$\min n \in \mathbb{Z}_+$ for which $(5\uparrow\uparrow n) \bmod 2^{32}$ is at a fixed point?

The values $5 \uparrow \uparrow n$ are known as tetrations of $5$ . For compactness, we will write $t_n$ to denote $5 \uparrow \uparrow n$ . Consider first the easier-to-compute sequence of residues $t_n \bmod 2^8$ . The first few values are $t_1 \bmod{2^8} = 5, \ \ \ \ \ t_2 \bmod{2^8} = 53, \ \ \ \ \ t_3 \bmod{2^8} = 117.$ So far, the sequence is not constant, but something important happens between $t_2$ and $t_3$ : $t_3 \equiv t_2 \pmod{64}. \tag{1}$ This matters because $64$ is the multiplicative order of $5$ modulo $2^8$ (i.e., $64$ is the smallest positive integer $m$ that satisfies $5^m \equiv 1 \pmod {2^8}$ ). By an elementary property of modular exponentiation, $a \equiv b \pmod{64}$ if and only if $5^a \equiv 5^b \pmod{2^8}$ . Once we see (1), we know $t_4 \equiv 5^{t_3} \equiv 5^{t_2} \equiv t_3 \pmod{2^8}.$ Of course, this means $t_4 \equiv t_3 \pmod{64}$ as well, so when we continue the exponentiation, we get $t_3 \equiv t_4 \equiv t_5 \equiv t_6 \equiv \cdots \pmod{2^8}.$ To summarize, the residues $t_n \bmod 2^8$ are constant for $n \geq 3$ because $t_{n + 1} \equiv t_n \pmod{64}$ , and $64$ is the multiplicative order of $5$ modulo $2^8$ . The residues become constant starting at $t_3$ because $t_3$ is the first tetration for which a congruence like (1) holds. We can apply this reasoning to help us answer the question posed, but first, here is a useful fact about the multiplicative order of $5$ modulo some power of $2$ .

Lemma. For any integer $c \geq 2$ , the multiplicative order of $5$ modulo $2^c$ is $2^{c - 2}$ .

Proof. When $c = 2$ , the lemma holds since $5 = 1 \cdot 2^2 + 1 \implies 5^{2^0} \equiv 1^1 \equiv 1 \pmod{2^2}.$ Notice that the right side of the first equation has an odd multiple of $2^2$ . By way of induction, suppose the lemma holds for some value $c \geq 2$ , so that $5^{2^{c - 2}} \equiv 1 \pmod{2^c}, \tag{2}$ and $2^{c - 2}$ is the smallest positive exponent for which this congruence holds. Further suppose that, for some odd integer $d$ , $5^{2^{c - 2}} = d \cdot 2^c + 1. \tag{3}$ Notice that $5^{2^{c - 1}} = (5^{2^{c - 2}})^2 = (d \cdot 2^c + 1)^2.$ We can rewrite the right side of this equation to show $5^{2^{c - 1}} = 2^{c + 1}(d^2 \cdot 2^{c - 1} + d) + 1.$ As in the inductive hypothesis, this equation has an odd multiple of $2^{c + 1}$ . It also shows $5^{2^{c - 1}} \equiv 1 \pmod{2^{c + 1}}.$ This means that the order of $5$ is some power of $2$ , say $2^b$ , with $b \leq c - 1$ . But $5^{2^b} \equiv 1 \pmod{2^{c + 1}}$ implies $5^{2^b} \equiv 1 \pmod{2^c}$ . This means $b \not < c - 2$ since otherwise $2^{c - 2}$ would not be the smallest exponent for which (2) holds. Also, $b \neq c - 2$ because in (3) we assumed $d$ was odd, which means $d \cdot 2^c$ is not divisible by $2^{c + 1}$ and thus $5^{2^{c - 2}} \not \equiv 1 \pmod{2^{c + 1}}$ . Therefore, we have shown that the order of $5$ is $2^{c - 1}$ , so the lemma holds for $c + 1$ . By induction, the lemma holds for all $c \geq 2$ .

We have shown that $t_{n + 1} \equiv t_n \pmod{2^8} \ \ \ \ \ \iff \ \ \ \ n \geq 3,$ and by the lemma, $2^8$ is the multiplicative order of $5$ modulo $2^{10}$ . This means $t_{n + 2} \equiv 5^{t_{n + 1}} \equiv 5^{t_n} \equiv t_{n + 1} \pmod{2^{10}} \ \ \ \ \iff \ \ \ \ n \geq 3.$ If we repeat this reasoning twelve times, we obtain $t_{n + 13} \equiv t_{n + 12} \pmod{2^{32}} \ \ \ \ \iff \ \ \ \ n \geq 3,$ or equivalently, $t_{n + 1} \equiv t_n \pmod{2^{32}} \ \ \ \ \iff \ \ \ \ n \geq 15.$ Therefore, we have shown that smallest value $n$ for which $t_n \bmod 2^{32}$ is constant is $\boxed{15}$ . Furthermore, we can say in general that $t_{n + 1} \equiv t_n \pmod{2^{2k}} \ \ \ \ \iff\ \ \ \ n \geq k - 1.$