# 2.Write a Pythagorean triplet whose one member is (i) 6 (ii) 14 (iii) 16 (iv) 18

For a natural number m greater than 1,Triplets are in the form 2m, m2 – 1, m2 + 1
(i) Let $$m^2 – 1 = 6$$

$$\Rightarrow m^2 = 6 + 1 = 7$$

So,here we observe that value of m is not integer.

Now, let us try for $$m^2 + 1 = 6$$

$$\Rightarrow m^2 = 6 – 1 = 5$$

So, here too we have no integer value of m.

Now, let 2m = 6

$$\Rightarrow m = 3\; which\; is \;an\; integer.$$

Other number are:

$$m^2 – 1 = 32 – 1 = 8$$ and $$m^2 + 1 = 32 + 1 = 10$$

Hence, the required triplets are 6, 8 and 10.

(ii) Let $$m^2 – 1 = 14$$

$$\Rightarrow m^2 = 1 + 14 = 15$$

So, we observe that the value of m will not be an integer.

Now take 2m = 14

$$\Rightarrow$$ m = 7 which is an integer.

The other numbers of triplets are 2m =$$2 \times 7$$ = 14

$$m^2 – 1 = (7)2 – 1 = 49 – 1 = 48$$

and $$m^2 + 1 = (7)2 + 1 = 49 + 1 = 50$$

Hence the triplets are 14, 48 and 50.

(iii) Let, 2m = 16

$$\Rightarrow$$m = 8

The other numbers are

$$m^2 – 1 = (8)2 – 1 = 64 – 1 = 63$$

$$m^2 + 1 = (8)2 + 1 = 64 + 1 = 65$$

Hence the triplets are 16, 63 and 65

(iv) Let 2m = 18

$$\Rightarrow$$ m = 9

The other triplets are:

$$m^2 – 1 = (9)2 – 1 = 81 – 1 = 80$$

and $$m^2 + 1 = (9)2 + 1 = 81 + 1 = 82$$

Hence we have the triplets as 18, 80 and 82.