Question

question 6 of 10
which combination of integers can be used to generate the pythagorean triple (8, 15, 17)?
a. x = 3, y = 2
b. x = 4, y = 1
c. x = 4, y = 3
d. x = 2, y = 2

Explanation:

Step1: Recall Pythagorean triple generation formula

The Pythagorean triple $(a,b,c)$ can be generated using the formulas $a = x^{2}-y^{2}$, $b = 2xy$, $c=x^{2}+y^{2}$ where $x>y>0$ are positive - integers.

Step2: Test option A

If $x = 3$ and $y = 2$, then $a=x^{2}-y^{2}=3^{2}-2^{2}=9 - 4=5$, $b = 2xy=2\times3\times2 = 12$, $c=x^{2}+y^{2}=3^{2}+2^{2}=9 + 4 = 13$. This is not the given triple.

Step3: Test option B

If $x = 4$ and $y = 1$, then $a=x^{2}-y^{2}=4^{2}-1^{2}=16 - 1 = 15$, $b = 2xy=2\times4\times1=8$, $c=x^{2}+y^{2}=4^{2}+1^{2}=16 + 1 = 17$. This is the given triple $(8,15,17)$.

Step4: Test option C

If $x = 4$ and $y = 3$, then $a=x^{2}-y^{2}=4^{2}-3^{2}=16 - 9 = 7$, $b = 2xy=2\times4\times3 = 24$, $c=x^{2}+y^{2}=4^{2}+3^{2}=16 + 9 = 25$. This is not the given triple.

Step5: Test option D

If $x = 2$ and $y = 2$, then $x=y$ which does not satisfy the condition $x>y>0$ for generating Pythagorean triples.

Answer:

B. $x = 4,y = 1$