Let [tex]x[/tex] and [tex]y[/tex] be two legs of a right triangle. By Pythagoras' theorem, the hypotenuse will have length [tex]\sqrt{x^2+y^2}[/tex]:
[tex]x^2+y^2=(\sqrt{x^2+y^2})^2[/tex]
What you should do is read the first two values in the given lists as [tex]x[/tex] and [tex]y[/tex], and the third as [tex]\sqrt{x^2+y^2}[/tex].
So for example, if I gave you the list 3, 4, 5, we take
[tex]x=3[/tex]
[tex]y=4[/tex]
[tex]\implies\sqrt{x^2+y^2}=\sqrt{9+16}=\sqrt{25}=5[/tex]
Since [tex]\sqrt{x^2+y^2}=x^2+y^2[/tex], it follows that (3, 4, 5) is indeed a Pythagorean triple.
Let's check with the first problem: If [tex]x=65[/tex] and [tex]y=72[/tex], we have
[tex]\sqrt{x^2+y^2}=\sqrt{9409}=97[/tex]
which means (65, 72, 97) is also a Pythagorean triple.
