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25 December 2021

In this paper I offer an algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial expansion, (a+b) n = a n + b Pn i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I show that a simple rewrite of the Fermat’s equation stating the theorem, A p + B p = (A + B − D) p , A, B, D and p positive integers, D < A < B, p ≥ 3 and prime, entails the contradiction, A(B − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 A i−1−j (A + B − D) j−1 # + B(A − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 B i−1−j (A + B − D) j−1 # = 0, the sum of two positive integers equal to zero. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat’s Last Theorem. AMS 2020 subject classification: 11A99, 11D41 Diophantine equations, Fermat’s equation ∗The corresponding author. E-mail: bookie@hevanet.com 1 1 Introduction To prove Fermat’s Last Theorem, it suffices to show that the equation A p + B p = C p (1In this paper I offer an algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial expansion, (a+b) n = a n + b Pn i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I show that a simple rewrite of the Fermat’s equation stating the theorem, A p + B p = (A + B − D) p , A, B, D and p positive integers, D < A < B, p ≥ 3 and prime, entails the contradiction, A(B − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 A i−1−j (A + B − D) j−1 # + B(A − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 B i−1−j (A + B − D) j−1 # = 0, the sum of two positive integers equal to zero. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat’s Last Theorem.

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