Smith’s determinant

Let \gcd(i,j) denote the greatest common divisor of i, j and \varphi the Euler’s totient function. Prove that:

\displaystyle \begin{vmatrix} \gcd(1,1) &\gcd(1, 2) &\cdots & \gcd(1,n)\\ \gcd(2,1)&\gcd(2,2) &\cdots & \gcd(2,n)\\ \vdots& \vdots & \ddots &\vdots \\ \gcd(n,1)&\gcd(n,2) &\cdots &\gcd(n,n) \end{vmatrix}= \prod_{j=1}^{n}\varphi(j)


Discuss this at the forum of JoM here.
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Integral and Inequality

Let f:\mathbb{R} \rightarrow \mathbb{R} be a positive real valued and continuous function such that it is periodic of period T=1. Prove that

\displaystyle \int_0^1 \frac{f(x)}{f \left(x + \frac{1}{2} \right)}\, {\rm d}x \geq 1


Since the function f is 1 periodic , then it holds that:

f(x+1)=f(x)=f(x-1) \quad \text{forall} \; x \in \mathbb{R}


\begin{aligned} \int_{0}^{1}\frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x &=\int_{0}^{1/2} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x + \int_{1/2}^{1} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x \\ &= \int_{0}^{1/2} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x + \int_{1/2}^{1} \frac{f(x)}{f\left ( x+1-1 + \frac{1}{2} \right )} \, {\rm d}x\\ &= \int_{0}^{1/2} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x + \int_{1/2}^{1} \frac{f(x)}{f\left ( x- \frac{1}{2} \right )} \, {\rm d}x \\ &\!\!\!\!\!\overset{u=x-1/2}{=\! =\! =\! =\! =\!} \int_{0}^{1/2} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x + \int_{0}^{1/2} \frac{f\left ( x + \frac{1}{2} \right )}{f(x)} \, {\rm d}x \\ &=\int_{0}^{1/2} \left [ \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} + \frac{f\left ( x+\frac{1}{2} \right )}{f(x)} \right ] \, {\rm d}x \\ &\geq \int_{0}^{1/2} 2 \, {\rm d}x \\ &=1 \end{aligned}

since x + \frac{1}{x} \geq 2 forall x>0.

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A sum on Beatty’s theorem

Let \alpha, \beta be positive irrational numbers such that \displaystyle \frac{1}{\alpha} + \frac{1}{\beta}=1. Evaluate the (pseudo) sum:

\displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{\lfloor n\alpha\rfloor^2}+\frac{1}{\lfloor n\beta\rfloor^2}\right)


We are using Beatty’s theorem .In brief, it states that for positive irrational numbers \alpha, \beta with \displaystyle \frac{1}{\alpha}+\frac{1}{\beta}=1 the sequences \lfloor \alpha\rfloor, \lfloor 2\alpha\rfloor, \lfloor 3\alpha\rfloor, \dots and \lfloor \beta\rfloor, \lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor, \dots are complementary. (i.e. disjoint and their union is \mathbb{N}). Thus our sum is nothing else than

\displaystyle \zeta(2)=\frac{\pi^2}{6}

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No continuous mapping

Prove that there does not exist an 1-1 and continuous mapping from \mathbb{R}^2 to \mathbb{R}.


Due to connectness we have that f \left(\mathbb{R}^2 \right)= \mathcal{I} where \mathcal{I} is an interval. Note that if we remove a point from the plane, it still remains connected. Having that in mind we observe that f \left ( \mathbb{R}^2 \setminus f^{-1}\left ( a \right ) \right ) is connected and equals \mathcal{I} \setminus \left \{ a \right \} whereas this is not connected leading to a contradiction.
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