<s>[INST] Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$? Suppose two numbers $a$ and $b$ as, $a=kq_1+r_1=3\times 17 + 1  = 52$ and $b = kq_2+r_2=3 \times 15 +1=46$.
It is clear that $52$ and $46$ leave the same reminder 1 when divided by $3$, because I designed them this way. But surprisingly however I design the numbers the largest $x$ which leaves the same reminder is $kq_1-kq_2=k(q_1-q_2)$. Why is that? In this case we have $52 = 6\times 8+ \color \red  4$ and $46 = 6\times 7 + \color \red 4$.
Now suppose there are three numbers $a$, $b$, $c$ and $x$(assuming $a>b>c \geq x$) such that $x$ leaves the same reminder when we divide each of $a,b$ and $c$ with it. $x$ is supposed to be the largest possible value that holds the assertion.  Now $x$ is given by the H.C.F of $a-b, a-c$ and $b-c$. Why is that? How can we prove this mathematically?
 [/INST] Answer for the first question:
we have $a = k q_{1} + r$ and $ b = k q_{2} + r$. (Where $a >b$, $r < k$) 
We deduce that $a - k q_{1} = b- k q_{2}$.
and then,
$$k = \frac{a - b}{q_{1} - q_{2}}$$ 
The highest value of $k$ is when $q_{1} - q_{2} = 1$.
Let prove that it is always possible.


*

*Suppose that $k = a - b$


we have $a = (a - b) q_{1} + r_{1}$ and $b = (a - b) q_{2} + r_{2}$
suppose that: $r_{1} = r_{2}$
Then: $(a - b) = (a - b) (q_{1} - q_{2})$
We get: $q_{1} - q_{2} = 1$
Now let suppose that $q_{1} - q_{2} = 1$
similarly we get $r_{1} = r_{2}$
So we have just proved that:
\begin{align}
 (k = a - b) \Rightarrow ( (r_{1} = r_{2}) \Leftrightarrow  (q_{1} - q_{2} = 1))
\end{align}
Now, let suppose that $k = a - b$, $r_{1} \neq r_{2}$ and $q_{1} - q_{2} \neq 1$
We get 
\begin{align}
a - b = \frac{r_{1} - r_{2}}{1 - q_{1} + q_{2}}
\end{align}
It is clear from this last formulae that:
\begin{align}
r_{1} \geq (a-b) \ \ \ \ \text {or } \ \ \ \ r_{2} \geq (a-b)
\end{align}
which is a contradiction.
This means that: 
\begin{align}
 (k = a - b) \Rightarrow ( (r_{1} = r_{2}) \ \ \ and  \ \ \  (q_{1} - q_{2} = 1))
\end{align}
</s>