How do you prove that the series 5, 25, 625, ... can be continued forever to give a 10-adic solution to $n^2=n$? [closed]

Question

How do you prove that the series 5, 25, 625, ... can be continued forever to give a 10-adic solution to $n^2=n$? Here's a proof for a different solution (...1787109376): https://oeis.org/A018248/a018248.pdf

Answer 1

$\mathbb{Z}_{10}\cong \mathbb{Z}_5 \times \mathbb{Z}_2$ by the chinese remainder theorem (taking the limits in $\mathbb{Z}/10^n\cong \mathbb{Z}/5^n \times \mathbb{Z}/2^n$). As $\mathbb{Z}_5$ and $\mathbb{Z}_2$ are integral domains, $x^2=x$ has four solutions in total, namely $(1,1)$, $(0,1)$, $(1,0)$ and $(0,0)$. The one you're referring to corresponds to $(0,1)$ (as $625 = 0$ mod $5^3$ and $1$ mod $2^3$), the one in the link to $(1,0)$. The other two are of course $0$ and $1$. You can also write yours as $1$ minus the one from the link.

Answer 2

If $A - B$ is divisible by $10^n$, and $A$ and $B$ share a common last digit of $5$, then $A^2 - B^2 = (A + B)(A - B)$ is divisible by $10^{n + 1}$ (as $A + B$ is divisible by $10$).

In other words, if $A$ and $B$ have the same last $n$ many digits, and share a common last digit of $5$, then $A^2$ and $B^2$ have the same last $n + 1$ many digits.

From this it inductively follows that in the sequence of values $5, 5^2 = 25, 25^2 = 625, \dotsc$, where each value is the square of the previous one, the last $n$ digits stabilize from the $n$th element onwards (with $5$ as the $1$st element), yielding a 10-adic number which is its own square.

As noted by Achim Krause, you can also understand what is going on as that this sequence is converging to $1$ in the $2$-adics and $0$ in the $5$-adics. (The convergence in the $5$-adics will be much faster, in that of course each squaring doubles the number of $0$s at the end of the base $5$ representation, but the convergence in the $2$-adics will only gain one newly stabilized digit at a time, so that the convergence in the $10$-adics is also by one digit at a time.)