For all positive integers a, b, and n > 2 , (an + (a + b)n)1 ⁄ n∉ℤ.
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 c = a + b then cn ≡ (a + b)n. The general formula for the binomial expansion of (a + b)n isFor example:
(a + b)n = an + kn1an − 1b + kn2an − 2b2 + ⋯ + knn − 1abn − 1 + bn
For c2 = (a + b)2 ≡ a2 + 2ab + b2
⇒ (a2 + 2ab + b2)1 ⁄ 2 = c
For c3 = (a + b)3 ≡ a3 + 3a2b + 3ab2 + b3
⇒ (a3 + 3a2b + 3ab2 + b3)1 ⁄ 3 = c
For c4 = (a + b)4 ≡ a4 + 4a3b + 6a2b2 + 4ab3 + b4
⇒ (a4 + 4a3b + 6a2b2 + 4ab3 + b4)1 ⁄ 4 = c... etc.
Evaluating the polynomial (the expanded binomial) results in an integer having the integer nth root = c. Note that for a binomial expansion of two variables (a & b) the resulting polynomial has n + 1 terms with coefficients totaling 2n and (to repeat) has an nth integer root = a + b = c. If a polynomial of two variables has more or less than n + 1 terms and other than 2n coefficients it cannot have an nth integer root simply because all integers raised to a power n as a binomial integer pair are of this general formula.For all positive integers a, b, and n; (an)1 ⁄ n ∈ ℤ, ((a + b)n)1 ⁄ n ∈ ℤ and for n > 2 the inequality can be generalized as
(an + (a + b)n)1 ⁄ n∉ℤ
an + (a + b)n = 2an + kn1an − 1b + kn2an − 2b2 + ⋯ + knn − 1abn − 1 + bnthe resulting polynomial is not equivalent to the general formula for the binomial expansion of an integer (as an integer pair); it now has coefficients totaling 2n + 1 and therefore will not have an integer nth root. This inequality proves our proposition.
However, the reader may find this result counter intuitive since for n = 2, (our proposition is for n > 2) we find
(32 + (3 + 1)2) = 52a prime Pythagorean triple of 3, 4, 5. In this case, we can express the Pythagorean equation as
a2 + (a + b)2 = (a + c)2and rearranging the equation, we find the difference of two squares identity
(a + b2) = (a + c)2 − a2reduces the equation to a linear (first order) expression
(a + b)(a + b) = (a + c + a)(a + c − a)which we can further simplify to
y⋅y = m⋅n
For the difference of two nth 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
an + (a + b)n = (a + c)n
Expanding both binomials and subtracting the an term from both sides, we obtain the equation
an + kn1an − 1b + kn2an − 2b2 + ⋯ + knn − 1abn − 1 + bn
= kn1an − 1c + kn2an − 2c2 + ⋯ + knn − 1acn − 1 + cn
The left (upper) side of our equation now is of the general form for a binomial (a + b)n (and has an integer nth root) whereas the right (lower) side, after subtracting the an term, no longer is of the general formula for a binomial integer pair (it now has n terms and coefficients totaling 2n − 1) and does not have an integer nth root; therefore for all positive integers a, b, and n > 2 we have
(an + (a + b)n)1 ⁄ n∉ℤand conversely
((a + b)n − an)1 ⁄ n∉ℤ