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Theorem of the month
Fermat's Last Theorem
For all [math]\displaystyle{ p \in {}^{\omega }{\mathbb{P}_{\ge 3}} }[/math] and [math]\displaystyle{ x, y, z \in {}^{\omega }{\mathbb{N}^{*}} }[/math], always [math]\displaystyle{ x^p + y^p \ne z^p }[/math] holds and thus for all [math]\displaystyle{ m \in {}^{\omega }{\mathbb{N}_{\ge 3}} }[/math] instead of [math]\displaystyle{ p }[/math].
Proof:
Because of Fermat's little theorem, rewritten, [math]\displaystyle{ f_{akp}(n) := (2n + a - kp)^p - n^p - (n + a)^p \ne 0 }[/math] is to show for [math]\displaystyle{ a, k, n \in {}^{\omega }{\mathbb{N}^{*}} }[/math] where [math]\displaystyle{ kp < n }[/math].
Induction for [math]\displaystyle{ n }[/math] implies the claim due to the case [math]\displaystyle{ m = 4 }[/math][1] and [math]\displaystyle{ y > x > p }[/math][2]:
Induction basis [math]\displaystyle{ (n \le p): f_{akp}(n) \ne 0 }[/math] for all [math]\displaystyle{ a, k }[/math] and [math]\displaystyle{ p }[/math]. Let [math]\displaystyle{ r \in {}^{\omega }{\mathbb{N}_{< p}} }[/math].
Induction step [math]\displaystyle{ \,(n = q + r \; \rightarrow \; n^{*} = n + p): }[/math] Let [math]\displaystyle{ f_{akp}(n^{*}) \ge 0 }[/math], but [math]\displaystyle{ f_{akp}(n) < 0 }[/math], since [math]\displaystyle{ f_{akp}(n) }[/math] is strictly monotonically increasing and otherwise nothing to prove.
It holds [math]\displaystyle{ f_{akp}(n^{*}) = (\int_0^{n^{*}}{f_{akp}(v)}dv)' \ne 0 }[/math], since [math]\displaystyle{ (n^{*})^{p + 1} + (n^{*} + a)^{p + 1} }[/math] does not divide [math]\displaystyle{ ((n^{*})^p + (n^{*} + a)^p)^2 }[/math] after separating the positive factor as polynomial division shows.[math]\displaystyle{ \square }[/math]
Recommended reading
References
- ↑ Ribenboim, Paulo: Thirteen Lectures on Fermat's Last Theorem : 1979; Springer; New York; ISBN 9780387904320, p. 35 - 38.
- ↑ loc. cit., p. 226.