Kapalci

2022-06-25

why this Diophantine equation $2ks=\left(5t+3\right)\left(16t+9\right)$ 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:
$\frac{\left(5t+3\right)\left(16t+9\right)}{2k}$
Any hint will be grateful.
P.S.: Without considering cases for $k\equiv 0,1,2,3,4\left(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}5\right)$

Judovh0

Note that $t$ will have to be odd, say $2w+1$, so we want $\left(5w+4\right)\left(32w+25\right)$ 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
$5w\equiv -4\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{2}^{a}\right),\phantom{\rule{2em}{0ex}}32w\equiv -25\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{5}^{b}l\right).$
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 $w\equiv c\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{2}^{a}\right)$, $w\equiv d\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}{5}^{b}l\right)$. By the Chinese Remainder Theorem, this system has a solution.

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