Given:
Consider the three number are given by:
[tex]a=x^2-y^2[/tex]
[tex]b=2xy[/tex]
[tex]c=x^2+y^2[/tex]
To find:
The three numbers are the Pythagorean triple generated by using 4 for x and 1 for y.
Solution:
We have,
[tex]a=x^2-y^2[/tex]
[tex]b=2xy[/tex]
[tex]c=x^2+y^2[/tex]
Substituting 4 for x and 1 for y, we get
[tex]a=4^2-1^2[/tex]
[tex]a=16-1[/tex]
[tex]a=15[/tex]
Similarly,
[tex]b=2(4)(1)[/tex]
[tex]b=8[/tex]
And,
[tex]c=4^2+1^2[/tex]
[tex]c=16+1[/tex]
[tex]c=17[/tex]
The three three numbers are the Pythagorean triple are 8,15 and 17.
Therefore, the correct option is C.
