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?

Answer for the second question.
We have 
\begin{align}
a = k q_{1} + r_{1} \\
a = k q_{2} + r_{2} \\
a = k q_{3} + r_{3} \\
\end{align}
Where $a > b > c \geq k$.
Let suppose that $r_{1} = r_{2} = r_{3}$
We get $a - b = k (q_{1} - q_{2})$; $a - c = k (q_{1} - q_{3})$; and $b - c = k (q_{2} - q_{3})$.
We deduce that $k$ is a devider for $(a-b)$, $(a-c)$, and $(b-c)$. And the highest value for $k$ is the H.C.F of these three values.
Let prove now if $k$ is the H.C.F of $(a-b)$, $(a-c)$, and $(b-c)$, we would get $r_{1} = r_{2} = r_{3}$.
We have:
\begin{align}
a - b = k (q_{1} - q_{2}) + (r_{1} - r_{2}) \\
a - c = k (q_{1} - q_{3}) + (r_{2} - r_{3}) \\
b - c = k (q_{2} - q_{3}) + (r_{2} - r_{3}) \\
\end{align}
$k$ is a devider of $(a-b)$, $(a-c)$, and $(b-c)$, so  $r_{1} - r_{2} = r_{2} - r_{3} = r_{1} - r_{3} = 0$ and consequently $r_{1} = r_{2} = r_{3}$ 
