Find pairs $(a,b,c,d)$

Question

Given that $\gcd(am + b, cm + d) = 2019$ for every positive integer m, find all possible values of $|ad-bc|$ with $(a,b,c,d) \in\ \mathbb{Z}_+ $

How can I solve this problem? I am very new to number theory and don't understand how to manipulate it to get a solution.

I considered making the equation into a Diophantine equation but don't know how to find the constant values of the Diophantine.

Any answer is appreciated.

Thank you for your patience!

Answer 1

With your stated condition, as you said in your question, you can create $2$ Diophantine equations. However, you don't need to directly solve for the constant values as the question only asks for what $|ad - bc|$ can be. Instead, note $\gcd(am+b,cm+d) = 2019$ means you have for every positive integer $m$ that

$$am + b = 2019e \tag{1}\label{eq1A}$$

$$cm + d = 2019f \tag{2}\label{eq2A}$$

for some integers $e$ and $f$, with $\gcd(e,f) = 1$. Multiply \eqref{eq1A} by $d$ and subtract \eqref{eq2A} multiplied by $b$ to get

$$(ad - bc)m = 2019(de - bf) \tag{3}\label{eq3A}$$

Next, let

$$\gcd(b,d) = g, \; b = gh, \; d = gi, \; \gcd(h,i) = 1 \tag{4}\label{eq4A}$$

Thus, \eqref{eq3A} then becomes

$$(ad - bc)m = 2019(g)(ie - hf) \tag{5}\label{eq5A}$$

Note that $ie - hf$, due to the variables involved between the $2$ terms not having any common factors, can be basically any integer. However, you do have that $2019(g)$ must divide $(ad - bc)m$. Since this has to be true for all positive integers $m$, including those which have no common factors with $2019(g)$, it can only always happen if

$$2019(g) \mid |ad - bc| \tag{6}\label{eq6A}$$

Next, multiply \eqref{eq1A} by $c$ and subtract \eqref{eq2A} multiplied by $a$ to get

$$cb - ad = 2019(ce - af) \tag{7}\label{eq7A}$$

Let

$$\gcd(a,c) = j, \; a = jk, \; c = jn, \; \gcd(k,n) = 1 \tag{8}\label{eq8A}$$

Thus, \eqref{eq7A} becomes

$$cb - ad = 2019(j)(ne - kf) \tag{9}\label{eq9A}$$

As before, this means

$$2019(j) \mid |ad - bc| \tag{10}\label{eq10A}$$

If you let

$$q = \text{lcm}(g,j) \tag{11}\label{eq11A}$$

Then combining \eqref{eq6A} and \eqref{eq10A}, you get overall that $|ad - bc|$ must be a non-negative integral multiple of $2019(q)$.