3 integers form an arithmetic progression with d=2 i.e. a, a+2, a+4.
The sum of their squares is a 4-digit number divisible by 1111.
(In reply to is there a mistake in the puzzle?
Why did you stop short of 9999, indeed you stopped short of 1111. If the number is to be divisible by 1111 it must be in this higher range.
Posted by Jer
on 2011-12-26 11:29:12