Let’s say that $p,q,r$ are positive integers, with $\gcd(p,q)=1$ and $r=kp+mq$ being the Bézout representation. When $(k,m)$ are “minimal”, we know that $\lvert{m}\rvert < \lvert{p}\rvert$.
Now let’s say that [by other means] I’ve proven that $p \mid (m^2-2)$. Is there a way that I can guarantee the size of the quotient? e.g., can I ever say “For one of the two minimal values of $m$, we must have $m^2-2=p$”? If that’s not always guaranteed, what are the ways I can go about finding a concrete quotient for $\tfrac{m^2-2}{p}$?
EDIT: Regarding the “by other means” comment… One relationship between the three variables is $$r^2=p^2+2pq+2q^2\!.$$ Substituting $r=kp+mq$ into that gives $p \mid (m^2-2)$ and $q \mid (k^2-1)$. Based on other relationships I know to be true of the numbers involved, I know that $p=7$ and $-3,4$ are the two minimal coefficients, so $m^2-2 = (-3)^2-2 = 7 = 1(7)$ and $m^2-2 = (4)^2-2 = 14 = 2(7)$ give $1$ and $2$ as the two related quotients.
Comment:
An experimental approach:
We can see that if (p, q)=1 and p, q and r are constant then there is only one linear relation between them. I show this with some examples:
$(r, p, q)=(452, 31, 47)\rightarrow (k=7)\times 31+(m=5)\times 47 =452$
$(r, p, q)=(2034, 35, 67)\rightarrow (k=16)\times 35+(m=22)\times 67=2034$
But if $(p, q)\neq1$ or p and q have a coommon factor greater than 1, then the linear relation is not unique, for example:
$(p, q, r)=(1990, 35, 65)\rightarrow (k, m)=(3,29), (16, 22), (29, 15)$
In this case better question is: when ($k+m$) is minimum. In this example $3+29=32$ is minimum.