0
$\begingroup$

Does there exists an integer solution (for every integer $m\geq 1$) for the following equation?

$$x_1x_2...x_n+(2y+1)z+y=4m+3$$ where, $1\leq x_1\leq x_2\leq...\leq x_n\leq l$,$0\leq y \leq (\frac{l}{2}-1)$, $1\leq z \leq \frac{l}{2}$ and $l\leq c(m^{\frac{1}{f(m)}})$ for some fixed $c>0$ and some function $f(m)$.

$\endgroup$
  • $\begingroup$ Ummm.. (1) this is not linear algebra; and (2) where does this problem come from? $\endgroup$ – anthonyquas Apr 17 '15 at 13:58
  • 2
    $\begingroup$ No motivation, problem looks detailed and involved $\endgroup$ – P Vanchinathan Apr 17 '15 at 15:14
  • $\begingroup$ I answered the problem in the negative. See below. $\endgroup$ – GH from MO Apr 17 '15 at 15:26
  • $\begingroup$ I wonder why this question was migrated away from MO. Judging from the answer by GH, it looks like an MO-worthy question to me. $\endgroup$ – Gerry Myerson Apr 17 '15 at 22:33
  • $\begingroup$ Can we expect the solution for new $l$? $\endgroup$ – Shivaraj Apr 21 '15 at 6:52
4
$\begingroup$

The answer is no. Consider a large dyadic interval $M\leq m<2M$. If every number $4m+3$ could be written in the form indicated, then a positive proportion of the numbers $n<8M$ could be written in the form $n=n_1+n_2$, where $n_1$ is free of prime factors exceeding $c\log^k(2M)$, and $n_2<c^2\log^{2k}(2M)$. It is known (see Theorem 1 in Section III.5 of Tenenbaum: Introduction to analytic and probabilistic number theory) that the number of possibilities for $n_1$ is $$ \leq\Psi(8M,c\log^{2k}(2M))\ll M\exp(-c'\log M/\log\log M), $$ with some $c'>0$ depending on $c$ and $k$. Hence the number of possibilities for the sum $n=n_1+n_2$ is $$ \ll M\exp(-c_1\log M/\log\log M)\log^{2k}(2M)=o(M).$$ Contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.