Given that $a,b$ are co-prime, we have infinitely many solutions for $x,y$ to the equation $$ax+by=c.$$ Furthermore, solutions have the form: $x=ca^{-1}+tb,y=cb^{-1}-ta$.
Given that $c$ can be chosen free (as can $t$), how many pairs of $x,y$ are co-prime?
i.e. How many pairs of $(ca^{-1}+tb,cb^{-1}-ta)$ are co-prime?
It feels like it should really be infinite because we have 2 degrees of freedom and we are trying to satisfy a linear system. Can we prove it?
Any help would be appreciated.
Yes, you are right. For such solutions $( x, y)$ to be co-prime, it is sufficient that $c = 1$ by Euclid's gcd algorithm. The number of such solutions are infinite as can be obtained by varying the values of $t$ in your setup.
As an example, consider the equation $2x - 3y = 1$. Then all pairs of the form $( -1 + 3t, -1 + 2t)$ are solutions to this equation and are co-primes.
Of course, if there is some further restriction on the values that $x$ and $y$ can take, then it might only be finitely many.