— Burt Kaliski Jr. (@modulomathy) September 3, 2013
The three ways are
The third way is an interesting one because the integers are consecutive. If we let x denote the middle integer, then the sum of the squares would be
(x-1)2+x2+(x+1)2 = x2-2x+1+x2+x2+2x+1 = 3x2+2 .
Any number that is two more than three times a square can thus be expressed as the sum of squares of three consecutive integers. To meet the constraint that the integers must be positive, the number must be at least 12+22+32 = 14. (The relationship also holds for the numbers 3 and 5, if non-positive integers are allowed.)
Note: If we relaxed the constraint that the integers must be positive in the present problem, one more way would be possible: 72+142. This is a nice consequence of the fact that 245 is the product of a sum of squares, 5, and a square, 49.