Question -
Answer -
(i)Let m2 тАУ 1 = 6
[Triplets are in the form 2m, m2 тАУ 1, m2 + 1]
m2 = 6 + 1 = 7
So, the value of m will not be an integer.
Now, let us try for m2 + 1 = 6
тЗТ m2 = 6 тАУ 1 = 5
Also, the value of m will not be an integer.
Now we let 2m = 6 тЗТ m = 3 which is aninteger.
Other members are:
m2 тАУ 1 = 32 тАУ 1 = 8 and m2 + 1 = 32 + 1 = 10
Hence, the required triplets are 6, 8 and 10
(ii)Let m2 тАУ 1 = 14 тЗТ m2 = 1 + 14 = 15
The value of m will not be an integer.
Now take 2m = 14 тЗТ m = 7 which is aninteger.
The member of triplets are 2m = 2 ├Ч 7 = 14
m2 тАУ 1 = (7)2 тАУ 1 = 49 тАУ 1 = 48
and m2 + 1 = (7)2 + 1 = 49 + 1 = 50
i.e., (14, 48, 50)
(iii)Let 2m = 16 m = 8
The required triplets are 2m = 2 ├Ч 8 = 16
m2 тАУ 1 = (8)2 тАУ 1 = 64 тАУ 1 = 63
m2 + 1 = (8)2 + 1 = 64 + 1 = 65
i.e., (16, 63, 65)
(iv)Let 2m = 18 тЗТ m = 9
Required triplets are:
2m = 2 ├Ч 9 = 18
m2 тАУ 1 = (9)2 тАУ 1 = 81 тАУ 1 = 80
and m2 + 1 = (9)2 + 1 = 81 + 1 = 82
i.e., (18, 80, 82)