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Years ago, I defined the "congruence speed" (radix-$10$) of the integer tetration $^{b}a$ as $V(a,b)$, which is the number of the new(!) rightmost digits that freeze when we move from $b \in \mathbb{Z}^+$ to $b+1$ (i.e., if exactly $d$ digits are frozen at height $b-1$, then $V(a,b) \in \mathbb{N}_0$ is such that $^{b}a \equiv \hspace{1mm} ^{b+1}a \pmod {10^{V(a,b)+d}} \wedge ^{b}a \not\equiv \hspace{1mm} ^{b+1}a \pmod {10^{V(a,b)+d+1}}$).
Then, I defined the "constant congruence speed" of tetration as $V(a)$, which corresponds to the non-negative integer number of the new rightmost digits that freeze when we move from a hyperexponent $\overline{b}:=a+1$ to $\overline{b}+c$, for every $c \in \mathbb{Z}^+$ (as long as $a$ is not a multiple of $10$, the congruence speed of $a$ does not depend on $b$ anymore since $\overline{b}=a+1$ represents a sufficient condition for the above).

Now, I observe that we can do roughly the same thing by taking into account exponentiation, instead of tetration. I mean, let me indicate as $V(a)$^ this (new) constant congruence speed of $a^b$; then we know that if $a$ is a multiple of $10$, $a=10 \cdot k, k \in \mathbb{Z}^+ : k \not\equiv 0 \pmod {10}$, and thus $V(a)$^$=k$ (trivial).

Question: Can we simply state that $V(a)$^$=0$ otherwise by observing that $\left(a^2 \equiv \hspace{1mm} a^3 \pmod {10^d} \wedge a^2 \not\equiv \hspace{1mm} a^3 \pmod {10^{d+1}}\right) \Rightarrow a^3 \not\equiv \hspace{1mm} a^4 \pmod {10^{d+1}}$ holds for any $a \geq 2$ which is not a multiple of $10$ (since the value of $V(1)$ is also trivial to be defined by its own)?

As an example, let me take $a=5$ and so $V(5)$^$=0$ since $5^2 \equiv 5^3 \pmod{10^2} \Rightarrow d(5,2) \geq 2$ and $5^3 \not\equiv 5^4 \pmod {10^3}$.

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  • $\begingroup$ This is likely very closely related to the fact that that $\phi(5)$ is just a power of 2. My guess is that the same behavior will happen for any base $b$ where $b$ is a power of 2 times a Fermat prime. Euler's theorem on exponentiation will be relevant here. $\endgroup$
    – JoshuaZ
    Commented Jan 25 at 14:00
  • $\begingroup$ Of course, it is... I mean, for the sake of simplicity, let us consider the decimal numeral system and assume $a \in \mathbb{Z}^+$. Then, if $a$ is not a multiple of $10$, $a^{20 \cdot k+1} \equiv a \pmod {25}$ holds for every $k \in \mathbb{N}_0$ (the proof is very simple and I wrote it in a previous paper)... it basically considers the fact that $\lambda(25)=\phi(25)=20$. $\endgroup$
    – Marco Ripà
    Commented Jan 25 at 15:58
  • $\begingroup$ Furthermore, if $a \equiv 5 \pmod{10}$, it follows that $a^2 \equiv a^3 \pmod{25}$, $a^2 \equiv a^4 \pmod{25}$, $\ldots$ $a^2 \equiv a^{20 \cdot k + 1} \pmod{25}$. $\endgroup$
    – Marco Ripà
    Commented Jan 25 at 16:01

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