Repeated sum of square of digits always arrives at $1$ or $89$ I found an astonishing result in a problem today here !
A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.
For example,
$44 \to 32 \to 13 \to 10 \to 1 \to 1$
$85 \to 89 \to 145 \to 42 \to 20 \to 4 \to 16 \to 37 \to 58 \to 89$ 
Therefore any chain that arrives at $1$ or $89$ will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at $1$ or $89$.
Why is this true? 
 A: I suppose you're excluding $0$.
If $x$ has $n$ digits, the sum of squares of the digits of $x$ is at most $81 n$, and that has at most $1 + \log_{10}(81 n)$ digits.  Now $1 + \log_{10}(81 n) < n$ if $n \ge 4$, so eventually you get to a number of at most $3$ digits.
Check all $999$ positive numbers of at most 3 digits, and you'll find that they all end up at $1$ or $89$.
EDIT: Although not so easy to see immediately, it turns out that the sum of squares of the digits of any $3$-digit number is less than the number, so we can restrict our attention to numbers of $2$ or fewer digits.  
A: Let $s(n)$ be the function that adds the square of the digits in $n$.
Note that for $r\geq 3$, and $d_k\in\{0,1,\dots 9\}$ for $k=1,2,\dots, r$ with $d_r\geq 1$, we have that
$$n=d_0+\sum_{k=1}^{r-1} d_k 10^k+d_r10^r >d_0^2+\sum_{k=1}^{r-1} d_k^2+d_r^2=s(n)$$
because $10>d_k$ and $d_0 +d_r10^r\geq 1000\geq 9^2+9^2\geq d_0^2+d_r^2$.
Therefore if $n$ has at least three digits then $n>s(n)$.  This implies that eventually we will obtain a number of at most three digits. So it suffices to check what happens for the integers in $[1,999]$: it can be verified that given $n\in [1,999]$, then $s^{(k)}(n)\in\{1,89\}$ with $k\leq 11$.
Hence, we may conclude that for any $n\geq 1$ the sequence, after a finite number of steps, will enter in one of the two cycles:
$$(1)\quad \mbox{or}\quad (89,145,42,20,4,16,37,58).$$
