<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] Unfortunately, your claim is incorrect.  The best you can say is that if three numbers $a$, $b$, and $c$ have the same remainder by division by $x$, then their differences are multiples of $x$, not equal to $x$.
There are many examples that show this, for example, when $a=15$, $b=9$, $c=3$ and $x=2$, then $a$ has remainder $1$ when divided by $2$, $b$ has remainder $1$ when divided by $2$, and $c$ has a remainder of $1$ when divided by $2$.  Their differences, however are $15-9=6$, $9-3=6$, and $15-3=12$, all of which are divisible by $6$, which is only a multiple of $2$, not equal to $2$.
One can prove the following claim: Suppose that $a$ and $b$ have the same remainder when divided by $x$.  Then $x$ divides $a-b$.
Proof: Suppose that the remainder of $a$ divided by $x$ is $r$.  Then $a=q_1x+r$.  By assumption $b$ has the same remainder, so $b=q_2x+r$.  Then $a-b=(q_1-q_2)x$, which is an integer times $x$, so the difference is a multiple of $x$, but not necessarily equal to $x$ (when $q_1-q_2\not=1$).
A corollary of this is that the HCF of $a-b$, $a-c$, and $b-c$ will be divisible by $x$, but still may be not equal to $x$, unless you choose the values in a lucky way.  The problem with the example above is that for $a$, $b$, and $c$, $q_a-q_b$, $q_a-q_c$, and $q_b-q_c$ are all divisible by $3$.
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