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}}