so for [A] [A] for n = 2 and the difference of two squares identityn > 2:
(a + b)2 − a2 = (a + b + a)(a + b − a)for which an infinite set of (Pythagorean) triples can be found which satisfy the equation.
((a + b)n±an)1 ⁄ n ∉ ℕ
an + (a + b)n ≠ (a + c)nand demonstrate the proof using the binomial identity [B] [B] A simple proof of Fermat’s Last Theorem is demonstrated by substituting binomial integer pairs for x, y, and z such that we can say there are no three positive integers a, b, and c that satisfy the equation an + (a + b)n = (a + c)n for any integer value greater than 2. Expanding our binomials
(a + b)n = an + kn1an − 1b + kn2an − 2b2 + ⋯ + knn − 1abn − 1 + bn
(a + c)n = an + kn1an − 1c + kn2an − 2c2 + ⋯ + knn − 1acn − 1 + cnwe see that by combining the an term with either polynomial (by addition or subtraction) results in a polynomial that no longer has an nth integer root equal to either (a + b) or (a + c).