Maaz-ul-Haq
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 Jan 6 comment A trivial application of Wilson's theorem to Brocard's Problem It is true that it solves for a very few $n$. But it strikes me as a surprise that the statement that "no solution exists for any non-Wilson prime less one." is not reported even if it is obvious. Jan 6 asked A trivial application of Wilson's theorem to Brocard's Problem Jun 30 comment An easy-to-state elusive combinatorial problem But as you go up in dimensions the gap between the cubes gets smaller and smaller. So while the 2D case was sort of a checkerboard square lattice (with gaps of the same width as the square), the 3D case has cubes with gaps two-thirds of the cube-edge length. Jun 30 comment An easy-to-state elusive combinatorial problem (revisited) If $ry$ increases by $3(y/z)\ge 3$ wouldn't that imply an increment in $rx$ by $3(x/y)\ge 3$ as well? Jun 30 asked An easy-to-state elusive combinatorial problem (revisited) Jun 30 comment An easy-to-state elusive combinatorial problem Can Case $1$ and $2$ with $m\ne 1$ be given an inductive flavor to cover for all dimensions? Or do more cases arise when you have an increment in dimensions? Jun 28 comment An easy-to-state elusive combinatorial problem It was inspired by the View Obstruction paper by Cusick, though this one is on a completely different route, more like scaling n-cubes to cover the entire totally positive orthant of $\mathbb{R}^n$. Jun 28 comment An easy-to-state elusive combinatorial problem I am actually interested in a general version of this problem where you have $a_1, a_2, a_3,... a_{\lambda-2} \in \mathbb{N}$ such that $a_i\geq 1$ for all $1\leq i\leq \lambda-2$ and it is conjectured that the bound is $x=\lambda-1$ such that $n\in[1,x]$ to ensure that all $a_i$ are contained within $\lambda-2$-dimensional hypercubes generated by $[\lambda (m-1)+1,\lambda m -1]$ in all dimensions. It would be quite a task to solve this in general as apparently the problem increases in difficulty as you increase the dimensions. Jun 28 accepted An easy-to-state elusive combinatorial problem Jun 28 comment An easy-to-state elusive combinatorial problem "It is easy to check that, any point in this triangle, we can rescale it by a factor of $\leq 3$ to land in the square $[9,11]\times[5,7]$." Or precisely this? Jun 27 awarded Commentator Jun 27 comment An easy-to-state elusive combinatorial problem @domotorp exactly. Jun 27 revised An easy-to-state elusive combinatorial problem added 19 characters in body Jun 27 revised An easy-to-state elusive combinatorial problem added 9 characters in body Jun 27 awarded Critic Jun 26 comment An easy-to-state elusive combinatorial problem And you'd have to keep track of all the infinitely many squares in the checkerboard? Seems implausible. Jun 26 comment An easy-to-state elusive combinatorial problem In the case $a_1=a_2=4$ we select $n=\frac{5}{4}$ and that would render both $a_1, a_2 \in [5,7]$. The question is that no matter what $a_1$ and $a_2$ you go for, I can always find an $1\le n\le 3$ such that $4k−3≤na_1≤4k−1$ and $4l−3≤na_2≤4l−1$ where $k,l \in \mathbb{N}$. Jun 26 comment An easy-to-state elusive combinatorial problem How do you calculate unions continuously with $n\in\mathbb{R}$? Geometrically you reckon? Jun 26 asked An easy-to-state elusive combinatorial problem Jun 3 revised Existence of a solution to a system of Diophantine Inequalities added 232 characters in body