What happens if you just keep squaring?
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it as a friendly introduction to p-adic numbers, both when answering and when asking questions about its claims such as ...999 = -1 and 10-adic idempotents.
Recommended by 1 people across 1 different months, 2023 to 2025.
What it is
A Veritasium video by Derek Muller and mathematician Alex Kontorovich that introduces p-adic numbers, starting from the idea of repeatedly squaring numbers like 5 and watching digits stabilise. It shows how ...999 can equal -1 in 10-adics, then uses p-adics to sketch how local-to-global reasoning solves Diophantine equations.
What people said
Recommends user700480 · 2023-06 · Mathematics Stack Exchange · on “$10$ adic numbers Does infinite equal $-1$”
In fact, I believe you are talking about [this video] (brilliant video by Veritasium BTW) - re-watch somewhere around 26:31, where (on the example of $3$-adics) they explain that those numbers _don't belong to a number line_ but actually make up an intricate fractal-like structure.
Cites Michal · 2023-06 · Mathematics Stack Exchange · on “How to prove that specific value (...5) is infinite in 10-adic”
I have watched: Veritasium [this video] The idea behind is that there exist value that satisfies: $n^2 = n$ . That value is its own square. We can create this value by: $$ 5^2 = 25 $$ $$ 25 ^ 2 = 625 $$ $$ 625^2 = 390625 $$ $$ 0625^2 = ... $$ $$ 90625^2 = 8212890625 $$ How we can prove that we can go creating this value for indefinitely ?
Cites Cincinnatus · 2023-07 · Mathematics Stack Exchange · on “are p-adic values infinite if using a limit?”
I watched a Veritasium video [this video] about p-adic numbers. In 2:30 [this video] they take the 10-adic number ...857142857143 , they multiply it by 7 and it happens that this number is designed in a way that all products except the first one result in a number ending in 0 and carrying some value. Since the 10-adic number has infinite digits, this results in a 1 with infinite zeros on the left, so in the video they then claim that this equals 1, and therefore this 10-adic number represents the number 1/7 ...857142857143 x 7 ____________ ...000000000001 Saying that the result is 1 makes some kind of logic (we have a non ending amount of zeros on the left, so what else could it be?) However there's a
Cites Inquisitive · 2023-10 · Mathematics Stack Exchange · on “Does ...999.999... = 0?”
Does the number $...999.999... = 0$ ? My reasoning for asking this is outlined below. First, for base 10 it can be shown [this video] that $...999 = -1$ . In addition it is common knowledge [LINK] that $0.999... = 1$ . This seems to imply that $...999.999... = (-1 + 1) = 0$ . This appears bizarre at first, but it seems that $...999.999...$ has the zero element [LINK] properties of being both an additive identity ( $...999.999... + x = x$ ) and an absorbing element ( $(...999.999...)\cdot x = ...999.999...$ ) with base 10 decimal numbers. Things get stranger. Dividing $...999.999...$ by 9 yields the number $...111.111...$ . If the logic above is correct, this would imply that $...111.111... = 0$ a
Cites Kurt · 2025-06 · Mathematics Stack Exchange · on “$p$-adic solution to $x^2 + x^4 + x^8 = y^2$”
So I found a rather introductory video of $p$ -adic numbers of Veritasium that you could find here [this video] . The video included an outline of a solution to the Diophantine equation $x^2 + x^4 + x^8 = y^2$ . However, I found the solution there very incomplete. It is true that we could "downgrade" many terms into linear modular equations, but at the inductive step one will usually get stuck. Here is my process following the solution. We work in $3$ -adic numbers (really, $3$ -adic integers). Write $x = \ldots x_2x_1x_{0(3)}$ and $y = \ldots y_2y_1y_{0(3)}$ and we first notice $y_0 = 0$ (no matter what $3$ -adic integer $x$ is $x^2 + x^4 + x^8 \equiv 0 \bmod 3$ , so $y \equiv 0 \bmod 3$ and thu