If $ax + by =$ prime, are then $a$ and $b$ relative prime?

Question

I'm stuck on the following question:

For $a, b \in \Bbb Z$, assume that $ax + by = 4$ and $as + bt = 7$ for $x, y, s, t \in \Bbb Z$. Show that then $a$ and $b$ are relative prime.

The following is what I tried, but I need verification or your help:

Adding up the equations $ax + by = 4$ and $as + bt = 7$ results in $a(s + x) + b(t + y) = 11$. Then, since $s + x, t + y \in \Bbb Z$ we have the following: $am + bn = 11$ for $m, n \in \Bbb Z$. And now my conclusion: Since $11$ is a prime, $a$ and $b$ must be relative prime.

Is my conclusion correct? If so, according to which theorem. Otherwise, what am I missing?

Answer 1

Your conclusion is not correct: $11$ might divide both $a$ and $b$.

If $d=\gcd(a,b)$, then $d\mid ax+by$ and $d\mid as+bt$. That is, $d$ divides both $4$ and $7$. But the only natural number which divides both $4$ and $7$ is $1$. In other words, $\gcd(4,7)=1$.

Answer 2

What if $a$ and $b$ are both multiples of $11$. You still have to argue to rule out that case.

$$2(ax+by)-(as+bt)=1$$

$$a(2x-s)+b(2y-t)=1$$

Hence $\gcd(a,b)=1$.

We can also directly use $\gcd(a,b) \mid 4$ and $\gcd(a,b)\mid 7$ to conclude that $\gcd(a,b)=1$ since $\gcd(a,b) \in \{1,7\}$ but since $\gcd(a,b) \le 4$, $\gcd(a,b)=1$.