Question

Write a Pythagorean triplet whose one member is 

1. 6 
2. 14 
3.  16 
4.  18

Answer 1

For any natural number \(m > 1\), \(2m\), \(m^2- 1\), \(m^2 + 1\) 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 \(2m = 6\)
\(m = 3\)
Therefore, the Pythagorean triplets are \(2 \times 3\), \(3^2- 1\), \(3^2+ 1\) 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 \(2m = 14\)
\(m = 7\)
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 \(2m = 16\)
\(m = 8\)
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 \(2m =18\)
\(m = 9\)
Thus, \(m^2- 1 = 81 - 1 \)= \(80\) and \(m^2+ 1 = 81 + 1\) = \(82\)
Therefore, the Pythagorean triplet is \(18, 80,\) and \(82\).