Timeline for Showing that the congruence speed of any integer exponentiation $a^b$ is constant and $\geq 1$ iff $a>1$ is a multiple of $10$

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Jan 25 at 16:01	comment	added	Marco Ripà		Furthermore, if $a \equiv 5 \pmod{10}$, it follows that $a^2 \equiv a^3 \pmod{25}$, $a^2 \equiv a^4 \pmod{25}$, $\ldots$ $a^2 \equiv a^{20 \cdot k + 1} \pmod{25}$.
Jan 25 at 15:58	comment	added	Marco Ripà		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 $\lambda(25)=\phi(25)=20$.
Jan 25 at 14:00	comment	added	JoshuaZ		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.
Jan 25 at 11:31	history	asked	Marco Ripà	CC BY-SA 4.0