Pythagorean triplet whose one member is 15

(8, 15, 17); (9, 12, 15); (15; 20; 25)

Step-by-step explanation:

If

for m > n, then a, b, c make a Pythagorean triplet.

We have the system of equations:

Substitute to the second equation:

Therefore we have:

 it's impossible, because 15 is not an even number.

Let's consider all possible sums of two numbers resulting in 15.

We will check which of the numbers are perfect squares.

1 + 14

2 + 13

3 + 12

4 + 11

5 + 10

6 + 9

7 + 8

(Bold not perfect squares)

There are no two perfect squares among the listed pairs of numbers.

Other:

15, 112, 113

We know the Egyptian triangle with sides of length 3, 4, 5.

By modifying this Pythagorean triplet by multiplying by 3 we get:

(3)(3) = 9; (3)(4) = 12; (3)(5) = 15

By modifying this Pythagorean triplet by multiplying by 5 we get:

(5)(3) = 15; (5)(4) = 20; (5)(5) = 25

Vertical shift is -2. Horizontal shift is 1. Check the box that says reflect over x-axis. Horizontal shrink should be 4. Then to check if it looks right go to Desmos.com and in the graphing calculator put in y=-4(x-1)^2-2 and see if it looks like the one you did in your work