Find parameter m so that the linear coungrence admits solutions

Question

How do i find values of parameter $m ∈ N$ so that the congruence

$ 5920x ≡ 160 (56m) $

admits solutions?

The only way i found to solve this problem is by looking at factors, and i found that $m$ must be different from $37k$ where $k$ is any positive integer, because the coungruence has solution if and only if $gcd(a,m)$ divides $b$, which is not true if $m =37k$

Better explained:

$5920 = 2^5*5*37$

$160 = 2^5*5$

$56m = 2^3*7*m$

so $gcd(5920,56m) | 160$ if $m$ does not have $37$ as factor.

Is that wrong? And which way is more straightforward for solving this kind of problem?

Answer 1

The conclusion is correct, but you have not proved it. Likely you are implicitly using properties of unique prime factorization, and those properties need to be explicitly invoked in order to judge correctness of the argument.

As for other ways it can also be proved (more generally) using basic gcd laws as below (let me know if you need links for proofs of these standard gcd properties).

$$\begin{align} \exists x\!:\ 5920x &\equiv 160\!\!\!\pmod{56m}\\ \iff\quad\,\ \exists x\!:\ \ \ 740x &\equiv\ \ 20\!\!\! \pmod{7m},\ \ \text{by cancelling }\, 8\\ \iff\qquad (740,7m)&\mid 20, \ \ \text{by congruence solvability criterion}\\ \iff\quad\ (20(37),m)&\mid 20,\ \ {\rm by}\ \ [\![1]\!]\,\ {\rm and}\,\ (740,7)=1\\ \iff (20,m)(37,m)&\mid 20,\ \ {\rm by}\ \ [\![2]\!]\,\ {\rm and}\,\ (20,37)=1\\ \iff\qquad\quad\, (37,m)&=1 \end{align}\qquad$$

Where we applied well-known basic gcd laws $[\![1]\!]$ = here and $[\![2]\!]$ = here.