Take natural numbers $A_1,B_1,A_2,B_2$ random pairwise coprime in $[n,2n]$ for $n$ large enough and consider the space of solutions to $A_1a+B_1b=0$ and $A_1A_2a+A_1B_2b+B_1A_2c+B_1B_2d=0$ spanned by $1\times 2$ and $3\times 4$ matrices $$N_1=\begin{bmatrix} -B_1&A_1 \end{bmatrix}$$ and $$N_2=\begin{bmatrix} -B_2&A_2&0&0\\ -B_1&0&A_1&0\\ 0&0&-B_2&A_2 \end{bmatrix}$$ respectively.

Denote the $\Bbb Q$-linear space spanned by rows of $N_1$ by $T_{A_1,B_1}\subseteq\Bbb Q^2$ and $N_2$ by $T_{A_1,B_1,A_2,B_2}\subseteq\Bbb Q^4$ and denote the set of non-zero $\Bbb Z$ vectors in $T_{A_1,B_1}$ by $T_{A_1,B_1}[\Bbb Z]^\star$ and in $T_{A_1,B_1,A_2,B_2}$ by $T_{A_1,B_1,A_2,B_2}[\Bbb Z]^\star$.

Consider the quantities $$\mu_1((A_1,B_1))=\min_{v\in T_{A_1,B_1}[\Bbb Z]^\star}\|v\|_\infty$$ $$\mu_2((A_1,B_1)\otimes(A_2,B_2))=\min_{v\in T_{A_1,B_1,A_2,B_2}[\Bbb Z]^\star}\|v\|_\infty$$ where $\|v\|_\infty$ is largest coordinate by magnitude of vector $v$. One can show with probability $1-o(1)$ value of $\mu((A_1,B_1))$ is at least $c_1\cdot n$ for a constant $c_1>0$ and of $\mu_2((A_1,B_1)\otimes(A_2,B_2))$ is at least $c_2\cdot n^{2/3}$ for a constant $c_2>0$.

Similarly define $$\mu_k((A_1,B_1)\otimes(A_2,B_2)\otimes\dots\otimes(A_k,B_k))$$ where $$A_1,B_1,A_2,B_2,\dots,A_k,B_k$$ are random pairwise coprime in $[n,2n]$ for $n$ large enough.

It is possible to show that the typical value of $\mu_k((A_1,B_1)\otimes(A_2,B_2)\otimes\dots\otimes(A_k,B_k))$ is at least $$c_k\cdot n^{k/(2^k-1)}$$ for some $c_k>0$ with probability $1-o(1)$ at every given $k\in\mathbb N$.

However what is the minimum $\gamma_k$ in $$\max(A_1,\dots,A_k,B_1,\dots,B_k)-\min(A_1,\dots,A_k,B_1,\dots,B_k)\leq n^{\gamma_k}$$ to get typical value of $\mu_k((A_1,B_1)\otimes(A_2,B_2)\otimes\dots\otimes(A_k,B_k))$ to be at least $$\Omega(n^{k/(2^k-1)})$$ with probability $1-o(1)$ at every given $k\in\mathbb N$? Is $n^{\gamma_k}$ as low as $n^{k/(2^k-1)}$ or does at least $2^k\gamma_k=2^{o(k)}$ hold (at $k=1$ we have $\gamma_1\rightarrow0$)?

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  • 1
    You delete more than half your questions. How about keeping this one open for a change. – Todd Trimble Oct 9 at 23:57
  • Ok I shall do that. – Freeman. Oct 10 at 0:47

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