91 is Pseudoprime to 35 Bases less than 91

Theorem

$91$ is a Fermat pseudoprime in $35$ bases less than itself:

$3, 4, 9, 10, 12, 16, 17, 22, 23, 25, 27, 29, 30, 36, 38, 40, 43, 48, 51, 53, 55, 61, 62, 64, 66, 68, 69, 74, 75, 79, 81, 82, 87, 88, 90$

Proof

By definition of a Fermat pseudoprime, we need to check for $a < 91$:

$a^{90} \equiv 1 \pmod {91}$

is satisfied or not.

Note that:

\(\displaystyle \paren {91 - a}^{90}\)

\(=\)

\(\displaystyle {-a}^{90}\)

Congruence of Powers

\(\displaystyle \)

\(=\)

\(\displaystyle a^{90}\)

So we essentially need to check $2 \le a \le 45$ only.

Since $91 = 7 \times 13$, we can also exclude all multiples of $7$ and $13$.

Historical Note

This result is attributed by David Wells in his $1997$ work Curious and Interesting Numbers, 2nd ed. to Tiger Redman, but no corroboration can be found for this on the internet.

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $91$