Is it possible to define $b$ in terms of $c$, $ d$, and $mk$ given that $a-b=m k=c-d$?

Question

Given the following: $$a-b=mk, \qquad c-d=mk$$ For some integers $a,b,c,d,m,k$ I need to express $b$ in terms of $c,d,$ and $mk$ in order for my proof to work, but no matter how I fiddle around with this I can’t do it. Is it possible?

Answer 1

Suppose you could and you had $b = f(c,d)$.

Let $c_0; d_0$ be a pair of numbers so that $c_0-d_0 = mk$. That would mean that if $a - b = mk$ that $b$ must be equal to $f(c_0, d_0)$. But notice $(a+1) - (b+1) = mk$ as well. So that means that $b+1 = f(c_0, d_0)$ as well. But $f(c_0, d_0)$ can't be equal to two different things.

If you think about it, this question makes no sense. There are an infinite number of pairs of $(a= mk + b, b)$ so that $a-b = mk$ and there are an infinite number of pairs of $(c,d)$ where $c-d = mk$. Which $b$ you pick to get $a-b = mk$ and which pair of $(c,d)$ you pick to get $c-d = mk$ are completely independent. There's no way picking a specific $(c,d)$ can determine one precise $b$ because... you can always just pick another $b$ instead.