Answer: FALSE
Step-by-step explanation:
[tex]\text{Pythagorean formula is } (x)^2+(y)^2=(hypotenuse)^2\\\text{Pythagorean triple states that each leg and the hypotenuse are integers.}\\\\\text{Given:}\\leg_1=x^2-y^2 \qquad \qquad leg_2=2xy\qquad \qquad hypotenuse=x^2+y^2\\\\\text{Let x = 16 and y = 2k + 1 (the odd leg).}\\\text{Need to show that the hypotenuse is an integer:}[/tex]
[tex](16)^2+(2k+1)^2=(hypotenuse)^2\\256+4k^2+4k+1=(hypotenuse)^2\\4k^2+4k+257=(hypotenuse)^2\\4(k^2+k+64.25)=(hypotenuse)^2\\k^2+k+64.25\text{ is not a perfect square because }\sqrt{64.25} \text{ is irrational}\\\text{Therefore, the hypotenuse is not an integer.}\\\\\text{The statement is FALSE.}[/tex]
