**Proposition:**

For all positive integers

*a*,*b*, and*n*> 2 , (*a*^{n}+ (*a*+*b*)^{n})^{1 ⁄ n}∉ℤ.**Proof**:

Central to our argument for this proof is an understanding of the
binomial expansion (also referred to as the binomial identity) for any
power

*n*and that*any*integer can be expressed as a binomial integer pair.
Let For example:

*c*=*a*+*b*then*c*^{n}≡ (*a*+*b*)^{n}. The general formula for the binomial expansion of (*a*+*b*)^{n}is
(

*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}
For

*c*^{2}= (*a*+*b*)^{2}≡*a*^{2}+ 2*ab*+*b*^{2}
⇒ (

*a*^{2}+ 2*ab*+*b*^{2})^{1 ⁄ 2}=*c*
For

*c*^{3}= (*a*+*b*)^{3}≡*a*^{3}+ 3*a*^{2}*b*+ 3*ab*^{2}+*b*^{3}
⇒ (

*a*^{3}+ 3*a*^{2}*b*+ 3*ab*^{2}+*b*^{3})^{1 ⁄ 3}=*c*
For

*c*^{4}= (*a*+*b*)^{4}≡*a*^{4}+ 4*a*^{3}*b*+ 6*a*^{2}*b*^{2}+ 4*ab*^{3}+*b*^{4}
⇒ (

... etc.*a*^{4}+ 4*a*^{3}*b*+ 6*a*^{2}*b*^{2}+ 4*ab*^{3}+*b*^{4})^{1 ⁄ 4}=*c*
Evaluating the polynomial (the expanded binomial) results in an integer having the integer For all positive integers

*n*^{th}root =*c*. Note that for a binomial expansion of two variables (*a & b*) the resulting polynomial has*n*+ 1 terms with coefficients totaling 2^{n}and (to repeat) has an*n*^{th}integer root =*a*+*b*=*c*. If a polynomial of two variables has more or less than*n*+ 1 terms and other than 2^{n}coefficients it cannot have an*n*^{th}integer root simply*because all integers raised to a power n as a binomial integer pair are of this general formula*.*a*,*b*, and*n;*(*a*^{n})^{1 ⁄ n}∈ ℤ, ((*a*+*b*)^{n})^{1 ⁄ n}∈ ℤ and for*n*> 2 the inequality can be generalized as
(

*a*^{n}+ (*a*+*b*)^{n})^{1 ⁄ n}∉ℤ*a*

^{n}+ (

*a*+

*b*)

^{n}= 2

*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}

^{n}+ 1 and therefore will not have an integer

*n*

^{th}root. This inequality proves our proposition.

However, the reader may find this result counter intuitive since for
and rearranging the equation, we find the difference of two squares identity

*n*= 2, (our proposition is for*n*> 2) we find
(3

a prime Pythagorean triple of 3, 4, 5. In this case, we can express the Pythagorean equation as ^{2}+ (3 + 1)^{2}) = 5^{2}*a*

^{2}+ (

*a*+

*b*)

^{2}= (

*a*+

*c*)

^{2}

(

reduces the equation to a linear (first order) expression
*a*+*b*^{2}) = (*a*+*c*)^{2}−*a*^{2}
(

which we can further simplify to
*a*+*b*)(*a*+*b*) = (*a*+*c*+*a*)(*a*+*c*−*a*)*y*⋅

*y*=

*m*⋅

*n*

For the difference of two

*n*^{th}powers for*n*> 2 and after factoring, we are left with at least one irreducible polynomial and cannot reduce it to a linear relationship.
We can check our proof for

*n*> 2 by assuming there may be some integers*a*,*b*, and*c*such that*a*

^{n}+ (

*a*+

*b*)

^{n}= (

*a*+

*c*)

^{n}

Expanding both binomials and subtracting the

*a*^{n}term from both sides, we obtain the equation*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}

=

*k*^{n}_{1}*a*^{n − 1}*c*+*k*^{n}_{2}*a*^{n − 2}*c*^{2}+ ⋯ +*k*^{n}_{n − 1}*ac*^{n − 1}+*c*^{n}
The left (upper) side of our equation now is of the general form for a binomial (

*a*+*b*)^{n}(and has an integer*n*^{th}root) whereas the right (lower) side, after subtracting the*a*^{n}term, no longer is of the general formula for a binomial integer pair (it now has*n*terms and coefficients totaling 2^{n}− 1) and does not have an integer*n*^{th}root; therefore for all positive integers*a*,*b*, and*n*> 2 we have
(

and conversely*a*^{n}+ (*a*+*b*)^{n})^{1 ⁄ n}∉ℤ
((

*a*+*b*)^{n}−*a*^{n})^{1 ⁄ n}∉ℤ