Answer to March 2012 Problem of the Month

Solution to the Problem:

The next pair of numbers possessing this quality is 102 and 201
(their squares are 10,404 and 40,401).

However, James Alarie sent in the following solution which adds
some completeness!!

I guess it's a bit too obvious to list (21,12) as the "next" pair; it
already exists as the (12,21) pair.

Those less than 100 are:
(1,1); (2,2); (3,3); (11,11); (12,21); (13,31); (21,12); (22,22);
(31,13); (101,101); (102,201); (103,301); (111,111); (112,211);
(113,311); (121,121); (122,221); (201,102); (202,202); (211,112);
(212,212); (221,122); (301,103); (311,113).

The non-reversed ones are:
(1,1); (2,2); (3,3); (11,11); (12,21); (13,31); (22,22); (101,101);
(102,201); (103,301); (111,111); (112,211); (113,311); (121,121);
(122,221); (201,102); (202,202); (212,212).

More up to 1000:
(1002,2001); (1003,3001); (1011,1101); (1012,2101); (1013,3101);
(1021,1201); (1022,2201); (1031,1301); (1102,2011); (1103,3011);
(1112,2111); (1113,3111); (1121,1211); (1122,2211); (1202,2021);
(1212,2121); (2012,2102); (2022,2202).

And Chad Fore also noticed the repetitions and that only the
digits 0, 1, 2, 3 work:

Without repetition of digits, the next pair I came up with is 102 and 201, it also
works for 103 and 301. I see it now. The last digit that it works for is 3 because
3 squared is the largest square that is a single digit. So only 2 and 3 work because
both of their squares are in the digits.

So it continues for any repetition of 1, 2 and 1, 3 separated by n zeros where n is a
whole number. So it would conceivably work for a very large number such as 100000000002
and 200000000001.   That is very cool.

So, thanks to James and Chad for seeing more in the problem than I did!