I’m trying to prove the following.
Lemma. Let $a,b,c$ be odd integers with $a>b>c\ge 1$. Then there exist positive integers $u$ and $v$ of opposite parity such that $a=bu-cv$. In particular, a solution $(u,v)$ can be found with $u>v \ge 1$ if and only if BLAH.
Proof (incomplete). Since $b$ and $c$ are relatively prime, Bézout’s identity shows that there exist integers $u_0$ and $v_0$ such that $$ bu_0-cv_0=1. \tag{$\star$}$$ For $u=au_0-b$ and $v=av_0-c$, we have $bu-cv = a(bu_0-cv_0)=a$, as desired. As Bézout's identity shows, there are an infinite number of pairs $(u_0,v_0)$ satisfying ($\star$); therefore, it is possible to select a solution such that $u$ and $v$ are both positive. Since $b$ and $c$ are both odd, ($\star$) implies that $u_0$ and $v_0$ are of opposite parity; since $a$ is odd, $u$ and $v$ are of opposite parity. Now note that $$a=bu-cv = b(u-v)+v(b-c) \quad\implies\quad \frac{a-v(b-c)}{b} = u-v.$$ If and only if $u>v$, we have $u-v \ge 1$, and thus $$\frac{a-v(b-c)}{b} \ge 1 \quad\implies\quad \frac{a-b}{b-c} \ge v \ge 1.$$
My question is: What final restrictions or characterizations (if any) can be placed on the numbers involved? Since $v$ is an integer, we evidently must have $(a-b)/(b-c) \ge 1$, which is to say $a \ge 2b-c$; therefore no solution with $u>v$ can be found if $a < 2b-c$. But is $a \ge 2b-c$ sufficient to guarantee that at least one solution $(u,v)$ exists with $u>v$?