Let $n$ be a $k$ digit integer in base $b$ Then $n$ is said to be a Kaprekar number in base $b$ if $n^2$ has the following property: when you add the number formed by its right hand digits to that formed by its left hand digits, you get $n$

Or to put it algebraically, an integer $n$ such that in a given base $b$ has $$n^2 = \sum_{i = 0}^{k - 1} d_ib^i$$ (where $d_x$ are digits, with $d_0$ the least significant digit and $d_{k - 1}$ the most significant) such that $$\sum_{i = {k \over 2} + 1}^k d_ib^{i - {k \over 2} - 1} + \sum_{i = 1}^{k \over 2} d_ib^{i - 1} = n$$ if $k$ is even or $$\sum_{i = \lceil {k \over 2} \rceil}^k d_ib^{i - \lfloor {k \over 2} \rfloor - 1} + \sum_{i = 1}^{k \over 2} d_ib^{i - 1} = n$$ if $k$ is odd.

$b^x - 1$ for a natural $x$ is always a Kaprekar number in base $b$

Bibliography

1

D. R. Kaprekar, ``On Kaprekar numbers" J. Rec. Math. 13 (1980-1981), 81 - 82.