$\gcd(a,b)=1$ from $2\min(a,b)-\max(a,b)=p$ where $p$ is odd prime

Question

Let $a,b$ be odd positive integers.

Suppose we have $$2\min(a,b)-\max(a,b)>0$$ $$2\min(a,b)-\max(a,b)=p$$ $$\min(a,b)\ne0\operatorname{mod}p$$ where $p$ is odd prime number.

Then I conjecture that $$\gcd(a,b)=1$$

Is there a way to prove it?

Answer 1

WLOG, assume $b \geq a$. Then your second condition says that $2a = b + p$. Now suppose that $q$ is a prime that divides $a$. If it also divided $b$, then it would also have to divide $p$; but then we would have to have $q = p$. But your third condition says that $p \not\mid a$. Thus there is no prime that divides both $a$ and $b$, so $\gcd(a, b) = 1$.

Note that you do not need the first requirement, nor the requirement that $a, b$ are odd.