6, 7, 8
Detailed Answer:
Assume the numbers are: n-1, n, n+1 ...... (1)
then, (n-1)2 + n2 + (n+1)2 = 149
3n2 + 2 = 149
3n2 = 147
n2 = 49
n = 7 (or -7 but not valid)
substitute in eq 1
the answer is as above 6, 7, 8
The numbers are 12 and 14.
12 and 13
The numbers are 8 and 9.
6,7,8 those numbers squared are 36,49,64 which add to 149
Consecutive numbers are whole numbers whose difference is 1.
The numbers are 12 and 14.
The numbers are 12 and 14.
8
The numbers are 13 and 14.
15 and 16
17 and 18
62 and 63
12 and 13
12 and 13 (and you may note that 12+13 = 25)
The numbers are 8 and 9.
6,7,8 those numbers squared are 36,49,64 which add to 149
The difference between the squares of two consecutive integers j and j+1 is |2j+1|. There are therefore two such pairs where this quantity is 17:-9 and -88 and 9