<s>[INST] How to show that $91$ is a pseudoprime to the base $3$? The given problem:

Use Lemma 2.3.3 together with Fermat's little theorem to show that 91
  is a pseudoprime to the base 3.

Lemma 2.3.3.
Let $m_1 \dots m_r \in $ N. If $a \equiv b \pmod {m_i}$, $\forall i =1, \dots, r$, then $a \equiv b \pmod {{\rm lcm}(m_1, \dots, m_r)}$. (If $\gcd(m_i, m_j)=1$ when $i \neq j$, so $a \equiv b \pmod {m_1 \dots m_r}$.) 
Theorem 2.4.5 (Fermat's little theorem II).
Let $p$ prime and $a \in$ Z such that $p \nmid a$. Then
$a^{p-1} \equiv 1 \pmod p$.

My own attempt: 
  Because $91=7 \cdot 13$, the number is composite. 
  According to Theorem 2.4.5 $3^6 \equiv 1 \pmod 7$. On the other hand $90 = 6 \cdot 15 $, so 
  $3^{90} = 3^{6 \cdot 15} = (3^6)^{15} \equiv 1^{15} \equiv 1 \pmod 7$
  Then let's assume according to Theorem 2.4.5 that 
  $3^{12} \equiv 1 \pmod {13}$. On the other hand $90=12 \cdot 7+6$.
  So $3^{90}=3^{12\cdot7+6}=3^{12\cdot7}\cdot 3^6 = (3^{12})^7\cdot3^6 \equiv 1^7 \cdot 3^6 \equiv 1 \pmod {13}$. So according to Lemma 2.3.3 we have $3^{90} \equiv 1 \pmod {91}$ so $3^{91} \equiv 3 \pmod {91}$.
  So 91 would be a pseudoprime to the base 3.

 [/INST] Powers of $3$ cycle as follows
$$
\begin{array}{|c|c|c|}
n&3^n&3^n \mod 91\\
\hline\\
1 & 3 & 3\\
2 & 9 & 9\\
3 & 27 & 27\\
4 & 81 & 81\\
5 & 243 & 61\\
6 & 729 & 1\\
\hline
\end{array}
$$
Therefore, since $90 = 6 \times 15$ 
$$
\begin{align*}
3^{90} &\equiv 1 \hspace{4pt} (\mod 91)\\
\Rightarrow 3^{91} &\equiv 3 \hspace{4pt} (\mod 91)
\end{align*}
$$
</s>