])=[\alpha (\phi _{p}),\alpha (\phi _{r}),\alpha (\phi _{pr}),\alpha (\phi _{\phi _{\alpha (p)}})]=[p,r,pr,\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}}
.
α
(
ϕ
p
ϕ
p
2
ϕ
r
ϕ
p
r
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
ϕ
p
ϕ
r
ϕ
p
r
⋅
ϕ
ϕ
α
(
p
)
)
{\displaystyle \alpha (\phi _{p}\phi _{p^{2}}\phi _{r}\phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (\phi _{p}\phi _{r}\phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})}
α
(
F
p
r
⋅
ϕ
p
2
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
F
p
r
⋅
ϕ
ϕ
α
(
p
)
)
{\displaystyle \alpha (F_{pr}\cdot \phi _{p^{2}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (F_{pr}\cdot \phi _{\phi _{\alpha (p)}})}
α
(
F
p
2
⋅
ϕ
r
⋅
ϕ
p
r
⋅
ϕ
ϕ
α
(
p
)
)
=
α
(
F
p
r
⋅
ϕ
ϕ
α
(
p
)
)
{\displaystyle \alpha (F_{p^{2}}\cdot \phi _{r}\cdot \phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (F_{pr}\cdot \phi _{\phi _{\alpha (p)}})}
F
ϕ
α
(
p
)
=
(
∏
p
i
λ
i
∣
ϕ
α
(
p
)
F
p
i
λ
i
)
⋅
(
∏
d
i
∣
ϕ
α
(
p
)
ϕ
d
i
)
{\displaystyle 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}})}
n
=
(
p
1
λ
1
p
2
λ
2
⋯
p
k
λ
k
)
=
[
p
1
λ
1
,
p
2
λ
2
,
.
.
.
,
p
k
λ
k
,
n
]
=
α
(
F
p
1
λ
1
F
p
2
λ
2
⋯
F
p
k
λ
k
⋅
ϕ
n
)
{\displaystyle n=(p_{1}^{\lambda _{1}}p_{2}^{\lambda _{2}}\cdots p_{k}^{\lambda _{k}})=[p_{1}^{\lambda _{1}},p_{2}^{\lambda _{2}},...,p_{k}^{\lambda _{k}},n]=\alpha (F_{p_{1}^{\lambda _{1}}}F_{p_{2}^{\lambda _{2}}}\cdots F_{p_{k}^{\lambda _{k}}}\cdot \phi _{n})}
n
=
(
p
1
λ
1
p
2
λ
2
⋯
p
k
λ
k
)
=
[
p
1
λ
1
,
p
2
λ
2
,
.
.
.
,
p
k
,
n
]
=
α
(
F
p
1
λ
1
F
p
2
λ
2
⋯
F
p
k
⋅
ϕ
n
)
{\displaystyle n=(p_{1}^{\lambda _{1}}p_{2}^{\lambda _{2}}\cdots p_{k}^{\lambda _{k}})=[p_{1}^{\lambda _{1}},p_{2}^{\lambda _{2}},...,p_{k}^{},n]=\alpha (F_{p_{1}^{\lambda _{1}}}F_{p_{2}^{\lambda _{2}}}\cdots F_{p_{k}^{}}\cdot \phi _{n})}
n
=
(
p
1
λ
1
p
2
λ
2
⋯
p
k
λ
k
)
=
[
p
1
λ
1
,
p
2
λ
2
,
.
.
.
,
p
k
λ
k
]
=
α
(
F
p
1
λ
1
F
p
2
λ
2
⋯
F
p
k
λ
k
)
{\displaystyle n=(p_{1}^{\lambda _{1}}p_{2}^{\lambda _{2}}\cdots p_{k}^{\lambda _{k}})=[p_{1}^{\lambda _{1}},p_{2}^{\lambda _{2}},...,p_{k}^{\lambda _{k}}]=\alpha (F_{p_{1}^{\lambda _{1}}}F_{p_{2}^{\lambda _{2}}}\cdots F_{p_{k}^{\lambda _{k}}})}
n
/
(
p
k
λ
k
−
1
)
=
(
p
1
λ
1
p
2
λ
2
⋯
p
k
)
=
[
p
1
λ
1
,
p
2
λ
2
,
.
.
.
,
p
k
]
=
α
(
F
p
1
λ
1
F
p
2
λ
2
⋯
F
p
k
)
{\displaystyle n/(p_{k}^{\lambda _{k}-1})=(p_{1}^{\lambda _{1}}p_{2}^{\lambda _{2}}\cdots p_{k}^{})=[p_{1}^{\lambda _{1}},p_{2}^{\lambda _{2}},...,p_{k}^{}]=\alpha (F_{p_{1}^{\lambda _{1}}}F_{p_{2}^{\lambda _{2}}}\cdots F_{p_{k}^{}})}
Constructing Fibonacci numbers
edit
Let
n
=
ϕ
α
(
p
)
{\displaystyle n=\phi _{\alpha (p)}}
.
