Application of Euclidian Division To A System of Equations

Question

Conjecture:

if $a,b,x,y,z \in \mathbb{Z}$, $a \neq 0$ and $0 < x$ then the system \begin{cases} xy=a & \text{(1)}\\ x(1-yz)=b & \text{(2)} \\ \end{cases} has a unique integral solution $\left( x,y,z \right)$.

-Is the conjecture true?

My work so-far:

Expand the LHS of $(2)$ to obtain that $x-\left( xy \right)z=b$; substitute $(1)$ into $(2)$ and rearrange to obatin that $b=\left( -z \right)a+x$. As both $x$ and $y$ are integers, obtain that $ x \le|a|$. By euclidian division obtain that the last equation is satisfied by a unique $-z$ and a unique $x$; use $(1)$ to obtain a unique $y$ and thus there's a unique integral solution $\left( x,y,z \right)$ to the above system.

-Is the above correct?

Answer 1

Your proof of uniqueness is correct.

Let's try a different approach.

The second equation, together with $xy=a$, entails $x=b+az$; this implies $\gcd(a,b)$ divides $x$; since also $x\mid a$ and $x\mid b$, we conclude that $x=\gcd(a,b)$.

Now let $a'=a/x$ and $b'=b/x$. Then we have $y=a'$ and $1-a'z=b'$, so we get a necessary condition for existence of a solution, namely $$ a'\mid(b'-1) $$

This condition is also sufficient: in this case the unique solution is $$ \begin{cases} x=\gcd(a,b) \\[4px] y=\dfrac{a}{\gcd(a,b)} \\[4px] z=\dfrac{\gcd(a,b)-b}{a} \end{cases} $$

Answer 2

The conjecture is false because of existence, not uniqueness (your proof is correct). Try with $ a = 3 $ and $ b = 5 $. The counterexample is based on the fact that you have a unique couple $ (-z,x) $ that comes from the euclidean division, but $ x $ must be a divisor of $ a $, from equation no. 1. Please test this counterexample, in order to check it.