# Smallest Strong Fibonacci Pseudoprime of Type I

## Theorem

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:

A **strong Fibonacci pseudoprime of type I** is a Carmichael number $N = \displaystyle \prod p_i$ such that an even number of the prime factors $p_i$ are of the form $4 m - 1$ where:

\(\text {(1)}: \quad\) | \(\displaystyle 2 \paren {p_i + 1}\) | \(\divides\) | \(\displaystyle \paren {N - 1}\) | for those $p_i$ of the form $4 m - 1$ | |||||||||

\(\text {(2)}: \quad\) | \(\displaystyle \paren {p_i + 1}\) | \(\divides\) | \(\displaystyle \paren {N \pm 1}\) | for those $p_i$ of the form $4 m + 1$ |

We have that:

- $N = 17 \times 31 \times 41 \times 43 \times 89 \times 97 \times 167 \times 331$

Of these, we see that:

\(\displaystyle 17\) | \(=\) | \(\displaystyle 4 \times 4 + 1\) | which is a prime number of the form $4 n + 1$ | ||||||||||

\(\displaystyle 31\) | \(=\) | \(\displaystyle 4 \times 8 - 1\) | which is a prime number of the form $4 n - 1$ | ||||||||||

\(\displaystyle 41\) | \(=\) | \(\displaystyle 4 \times 10 + 1\) | which is a prime number of the form $4 n + 1$ | ||||||||||

\(\displaystyle 43\) | \(=\) | \(\displaystyle 4 \times 11 - 1\) | which is a prime number of the form $4 n - 1$ | ||||||||||

\(\displaystyle 89\) | \(=\) | \(\displaystyle 4 \times 22 + 1\) | which is a prime number of the form $4 n + 1$ | ||||||||||

\(\displaystyle 97\) | \(=\) | \(\displaystyle 4 \times 24 + 1\) | which is a prime number of the form $4 n + 1$ | ||||||||||

\(\displaystyle 167\) | \(=\) | \(\displaystyle 4 \times 42 - 1\) | which is a prime number of the form $4 n - 1$ | ||||||||||

\(\displaystyle 331\) | \(=\) | \(\displaystyle 4 \times 83 - 1\) | which is a prime number of the form $4 n - 1$ |

Thus there are $4$ (an even number) of prime factors of $N$ of the form $4 n - 1$.

We note that:

\(\displaystyle N - 1\) | \(=\) | \(\displaystyle 288 \times 1 \, 539 \, 489 \, 196 \, 630\) | and so $\paren {17^2 - 1} \divides \paren {N - 1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 960 \times 461 \, 846 \, 758 \, 989\) | and so $\paren {31^2 - 1} \divides \paren {N - 1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1680 \times 263 \, 912 \, 433 \, 708\) | and so $\paren {41^2 - 1} \divides \paren {N - 1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1848 \times 239 \, 920 \, 394 \, 280\) | and so $\paren {43^2 - 1} \divides \paren {N - 1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 7920 \times 55 \, 981 \, 425 \, 332\) | and so $\paren {89^2 - 1} \divides \paren {N - 1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 9408 \times 47 \, 127 \, 220 \, 305\) | and so $\paren {97^2 - 1} \divides \paren {N - 1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 27 \, 888 \times 15 \, 898 \, 339 \, 380\) | and so $\paren {167^2 - 1} \divides \paren {N - 1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 109 \, 560 \times 4 \, 046 \, 850 \, 024\) | and so $\paren {331^2 - 1} \divides \paren {N - 1}$ |

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:

\(\displaystyle \paren {p^2 - 1}\) | \(\divides\) | \(\displaystyle \paren {N - 1}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {p + 1} \paren {p - 1}\) | \(\divides\) | \(\displaystyle \paren {N - 1}\) | Difference of Two Squares | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 2 \paren {p + 1} \paren {\frac {p - 1} 2}\) | \(\divides\) | \(\displaystyle \paren {N - 1}\) | as $p - 1$ is even | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 2 \paren {p + 1}\) | \(\divides\) | \(\displaystyle \paren {N - 1}\) |

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:

A **strong Fibonacci pseudoprime of type I** is a Carmichael number $N = \displaystyle \prod p_i$ such that an even number of the prime factors $p_i$ are of the form $4 m - 1$ where:

\(\text {(1)}: \quad\) | \(\displaystyle 2 \paren {p_i + 1}\) | \(\divides\) | \(\displaystyle \paren {N - 1}\) | for those $p_i$ of the form $4 m - 1$ | |||||||||

\(\text {(2)}: \quad\) | \(\displaystyle \paren {p_i + 1}\) | \(\divides\) | \(\displaystyle \paren {N \pm 1}\) | for those $p_i$ of the form $4 m + 1$ |

It can be established by an exhaustive search that there are no smaller Carmichael numbers with this property.

Hence the result.

$\blacksquare$

## Sources

- Jul. 1993: R.G.E. Pinch:
*The Carmichael Numbers up to $10^{15}$*(*Math. Comp.***Vol. 61**,*no. 203*: pp. 381 – 391) www.jstor.org/stable/2152963

- 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $443,372,888,629,441$