Let
d
1
,
d
2
,
…
,
d
j
{\displaystyle d_{1},d_{2},\ldots ,d_{j}}
be proper divisors of n, composed of at least two distinct prime divisors.
n
=
(
p
1
p
2
)
=
[
p
1
,
p
2
,
n
]
=
α
(
F
p
1
F
p
2
⋅
ϕ
n
)
=
α
(
ϕ
p
1
ϕ
p
2
⋅
ϕ
n
)
{\displaystyle n=(p_{1}^{}p_{2}^{})=[p_{1}^{},p_{2}^{},n]=\alpha (F_{p_{1}^{}}F_{p_{2}^{}}\cdot \phi _{n})=\alpha (\phi _{p_{1}^{}}\phi _{p_{2}^{}}\cdot \phi _{n})}
n
=
(
p
1
p
2
2
)
=
[
p
1
,
p
2
,
p
2
2
,
p
1
p
2
,
n
]
=
α
(
F
p
1
F
p
2
2
ϕ
p
1
p
2
⋅
ϕ
n
)
{\displaystyle n=(p_{1}^{}p_{2}^{2})=[p_{1}^{},p_{2}^{},p_{2}^{2},p_{1}^{}p_{2}^{},n]=\alpha (F_{p_{1}^{}}F_{p_{2}^{2}}\phi _{p_{1}^{}p_{2}^{}}\cdot \phi _{n})}
n
=
(
p
1
λ
p
2
λ
)
=
[
p
1
y
≤
λ
,
p
2
z
≤
λ
,
d
1
,
d
2
,
…
,
d
j
,
n
]
=
α
(
F
p
1
λ
F
p
2
λ
⋅
ϕ
d
1
ϕ
d
2
⋯
ϕ
d
j
⋅
ϕ
n
)
{\displaystyle n=(p_{1}^{\lambda }p_{2}^{\lambda })=[p_{1}^{y\leq \lambda },p_{2}^{z\leq \lambda },d_{1},d_{2},\ldots ,d_{j},n]=\alpha (F_{p_{1}^{\lambda }}F_{p_{2}^{\lambda }}\cdot \phi _{d_{1}}\phi _{d_{2}}\cdots \phi _{d_{j}}\cdot \phi _{n})}
n
=
(
p
1
p
2
p
3
)
=
{\displaystyle n=(p_{1}^{}p_{2}^{}p_{3}^{})=}
[
p
1
,
p
2
,
p
3
,
p
1
p
2
,
p
2
p
3
,
p
1
p
3
,
n
]
=
{\displaystyle [p_{1}^{},p_{2}^{},p_{3}^{},p_{1}^{}p_{2}^{},p_{2}^{}p_{3}^{},p_{1}^{}p_{3}^{},n]=}
α
(
F
p
1
F
p
2
F
p
3
⋅
ϕ
p
1
p
2
ϕ
p
2
p
3
ϕ
p
1
p
3
⋅
ϕ
n
)
{\displaystyle \alpha (F_{p_{1}^{}}F_{p_{2}^{}}F_{p_{3}^{}}\cdot \phi _{p_{1}^{}p_{2}^{}}\phi _{p_{2}^{}p_{3}^{}}\phi _{p_{1}^{}p_{3}^{}}\cdot \phi _{n})}
n
=
(
p
1
p
2
p
3
p
4
)
=
{\displaystyle n=(p_{1}^{}p_{2}^{}p_{3}^{}p_{4}^{})=}
[
p
1
,
p
2
,
p
3
,
p
4
,
p
1
p
2
,
p
1
p
3
,
p
1
p
4
,
p
2
p
3
,
p
2
p
4
,
p
3
p
4
,
p
1
p
2
p
3
,
p
1
p
2
p
4
,
p
2
p
3
p
4
,
p
3
p
1
p
4
,
n
]
=
{\displaystyle [p_{1}^{},p_{2}^{},p_{3}^{},p_{4}^{},p_{1}^{}p_{2}^{},p_{1}^{}p_{3}^{},p_{1}^{}p_{4}^{},p_{2}^{}p_{3}^{},p_{2}^{}p_{4}^{},p_{3}^{}p_{4}^{},p_{1}^{}p_{2}^{}p_{3}^{},p_{1}^{}p_{2}^{}p_{4}^{},p_{2}^{}p_{3}^{}p_{4}^{},p_{3}^{}p_{1}^{}p_{4}^{},n]=}
α
(
F
p
1
F
p
2
F
p
3
F
p
4
⋅
ϕ
p
1
p
2
ϕ
p
1
p
3
ϕ
p
1
p
4
ϕ
p
2
p
3
ϕ
p
2
p
4
ϕ
p
3
p
4
ϕ
p
1
p
2
p
3
ϕ
p
1
p
2
p
4
ϕ
p
2
p
3
p
4
ϕ
p
3
p
1
p
4
⋅
ϕ
n
)
{\displaystyle \alpha (F_{p_{1}^{}}F_{p_{2}^{}}F_{p_{3}^{}}F_{p_{4}^{}}\cdot \phi _{p_{1}^{}p_{2}^{}}\phi _{p_{1}^{}p_{3}^{}}\phi _{p_{1}^{}p_{4}^{}}\phi _{p_{2}^{}p_{3}^{}}\phi _{p_{2}^{}p_{4}^{}}\phi _{p_{3}^{}p_{4}^{}}\phi _{p_{1}^{}p_{2}^{}p_{3}^{}}\phi _{p_{1}^{}p_{2}^{}p_{4}^{}}\phi _{p_{2}^{}p_{3}^{}p_{4}^{}}\phi _{p_{3}^{}p_{1}^{}p_{4}^{}}\cdot \phi _{n})}
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{\displaystyle n=(p_{1}^{}p_{2}^{}\cdots p_{k}^{})=[p_{1}^{},p_{2}^{},...,p_{k}^{},n]=\alpha (F_{p_{1}^{}}F_{p_{2}^{}}\cdots F_{p_{k}^{}}\cdot \phi _{n})}