Integer solution to the equation

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 $f(m)$.

• Ummm.. (1) this is not linear algebra; and (2) where does this problem come from? – anthonyquas Apr 17 '15 at 13:58
• No motivation, problem looks detailed and involved – P Vanchinathan Apr 17 '15 at 15:14
• I answered the problem in the negative. See below. – GH from MO Apr 17 '15 at 15:26
• 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. – Gerry Myerson Apr 17 '15 at 22:33
• Can we expect the solution for new $l$? – Shivaraj Apr 21 '15 at 6:52

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.