Explanation:
By Pythagoras' Theorem,
![{c}^(2) = {a}^(2) + {b}^(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/mai1vdvhlepgkxrrqqwut73wx0o6m5wnep.png)
where c is always the largest number.
a and b can be interchangeable between the 2nd largest and the 3rd largest numbers.
Given a = 8, b = 15 and c = 17,
![{a}^(2) + b {}^(2) = {8}^(2) + {15}^(2) \\ = 64 + 225 \\ = 289 \\ \\ {c}^(2) = {17}^(2) \\ = 289](https://img.qammunity.org/2022/formulas/mathematics/college/bxrpshmbr6bbzlq8rp6avuxeitkupx8weq.png)
Since c^2 = a^2 + b^2 , 8 , 15 and 17 are pythagorean triplets.
Now let's move on to 9, 40 and 41.
![{a}^(2) + {b}^(2) = {9}^(2) + {40}^(2) \\ = 81 + 1600 \\ = 1681 \\ \\ {c}^(2) = {41}^(2) \\ = 1681](https://img.qammunity.org/2022/formulas/mathematics/college/l7ftvld8emxudwui03u6ybs2gbjwb58d3l.png)
Since c^2 = a^2 + b^2 , 9 , 40 and 41 are pythagorean triplets.
Last let's move on to 4,7 and 8.
![{a}^(2) + {b}^(2) = {4}^(2) + {7}^(2) \\ = 16 + 49 \\ = 65 \\ \\ {c}^(2) = {8}^(2) \\ = 64](https://img.qammunity.org/2022/formulas/mathematics/college/duu5q4c1inwhuv2aqmok7in9s32w469kxa.png)
Since a^2+b^2 IS NOT EQUAL to c^2, 4,7 and 8 ARE NOT pythagorean triplets.