Interesting property of idempotent 10-adic number

[Happiness]

We build a special number from $5$, by squaring it, appending the next digit of the square to it and repeating the steps.

$5^2=25.$ The next digit is 2, which is added to 5 to give 25.

$25^2=625.$ The next digit is 6, which is added to 25 to give 625.

$625^2=390625.$ The next digit is 0, which is added to 625 to give 0625.

$0625^2=390625.$ The next digit is 9, which is added to 0625 to give 90625.

$90625^2=8212890625.$ The next digit is 8, which is added to 90625 to give 890625.

We end up with the special number $F=\,...\,4106619977392256259918212890625.$

Why is it that the number in every step shares the same digits as its square? In other words, prove the existence of this number.

Source: https://divisbyzero.com/2008/12/29/a-10-adic-number-that-is-a-zero-divisor/
(The article mentioned that the digit 6 can be used. But I don't think so because $36\times36=1296$, which doesn't end with 36.)

 

[micromass]

The number seems to be the limit $L=\lim_{n\rightarrow +\infty} 5^{2^n}$. We see that
$$L^2 = \lim_{n\rightarrow +\infty} (5^{2^n})^2 = \lim_{n\rightarrow +\infty} 5^{2^{n+1}} = L$$
So $L^2 = L$ meaning that your number is $L=0$ or $L = 1$.

We see that
$$|5^{2^n}|_{10} = 1~\text{and}~|5^{2^n} - 1|_{10} = 1$$
so the sequence doesn't converge. It does converge in the $5$-adic numbers:

$$|5^{2^n}|_5 = 2^{-n}$$
So your number equals $0$ in the $5$-adic numbers.

EDIT: the number does exist in the $10$-adic numbers. The $10$-adics are not an integral domain so $L^2 = L~\Rightarrow~L\in \{0,1\}$ is not valid. The number doesn't seem to be a rational number though

 

[Happiness]

The number seems to be the limit $L=\lim_{n\rightarrow +\infty} 5^{2^n}$. We see that
$$L^2 = \lim_{n\rightarrow +\infty} (5^{2^n})^2 = \lim_{n\rightarrow +\infty} 5^{2^{n+1}} = L$$
So $L^2 = L$ meaning that your number is $L=0$ or $L = 1$.

We see that
$$|5^{2^n}|_{10} = 1~\text{and}~|5^{2^n} - 1|_{10} = 1$$
so the sequence doesn't converge. It does converge in the $5$-adic numbers:

$$|5^{2^n}|_5 = 2^{-n}$$
So your number equals $0$ in the $5$-adic numbers.

EDIT: the number does exist in the $10$-adic numbers. The $10$-adics are not an integral domain so $L^2 = L~\Rightarrow~L\in \{0,1\}$ is not valid. The number doesn't seem to be a rational number though

I am not familiar with p-adic numbers. Could this be proved without using p-adic numbers?

 

[Happiness]

The number seems to be the limit $L=\lim_{n\rightarrow +\infty} 5^{2^n}$.

$625^2=390625$ but $F$ ends with $890625$. So I think $F$ may not be the limit $L$.

 

[micromass]

$625^2=390625$ but $F$ ends with $890625$. So I think $F$ may not be the limit $L$.

Yes, you're right. I misunderstood the OP.

 

[micromass]

OK, so let $a_n$ be the number formed by the first $n$ digits of $F$. We know by construction that the $nth$ digit of $a_n$ is the $n$th digit of $a_{n-1}^2$. We prove by induction that the first $n$ digits of $a_n^2$ equal $a_n$.

We can obviously write $a_n = x 10^n + a_{n-1}$. Squaring gives us
$$a_n^2 = x^2 10^{2n} + 2x10^n a_{n-1} + a_{n-1}^2$$
We are interested in the first $n$ digits of this number. The number $x^2 10^{2n}$ has a zero on its first $2n$ digits, so it plays no role. The first $n-1$ digits of $2x10^n a_{n-1}$ are zero. The $n$th digit of this number is the first digit of $2xa_{n-1}$. But since the first digit of $a_{n-1}$ is $5$, we get that this first digit is the first digit of $2\cdot 5\cdot x$, that is: the first digit is $0$.

So the first $n$ digits of $a_n^2$ are the first $n$ digits of $a_{n-1}^2$, like we wanted.

 

[Happiness]

We can obviously write $a_n = x 10^n + a_{n-1}$.

$a_2=25=2(10)+5=2(10^1)+a_1$.

So $a_n = x 10^{n-1} + a_{n-1}$.

Right?

 

[micromass]

Right sorry, I should have proof read it. But the proof still goes through modulo these notational errors.

 

[micromass]

Using the same proof, I discovered a similar number:

$$2^5 = \mathbf{32},~32^5 = 33554\mathbf{432},~432^5 = 1504591950\mathbf{6432}, ...$$

 

[mfb]

The proof needs a slight modification, and relies on the ending number being 5 to make 2x*5 a multiple of 10.
That also explains why it works with 5 but not with 6.

The fifth power, starting with 2, also works because then 5x*2 is a multiple of 10. Apart from higher powers (^10, ^15, ... with 2, or ^4, ^6, ... with 5) that should be all we have in the decimal system. Other bases lead to similar results. Start with a divisor in this system, then use a power such that starting number * power is a multiple of the base.