Sum of squares
Un blog matematico
For every natural number n, the quantity n^2-n is divisible by 2.
Credits: Benjamin Dickman's answer in MO60457Viviani's Theorem: In an equilateral triangle, the sum of the distances of an interior point to the three sides equals the altitude of the triangle.
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The following proof can be found in the MO thread [1].
Assume that the harmonic series \sum_{n=1}^{+ \infty}\frac{1}{n} converges. Then the sequence of functions \{f_n\} defined by f_n = \frac{1}{n} \chi_{[0, \, n]} is dominated by the function g = \chi_{[0, \, 1]} + \frac{1}{2} \chi_{[1, \, 2]}+\frac{1}{3} \chi_{[2, \, 3]}+ \frac{1}{4} \chi_{[3, \, 4]}+\ldots, which is by assumption absolutely integrable over \mathbb{R}.
By applying Lebesgue's Dominate Convergence Theorem [2] we get: 1 = \lim_n \int_{\mathbb{R}} f_n(x) dx = \int_{\mathbb{R}} \lim_n f_n(x)dx = 0, a contradiction. \Box
References
[1] https://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts
[2] https://en.wikipedia.org/wiki/Dominated_convergence_theorem
Theorem. The only field automorphism of \mathbb{R} is the identity.
Theorem. Every non-constant polynomial P \in \mathbb{C}[z] must have a zero.
Let us consider the linear operator of Banach spaces F \colon C[0, \, 1/2] \to C[0, \, 1/2], \quad F(f(x)):= \int_0^x f(t)dt. Then the identity \int_0^x e^tdt=e^x-1 can be rewritten as (I-F)(e^x)=1. \quad (\sharp) The operator F is bounded and we have ||F|| = \sup_{f \in C[0, \, 1/2]} \frac{||F(f)||}{||f||} \leq \frac{1/2 \, ||f||}{||f||} = \frac{1}{2} <1. This implies that I-F is invertible as a bounded linear operator and moreover (I-F)^{-1} = \sum_{k=0}^{+ \infty} F^k. Substituting in (\sharp), we get e^x=(I-F)^{-1}(1)=\left(\sum_{k=0}^{+ \infty} F^k \right) (1) = \sum_{k=0}^{+ \infty} F^k(1) = \sum_{k=0} ^{+ \infty}\frac{x^k}{k!}, which is the well-known Taylor series expansion of e^x.