Exercise 2.3.37* (Description of the sequence $a_1 = 3, a_{i+1} = 3^{a_i}$ modulo $100$)

Question

Let $a_1 = 3, a_{i+1} = 3^{a_i}$ Describe this sequence (mod $100$).

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
By Problem 36, the multiplicative order of $3$ modulo $100$ is a divisor of $40$.
In fact  $3^{20} = 3486784401 \equiv 1 \pmod {100}$.

Then 
\begin{align*}
a_1 &= 3, \
a_2 &= 3^3 = 27,\
a_3 &= 3^{27} \equiv 3^7 \equiv 87 \pmod {100}\
\end{align*}
A fortiori, since $20 \mid 100$,   $a_3 \equiv 87 \equiv 7 \pmod{20}$, so $a_3 = 7 + 20k,\ k\in \N$. Therefore
$$a_4 = 3^{a_3} = (3^{20})^k\cdot 3^7\equiv 3^7 \equiv 87 \pmod{100}.$$

Suppose that $a_i \equiv 87 \pmod {100}$ for some $i \geq 3$.

Then $a_i \equiv 87 \equiv 7 \pmod {20}$, therefore
$$a_{i+1} = 3^{a_i} \equiv 3^{7} \equiv 87 \pmod {100}.$$

This induction shows that $a_i \equiv 87 \pmod {100}$ for all $i\geq 3$.
\begin{align*}
&a_1 \equiv 3 \pmod{100}, a_2 \equiv 27 \pmod{100},\
&\forall i\geq 3,\ a_i \equiv 87 \pmod{100}.
\end{align*}
The sequence $([a_i]_{100}])$ is constant from rank $3$.
\end{proof}

Exercise 2.3.37* (Description of the sequence $a_1 = 3, a_{i+1} = 3^{a_i}$ modulo $100$)

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Exercise 2.3.37* (Description of the sequence $a_1 = 3, a_{i+1} = 3^{a_i}$ modulo $100$)

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