Square and reversed integer For all $n=\overline{a_k a_{k-1}\ldots a_1 a_0} := \sum_{i=0}^k a_i 10^i\in \mathbb{N}$, where $a_i \in \{0,...,9\}$ and $a_k \neq 0$,
we define $f(n)=\overline{a_0 a_1 \ldots a_{k-1} a_k}= \sum_{i=0}^k a_{k-i}10^i$.
Is it true that, for all $m=\overline{a_k a_{k-1}\ldots a_1 a_0} \in \mathbb{N}$, we have
$f(m\times m)=f(m)\times f(m) \implies$$\forall i \in \{0, \ldots, k\}, a_i \in \{0,1,2,3\}$ ?
Example: $f(201)\times f(201)=102 \times 102=10404$
$=f(40401)=f(201\times 201)$.
It's true for $m \leq 10^8$.
Thanks in advance.
 A: At OEIS we have "Skinny numbers: numbers $n$ such that there are no carries when $n$ is squared by long multiplication." It gives as an equivalent formulation, $R(n^2) = R(n)^2$, where $R(n)$ means the digit reversal of $n$, and it says, "The decimal expansion of a skinny number $n$ may contain only 0's, 1's, 2's and 3's." However, it niether gives nor cites a proof. 
A: Way back when I was in school, I had investigated number with these properties. These numbers are a special case of a more general property of numbers satisfying $R(mn)  =R(m)R(n)$ where $R(n)$ means the digits reversal of $n$.
Examples
12*13 = 156, 21*31 = 651
101*102 = 10302, 101*103 = 20301
12012*11212 = 134678544, 21021*21211 = 445876431
11013*10212 = 112464756, 31011*21201 = 657464211
Infinitely many such numbers can be constructed but as Gerry Myerson said, the digits will be 0's, 1's, 2's or 3's.
A more challenging problem would be to study numbers with the property that 
$R(n_1 n_2 \ldots n_k)  =R(n_1)R(n_2)\ldots R(n_k)$.
Some interesting question in this direction:


*

*What is the largest $k$ for which we will find a solution?

*What is the largest $k$ for which we will find a solution if all $n_i$'s are equal? i.e. what is the largest $k$ for which $R(n^k)  =R(n)^k$ has a solution?
For some reason I stopped at $k=2$. May be someone might want to revisit the problem.
