Question

Pythagorean triplet whose one member is 15​

Answer 1

8,15 and 17 are Pythagorean triplets,the Pythagorean triplets of 15 are: (15,112 and113),(15,8 and 17),(15,20 and 25),(15,9and 12),(15,36 and 39)

Answer 2

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

Explanation:

If

`a=m^2-n^2;\ b=2mn;\ c=m^2+n^2`

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

`m^2-n^2=15\to(m-n)(m+n)=15\\\\(m-n)(m+n)=(3)(5)\to m-n=3\ \wedge\ m+n=5`

We have the system of equations:

`\underline{+\left\{\begin{array}{ccc}m-n=3\\m+n=5\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2m=8\qquad\text{divide both sides by 2}\\.\qquad\boxed{m=4}`

Substitute to the second equation:

Therefore we have:

`a=4^2-1^2=16-1=15\\b=2(4)(1)=8\\c=4^2+1^2=16+1=17`

`2mn=15` it's impossible, because 15 is not an even number.

`m^2+n^2=15`

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