If for two integers $a,b$, $a=b \mod x$, what is $x$ in terms of $a$ and $b$?.
I think the answer is $a-b$, but I'm not sure how to prove it without modular arithmetic, which I don't really understand.
Is there a simple way to prove and this without mods?
This follows directly from the definition of congruence with respect to modular arithmetic (there's not much you need to know):
$$a \equiv b \pmod{x} \;\;\; \Leftrightarrow \;\;\; x | (a-b)$$
(This latter expression, if you're not familiar, just means "$x$ evenly divides $(a-b)$," or, equivalently, "$(a-b)$ has no remainder when divided by $x$.")
The problem is that, in reality, this means that, for some integer $k$,
$$\frac{a-b}{x} = k \;\;\; \Leftrightarrow \;\;\; x = \frac{a-b}{k}$$
To say that $x = a-b$ (i.e. that $k=1$) means that there's some sort of context we're missing that would apply in this context. As presented, however, I think this is as far as one can get.