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Theorems of the month

Counting theorem for algebraic numbers

The number [math]\displaystyle{ \mathbb{A}(m, n) }[/math] of algebraic numbers of polynomial or series degree [math]\displaystyle{ m }[/math] and thus in general for the Riemann zeta function [math]\displaystyle{ \zeta }[/math] asymptotically satisfies the equation [math]\displaystyle{ \mathbb{A}(m, n) = \widetilde{\zeta(\grave{m})}\,z(m){{(2n+1)}^{m}}\left( n+\mathcal{O}({_e}n) \right) }[/math], where [math]\displaystyle{ z(m) }[/math] is the average number of zeros of a polynomial or series.

Proof:

The case [math]\displaystyle{ m = 1 }[/math] requires by the error term [math]\displaystyle{ \mathcal{O}({_e}n n) }[/math] and represents the number [math]\displaystyle{ 4{+}_{k=1}^{n}{\varphi (k)}-1 }[/math] by the [math]\displaystyle{ \varphi }[/math]-function. For [math]\displaystyle{ m \gt 1 }[/math], the divisibility conditions neither change the error term [math]\displaystyle{ \mathcal{O}({_e}n) }[/math] nor the leading term. Polynomials or series such that [math]\displaystyle{ \text{gcd}({a}_{0}, {a}_{1}, ..., {a}_{m}) \ne 1 }[/math] are excluded by [math]\displaystyle{ 1/\zeta(\grave{m}) }[/math]: The latter is given by taking the product over the prime numbers [math]\displaystyle{ p }[/math] of all [math]\displaystyle{ (1 - {p}^{-\grave{m}}) }[/math] absorbing here multiples of [math]\displaystyle{ p }[/math] and representing sums of geometric series.[math]\displaystyle{ \square }[/math]

Brocard's theorem

It holds that [math]\displaystyle{ \{(m, n) \in {}^{\omega} \mathbb{N}^2 : n! + 1 = m^2\} = \{(5, 4), (11, 5), (71, 7)\}. }[/math]

Proof:

From [math]\displaystyle{ n! = \acute{m}\grave{m} }[/math], it follows that [math]\displaystyle{ m = \hat{r} \pm 1 }[/math] für [math]\displaystyle{ r \in {}^{\omega} \mathbb{N}^{*} }[/math] and [math]\displaystyle{ n \ge 3 }[/math]. Thus [math]\displaystyle{ n! = \hat{r}(\hat{r}\pm2) = 8s(\hat{s} \pm 1) }[/math] holds for [math]\displaystyle{ s \in {}^{\omega} \mathbb{N}^{*} }[/math]. Let [math]\displaystyle{ 2^q \mid n! }[/math] and [math]\displaystyle{ 2^{\grave{q}} \nmid n! }[/math] for maximal [math]\displaystyle{ q \in {}^{\omega} \mathbb{N}^{*} }[/math]. Therefore [math]\displaystyle{ n! = 2^q(\hat{u} + 1) }[/math] holds for [math]\displaystyle{ u \in {}^{\omega} \mathbb{N}^{*} }[/math] and necessarily [math]\displaystyle{ n! = 2^q(2^{q-2} \pm 1) }[/math]. Then the prime factorisation of [math]\displaystyle{ n! }[/math] requires [math]\displaystyle{ n \le 7 }[/math] giving the claim.[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics