The problem asks to find a Pythagorean triplet whose smallest member is 8. The general form of the Pythagorean triplet to be used is $2m$, $m^2 - 1$, and $m^2 + 1$.

1. Problem Description

The problem asks to find a Pythagorean triplet whose smallest member is

8. The general form of the Pythagorean triplet to be used is $2m$, $m^2 - 1$, and $m^2 + 1$.

2. Solution Steps

First, the example tries to equate the smallest member, 8, to $m^2 - 1$.

$m^2 - 1 = 8$

$m^2 = 8 + 1 = 9$

$m = 3$

Then, the other two members are found as $2m$ and $m^2 + 1$.

$2m = 2(3) = 6$

$m^2 + 1 = 3^2 + 1 = 9 + 1 = 10$

However, the smallest member is given as 8, and here we have

6. Hence this initial assumption does not work.

Now, we will equate 8 to the term $2m$:

$2m = 8$

$m = 4$

Then $m^2 - 1 = 4^2 - 1 = 16 - 1 = 15$

and $m^2 + 1 = 4^2 + 1 = 16 + 1 = 17$

So the triplet is 8, 15,

1

7. We check if this is a Pythagorean triplet: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$.

3. Final Answer

The Pythagorean triplet whose smallest member is 8 is 8, 15,

1

7.