Kapalci

2022-06-25

why this Diophantine equation $2ks=(5t+3)(16t+9)$ has always a solution for every $k$?

I would like to solve the following Diophantine equation and show that it has always a solution; i.e. for every positive integer $k$, there exists an integer $t$ such that the fraction is an integer:

$\frac{(5t+3)(16t+9)}{2k}$

Any hint will be grateful.

P.S.: Without considering cases for $k\equiv 0,1,2,3,4(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}5)$

I would like to solve the following Diophantine equation and show that it has always a solution; i.e. for every positive integer $k$, there exists an integer $t$ such that the fraction is an integer:

$\frac{(5t+3)(16t+9)}{2k}$

Any hint will be grateful.

P.S.: Without considering cases for $k\equiv 0,1,2,3,4(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}5)$

Judovh0

Beginner2022-06-26Added 16 answers

Note that $t$ will have to be odd, say $2w+1$, so we want $(5w+4)(32w+25)$ to be divisible by $k$. Let $k={2}^{a}{5}^{b}l$ where $l$ is divisible neither by $2$ nor by $5$. We will succeed if we can find a value of $w$ such that $5w+4$ is divisible by ${2}^{a}$ and $32w+25$ is divisible by ${5}^{b}l$. So we want to solve the system of congruences

$5w\equiv -4\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{2}^{a}),\phantom{\rule{2em}{0ex}}32w\equiv -25\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{5}^{b}l).$

By multiplying the first congruence through by the inverse of $5$ modulo ${2}^{a}$, and the second congruence by the inverse of $32$ modulo ${5}^{b}l$, we obtain a system of congruences of the shape $w\equiv c\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{2}^{a})$, $w\equiv d\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{5}^{b}l)$. By the Chinese Remainder Theorem, this system has a solution.

$5w\equiv -4\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{2}^{a}),\phantom{\rule{2em}{0ex}}32w\equiv -25\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{5}^{b}l).$

By multiplying the first congruence through by the inverse of $5$ modulo ${2}^{a}$, and the second congruence by the inverse of $32$ modulo ${5}^{b}l$, we obtain a system of congruences of the shape $w\equiv c\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{2}^{a})$, $w\equiv d\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{5}^{b}l)$. By the Chinese Remainder Theorem, this system has a solution.

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