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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].

Proof details

From [math]\displaystyle{ x := n, y:= n + a }[/math] and [math]\displaystyle{ z := 2n + a + d }[/math] where [math]\displaystyle{ d \in {}^{\omega }{\mathbb{N}^{*}} }[/math], it follows due to [math]\displaystyle{ z^p \equiv y, y^p \equiv y }[/math] and [math]\displaystyle{ z^p \equiv z }[/math] first [math]\displaystyle{ d \equiv 0 \mod p }[/math], then [math]\displaystyle{ d = \pm kp }[/math]. Since [math]\displaystyle{ x + y = 2n + a > z }[/math] is required, [math]\displaystyle{ f_{akp}(n) }[/math] is chosen properly.

Induction for [math]\displaystyle{ n }[/math] implies the claim due to the case [math]\displaystyle{ m = 4 }[/math] and [math]\displaystyle{ y > x > p }[/math]:

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.

Proof details

The strict monotonicity follows from (continuously) differentiating by [math]\displaystyle{ n }[/math] such that [math]\displaystyle{ f_{akp}(n)' = p(2(2n + a - kp)^{p - 1} - n^{p - 1} - (n + a)^{p - 1}) > 0 }[/math].

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]

Proof details

[math]\displaystyle{ \int_0^{n^{*}}{f_{akp}(v)}dv = ((2n^{*} + a - kp)^{p + 1} / 2 - (n^{*})^{p + 1} - (n^{*} + a)^{p + 1})/(p + 1) + t = ((2n^{*} + a - kp)^{(p + 1)/2} \pm \sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})^2/(2p + 2) + t }[/math] for [math]\displaystyle{ t \in {}^{\omega}{\mathbb{Q}} }[/math] where the third binomial formula [math]\displaystyle{ r^2 - s^2 = (r \pm s)^2 := (r + s)(r - s) }[/math] was used. After separating the negligible factor [math]\displaystyle{ \hat{2}((2n^{*} + a - kp)^{(p + 1)/2} + \sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})/(p + 1) }[/math], the derivative is just [math]\displaystyle{ (\hat{2}(2n^{*} + a - kp)^{(p - 1)/2} - \hat{2}((n^{*})^p + (n^{*} + a)^p)/\sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}}) }[/math]. After squaring the terms, the polynomial division gives [math]\displaystyle{ (n^{*})^{p - 1} + (n^{*} + a)^{p - 1} + a^2(n^{*})^{p - 1}(n^{*} + a)^{p - 1}/((n^{*})^{p + 1} + (n^{*} + a)^{p + 1}) }[/math] as recalculating by multiplication confirms.

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Nonstandard Mathematics