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Of course there is. For example take $\varphi(\pi) = (1\,2)(3)(4)(5)$ if $\pi$ is an odd permutation and $\varphi(\pi) = (1)(2)(3)(4)(5)$ if $\pi$ is an even permutation.

Similar example works for the other question.

Here I am referring to jacks's reply. Apologies for not keeping the thread. Indeed as Apostolos says, the integral diverges. What jacks's proof shows is that for every \(0 < \varepsilon < \pi/2\) we have that \(\int_{\varepsilon}^{\pi-\varepsilon} \lfloor \cot{x} \rfloor \; dx = -\pi + 2\varepsilon....

Let us denote by \(S =\{x_1,\ldots,x_n\}\) our set. We need to define all \(x_i \ast x_j\) for each \(1 \leqslant i,j \leqslant n\) and the only restrictions are that \(x_i \ast x_j = x_j \ast x_i\). So \(\ast\) is uniquely determined by the definitions of \( x_i \ast x_j\) for each pair \( (i,j) \)...

I am assuming that the order in which we order the cones does not matter. I am also assuming that the order in which the flavours are placed in the cone also does not matter. (Although it did matter to me when I was a kid. :)) Furthermore, since it says "2 different flavours" I am assuming...

Let \(G\) be a graph on \(n\) vertices and let \(\bar{G}\) denote its complement. Show that \[\chi(G) + \chi(\bar{G}) \geqslant 2\sqrt{n}.\] Show also that equality can be achieved whenever \(n\) is a perfect square.

Let us denote the polynomial by \(f\). It is not too difficult to see that the roots of \(f\) are \(\pm \sqrt{2} \pm i\). It follows that the splitting field \(K\) of \(f\) is contained in \(\mathbb{Q}(i,\sqrt{2})\) and in fact it must be equal to it. Indeed since \(\sqrt{2} = \frac{1}{2}((\sqrt{2} ...

Let us write \(a_1,\ldots,a_n\) for the column vectors of the original matrix \(A\). Then the new matrix is \(B = (a_1+a_2|a_2+a_3|\cdots|a_n+a_1)\) and by linearity we have \[ \det(B) = \sum_{i_1,\ldots,i_n \in\{0,1\}} \det(a_{1+i_1}|\cdots |a_{n+i_n})\] where addition in indices is done modulo \(n...

Changing variables we have \[ I = \int_{[0,1]^n} \lfloor n-(y_1+\cdots+y_n)\rfloor \, dy_1 \, dy_2 \, \cdots \, dy_n\] The set of points $(y_1,\ldots,y_n) \in [0,1]^n$ for which $y_1+\cdots + y_n$ is an integer has measure zero. For all other points we have \[ \lfloor n-(y_1+\cdots+y_n)\rfloor = n-1...

This was problem 10213 from the Monthly. I will update my original post with the details. In the proposed solution by Robin Chapman it shows that the negative Pell equation $X^2 - abY^2 = -1$. has a solution. For example, with Vangelis' notation $(aus+bvt,uv+st)$ is such a solution. Since now $(2x+1...

Let \(x\) and \(y\) be positive integers such that \(x+xy\) and \(y+xy\) are both perfect squares. (a) Show that exactly one of \(x\) and \(y\) is a perfect square. (b) Can you characterise all such pairs? Edit after the solution: This was problem 10213 from the American Mathematical Monthly propose...