Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14

(iii) 16 (iv) 18

For any natural number *m* >
1, 2*m*, *m*^{2} − 1, *m*^{2} +
1 forms a Pythagorean triplet.

(i) If we take *m*^{2}
+ 1 = 6, then *m*^{2} = 5

The
value of *m* will not be an integer.

If
we take *m*^{2} − 1 = 6, then *m*^{2}
= 7

Again
the value of *m* is not an integer.

Let
2*m* = 6

*m*
= 3

Therefore,
the Pythagorean triplets are 2 × 3, 3^{2} − 1, 3^{2}
+ 1 or 6, 8, and 10.

(ii) If we take *m*^{2}
+ 1 = 14, then *m*^{2} = 13

The
value of *m* will not be an integer.

If
we take *m*^{2} − 1 = 14, then *m*^{2}
= 15

Again
the value of *m* is not an integer.

Let
2*m *= 14

*m*
= 7

Thus,
*m*^{2} − 1 = 49 − 1 = 48 and *m*^{2}
+ 1 = 49 + 1 = 50

Therefore, the required triplet is 14, 48, and 50.

(iii) If we take *m*^{2}
+ 1 = 16, then *m*^{2} = 15

The
value of *m* will not be an integer.

If
we take *m*^{2} − 1= 16, then *m*^{2}
= 17

Again
the value of *m* is not an integer.

Let
2*m* = 16

*m
*= 8

Thus,
*m*^{2} − 1 = 64 − 1 = 63 and *m*^{2}
+ 1 = 64 + 1 = 65

Therefore, the Pythagorean triplet is 16, 63, and 65.

(iv) If we take *m*^{2}
+ 1 = 18,

*m*^{2}
= 17

The
value of *m* will not be an integer.

If
we take *m*^{2} − 1 = 18, then *m*^{2}
= 19

Again
the value of *m* is not an integer.

Let
2*m *=18

*m
*= 9

Thus,
*m*^{2} − 1 = 81 − 1 = 80 and *m*^{2}
+ 1 = 81 + 1 = 82

Therefore, the Pythagorean triplet is 18, 80, and 82.

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