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 $82$.