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Throughout, n > 2.

Case 1:

Part (a) - Let the equality relation, x^2 + y^2 = z^2, be given and assume, x^n + y^n = z^n, is true. Factor, x^n + y^n = z^n, thus: x^n-2(x^2) + y^n-2(y^2) = z^n-2(z^2). Substituting equals-for-equals, (x^2 + y^2, for, z^2), we have, x^n-2(x^2) + y^n-2(y^2) = z^n-2(x^2 + y^2). Distributing, we get, x^n-2(x^2) + y^n-2(y^2) = z^n-2(x^2) + z^n-2(y^2). However, it is given that x and y are < z, making the left hand side of the equation ≠ to the right hand side of the equation. Therefore, x^n + y^n ≠ z^n.

Part (b) - Inversely, assume, x^n + y^n = z^n, and divide equals-by-equals thus: x^n/z^n-2 + y^n/z^n-2 = z^n/z^n-2 = z^2. Then the left hand side of the equation must be a member of the Set of two squares, rational and/or irrational, whose sum is equal to

z^2. However, by Part (a), any members of the Set such that x^2 + y^2 = z^2, results in the inequality, x^n + y^n ≠ z^n.

Consequently, whenever, x^2 + y^2 = z^2, then, x^n + y^n ≠ z^n, and FLT is true for Case 1.

Case 2: Let the inequality relation, x^2 + y^2 < z^2, be given and assume, x^n + y^n = z^n. Let x^2 + y^2 + u^2 = z^2, where, u^2 = z^2 - (x^2 + y^2), and let x^2 + y^2 = q^2 < z^2. Then, since q^2 < z^2, therefore, q^n < z^n. And since, x^2 + y^2 < q^2, then by Case 1, x^n + y^n < q^n. Therefore, we have, x^n + y^n < q^n < z^n, or, x^n + y^n < z^n. Consequently, whenever x^2 + y^2 < z^2, then, x^n + y^n ≠ z^n, and FLT is true for Case 2.

Case 3 - Let the inequality relation, x^2 + y^2 > z^2, be given and assume x^n + y^n = z^n. Let, x^2 + y^2 - u^2 = z^2, where, u^2 = (x^2 + y^2) - z^2. Then by the Trichotomy Law, x^n + y^n < = > z^n + u^n. If, x^n + y^n = z^n + u^n, then, x^n + y^n ≠ z^n. If, x^n + y^n < > z^n + u^n, then, by the Trichotomy Law, x^n + y^n < = >

z^n. If, x^n + y^n < > z^n, there is agreement with FLT. If we assume, x^n + y^n = z^n, then by Case 1 the assumption is false and, x^n + y^n ≠ z^n.

Consequently, FLT is true for Case 3.

Therefore, since Case 1, 2, and 3 exhaust all whole number possibilities, it has been shown what is necessary and sufficient to prove FLT, namely, x^n + y^n ≠ z^n, n>2, in whole numbers, is always true.

QED

willacaleb (LKS)