*a*,

*b*, and

*n*, ((

*a*+

*b*)

^{n})

^{1 ⁄ n}∈ ℕ

so for [A] [A] for

*n*= 2 and the*difference of two squares*identity
(

for which an infinite set of (Pythagorean) triples can be found which satisfy the equation.
*a*+*b*)^{2}−*a*^{2}= (*a*+*b*+*a*)(*a*+*b*−*a*)*n*> 2:
((

*a*+*b*)^{n}±*a*^{n})^{1 ⁄ n}∉ ℕ*x*,

*y*, and

*z*satisfy the equation

*x*

^{n}+

*y*

^{n}=

*z*

^{n}for any integer value of n greater than 2” we can substitute in the positive integers

*a*,

*b*, and

*c*for

*x*,

*y*, and

*z*:

*a*

^{n}+ (

*a*+

*b*)

^{n}≠ (

*a*+

*c*)

^{n}

*x*,

*y*, and

*z*such that we can say there are no three positive integers

*a*,

*b*, and

*c*that satisfy the equation

*a*

^{n}+ (

*a*+

*b*)

^{n}= (

*a*+

*c*)

^{n}for any integer value greater than 2. Expanding our binomials

(

*a*+*b*)^{n}=*a*^{n}+*k*^{n}_{1}*a*^{n − 1}*b*+*k*^{n}_{2}*a*^{n − 2}*b*^{2}+ ⋯ +*k*^{n}_{n − 1}*ab*^{n − 1}+*b*^{n}
(

we see that by combining the *a*+*c*)^{n}=*a*^{n}+*k*^{n}_{1}*a*^{n − 1}*c*+*k*^{n}_{2}*a*^{n − 2}*c*^{2}+ ⋯ +*k*^{n}_{n − 1}*ac*^{n − 1}+*c*^{n}*a*

^{n}term with either polynomial (by addition or subtraction) results in a polynomial that no longer has an

*n*

^{th}integer root equal to either (

*a*+

*b*) or (

*a*+

*c*).