which, when multiplied
by itself, yields a number whose decimal
representation, when cut in half down the middle, yields two integers whose sum
is the original number
OK, a simple example should clarify the muddy explanation above: if we multiply 45 by itself, we get 2025; cut 2025 in half to get 20 and 25, whose sum is the number we began with. Similarly, 99 times 99 = 9801, and 98+01 = 99 -- indeed, any integer containing only 9's is Kaprekar.
More generally, an integer is n-Kaprekar if when we split its square just left of its rightmost n digits and add the "halves", we obtain the original integer. Thus 4879 is 5-Kaprekar since 48792 is 23804641 and 238+04641=4879. The set of n-Kaprekar integers is in one-to-one correspondence with the set of unitary divisors of 10n-1.
If instead we work in binary, it turns out that every even perfect number is n-Kaprekar for some n, for example: 1102=100100