I’ve never tried approximating Diophantine solutions before, and have a beginner question.
In a certain system of equations, I have positive integers $a,b,c,d$ that by assumption satisfy the system. I’ve proven that $a > c$ and $d > c$ and $ad>bc$ [and a few other restrictions and characteristics that I believe don’t affect my question]. I’ve also found some upper and lower bounds on possible solutions: $2 \ge d/c > 91/50$ and $2 + \sqrt{2} > (ac+bd)/(ad-bc) \ge 3$ and so on. As a result of combining these bounds, I’ve determined that “$b \approx d$”, and as a result found that one of the equations in the system can be “factored” as: $$(a-c)(2b-a-c) \approx 1.$$
QUESTION: Can I now say $a-c \approx 1$ and $2b-a-c \approx 1$, or something similar? What ways can I qualify the $\approx$, both in the $b \approx d$ case and my “factorization”, so that I can proceed in a valid way with my [attempted] solution of the problem I’m tackling?
(Note: I’m trying to prove that the only positive integer solution to the equation I’m investigating is $(a,b,c,d)=(2,2,1,2)$, so I know a priori that in $b \approx d$ and $a-c \approx 1$ and $2b-a-c \approx 1$ the $\approx$ are true equalities.)