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.
For simplicity call the numbers a-2, a, a+2
Then the sum of squares becomes 3a^2 + 8
We are seeking a solution to 3a^2 + 8 = 1111n where n is a single digit.
Solving for a:
a = ±√((1111n-8)/3)
and trying the 9 possibilities of n finds the integer solution
So the triplets are
(41,43,45) and (-45,-43,-41)
Posted by Jer
on 2011-12-26 11:25:28