Theorem: Smallest Strong Fibonacci Pseudoprime of Type I

The smallest strong Fibonacci pseudoprime of type I is $443 \, 372 \, 888 \, 629 \, 441$.

Proof:
Let $N := 443 \, 372 \, 888 \, 629 \, 441$, to save writing it out in full each time.
From the definition of a strong Fibonacci pseudoprime of type I:
{{:Definition:Strong Fibonacci Pseudoprime of Type I}}
We have that:
:$N = 17 \times 31 \times 41 \times 43 \times 89 \times 97 \times 167 \times 331$
Of these, we see that:
{{begin-eqn}}
{{eqn | l = 17
      | r = 4 \times 4 + 1
      | c = which is a prime number of the form $4 n + 1$
}}
{{eqn | l = 31
      | r = 4 \times 8 - 1
      | c = which is a prime number of the form $4 n - 1$
}}
{{eqn | l = 41
      | r = 4 \times 10 + 1
      | c = which is a prime number of the form $4 n + 1$
}}
{{eqn | l = 43
      | r = 4 \times 11 - 1
      | c = which is a prime number of the form $4 n - 1$
}}
{{eqn | l = 89
      | r = 4 \times 22 + 1
      | c = which is a prime number of the form $4 n + 1$
}}
{{eqn | l = 97
      | r = 4 \times 24 + 1
      | c = which is a prime number of the form $4 n + 1$
}}
{{eqn | l = 167
      | r = 4 \times 42 - 1
      | c = which is a prime number of the form $4 n - 1$
}}
{{eqn | l = 331
      | r = 4 \times 83 - 1
      | c = which is a prime number of the form $4 n - 1$
}}
{{end-eqn}}
Thus there are $4$ (an even number) of prime factors of $N$ of the form $4 n - 1$.
We note that:
{{begin-eqn}}
{{eqn | l = N - 1
      | r = 288 \times 1 \, 539 \, 489 \, 196 \, 630
      | c = and so $\paren {17^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 960 \times 461 \, 846 \, 758 \, 989
      | c = and so $\paren {31^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 1680 \times 263 \, 912 \, 433 \, 708
      | c = and so $\paren {41^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 1848 \times 239 \, 920 \, 394 \, 280
      | c = and so $\paren {43^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 7920 \times 55 \, 981 \, 425 \, 332
      | c = and so $\paren {89^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 9408 \times 47 \, 127 \, 220 \, 305
      | c = and so $\paren {97^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 27 \, 888 \times 15 \, 898 \, 339 \, 380
      | c = and so $\paren {167^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 109 \, 560 \times 4 \, 046 \, 850 \, 024
      | c = and so $\paren {331^2 - 1} \divides \paren {N - 1}$
}}
{{end-eqn}}
Thus for all $p \divides N$, we have that $\paren {p^2 - 1} \divides \paren {N - 1}$.
From Difference of Two Squares we have that:
:$p^2 - 1 = \paren {p + 1} \paren {p - 1}$
and so for all $p \divides N$:
:$\paren {p - 1} \divides \paren {N - 1}$
We also have that $N$ is square-free.
Thus by Korselt's Theorem, $N$ is a Carmichael number.
Now we also have that:
{{begin-eqn}}
{{eqn | l = \paren {p^2 - 1}
      | o = \divides
      | r = \paren {N - 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {p + 1} \paren {p - 1}
      | o = \divides
      | r = \paren {N - 1}
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = 2 \paren {p + 1} \paren {\frac {p - 1} 2}
      | o = \divides
      | r = \paren {N - 1}
      | c = as $p - 1$ is even
}}
{{eqn | ll= \leadsto
      | l = 2 \paren {p + 1}
      | o = \divides
      | r = \paren {N - 1}
      | c = 
}}
{{end-eqn}}
Thus we have that:
:$\forall p \divides N: 2 \paren {p + 1} \divides \paren {N - 1}$
This is actually stronger than the conditions for which $N$ is a strong Fibonacci pseudoprime of type I:
{{:Definition:Strong Fibonacci Pseudoprime of Type I}}
It can be established by an exhaustive search that there are no smaller Carmichael numbers with this property.
Hence the result.
{{qed}}