Infinitely many values of $x$ that satisfy the equations

Question

I was working on number theory homework, and this problem was the second part to this problem.

Let $a,b,c,d,m$ be constants and $n$ and $k$ be coprime such that there are infinitely many values of $x$ that satisfy

$$ax+b=mn$$ $$cx+d=mk$$

Find the values of $ad-bc$ such that the above is true.

Now, according to the answer suggested above, these equations are true if $n \equiv b/m\pmod a,$ and $k \equiv d/m\pmod c,$ but I can't really find any insight on how to find the values of $x$ that would help with this part of the problem.

Answer 1

For constants $a,b,c,d,m$, you're asking about finding an infinite number of coprime $n$ and $k$ for which combination provides a solution for $x$ in

$$ax+b=mn \tag{1}\label{eq1B}$$

$$cx+d=mk \tag{2}\label{eq2B}$$

In particular, you're asking about any restrictions or conditions on $ad - bc$. Multiply \eqref{eq2B} by $a$ and subtract \eqref{eq1B} multiplied by $c$ to get

$$ad - cb = mak - mcn = m(ak - cn) \tag{3}\label{eq3A}$$

This means that $m \mid ad - bc$, i.e., you require for some $e \in \mathbb{Z}$ that

$$ad - cb = em \tag{4}\label{eq4A}$$

This is the requirement on $ad - bc$ for there to possibly be any solutions.

Next, multiply \eqref{eq1B} by $d$ and subtract \eqref{eq2B} multiplied by $b$, and use \eqref{eq4A}, to get

$$adx - cbx = dmn - bmk \implies (ad - cb)x = m(dn - bk) \implies ex = dn - bk \tag{5}\label{eq5A}$$

Let

$$f = \gcd(b,d) \implies b = fg, \; d = fh, \; \gcd(g,h) = 1 \tag{6}\label{eq6A}$$

Substituting \eqref{eq6A} in \eqref{eq5A} gives

$$ex = f(hn - gk) \tag{7}\label{eq7A}$$

Since $\gcd(g,h) = 1$, Bézout's identity states there exists integers $n_0$ and $k_0$ such that

$$1 = hn_0 - gk_0 \implies qe = h(qen_0) - g(qek_0), \; q \in \mathbb{Z} \tag{8}\label{eq8A}$$

Having $n = qen_0$ and $g = qek_0$ means, using \eqref{eq8A} in \eqref{eq7A}, that

$$ex = feq \implies x = fq \tag{9}\label{eq9A}$$

resulting in an infinite # of $x$. However, there's a requirement that $n$ and $k$ be coprime. You also have

$$h(qen_0 + rg) - g(qek_0 + rh) = h(qen_0) - g(qek_0) = qe, \; r \in \mathbb{Z} \tag{10}\label{eq10A}$$

Note if $\gcd(qen_0 + rg, qek_0 + rh) = s$ then, using \eqref{eq8A}, you get

$$s \mid n_0(qek_0 + rh) - k_0(qen_0 + rg) = r(hn_0 - gk_0) = r \tag{11}\label{eq11A}$$

This means for any $r$ chosen so that $\gcd(qek_0,r) = \gcd(qen_0,r) = 1$ , you get $s = 1$. Thus, for any such $r$, which can always be found (e.g., $r = tqek_0 n_0 + 1, \; t \in \mathbb{Z}$), you can use $n = qen_0 + rg$ and $k = qek_0 + rh$ since they are coprime which, along with $x$ given by \eqref{eq9A}, result in a solution for \eqref{eq1B} and \eqref{eq2B}. Varying $q$, as used in \eqref{eq8A}, means you can generate an infinite number of solutions for $x$.

Answer 2

Above equation shown below:

$ax+b=mn \tag{1}\label{eq1A}$

$cx+d=mk \tag{2}\label{eq2A}$

Taking constant's as , $(a,b,c,d,m)=(16,8,2,13,3)$ we get:

$(ad-bc)=192$

Also,

$3n-16x=8$ -----$(1)$

$3k-2x=13$ ----$(2)$

Equation (1) & (2) has parametric solution shown below:

$(x,n,k)=[(3w+1),(16w+8),(2w+1)]$

For, $w=2$ we have:

$(x,n,k)=(7,24,9)$

Answer 3

The example serves well as it is quite intuitive.

However, two numerical errors are involved: ...... Also,

3n−16x=8 (1)

3k−2x=13 (2)

Equation (1) & (2) has parametric solution shown below:

(x,n,k)=[(3w+1),(16w+8),(2w+5)]

For, w=2 we have: (x,n,k)=(7,40,9)