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.

By Chinese Remainder Theorem, this is equivalent to checking whether:
 * $a^{90} \equiv 1 \pmod 7$

and:
 * $a^{90} \equiv 1 \pmod {13}$

are both satisfied.

If $a$ is a multiple of $7$ or $13$, $a^{90} \not \equiv 1 \pmod {91}$.

Therefore we consider $a$ not divisible by $7$ or $13$.

By Fermat's Little Theorem, we have:
 * $a^6 \equiv 1 \pmod 7$

and thus:


 * $a^{90} \equiv 1^{15} \equiv 1 \pmod 7$

Now by Fermat's Little Theorem again:


 * $a^{12} \equiv 1 \pmod {13}$

and thus:


 * $a^{90} \equiv a^6 \paren{1^7} \equiv a^6 \pmod {13}$

We have:

and thus $a$ must be equivalent to $1, 3, 4, 9, 10, 12 \pmod {13}$.

This gives $1$ and the $35$ bases less than $91$ listed above.