BOMC-2012 Test Day 2 Problem 1

[*Mahi*]

*Mahi* vai ...


Please check this and show me the right path ...

Let $a_1-a_0=d_1$ , $b_1-b_0=d_2$

In the progressions $a_0,a_1...$ and $b_0,b_1...$
$a_i=a_0+d_1i$ ,$b_j=b_0+d_2j$; $(i,j=0,1,2,3...)$

According to the statement ,
If $n$ be an integer , then we can find $i,j$ such that
$n=a_i-b_j=a_0+d_1i-b_0-d_2j$ $.... (i)$

Let $d_2$ be negative , And $d_1$ positive .
then $d_2=-|d_2|$
So $|d_1|i+|d_2|j=n+b_0-a_0$ $....(ii)$

Since $i,j \geq 0$ , $|d_1|i+|d_2|j \geq 0$ .
Put $n=a_0-b_0-k$
For any natural values of $k$ ,
$n+b_0-a_0=-k<0$ (contradiction with $(ii)$)

So not for all but for some $n$'s there exist $i,j$ such that $a_i - b_j=n$

The problem is in the red part. Notice that $|d_1|x+|d_2|y=1$ is a linear Diophantine equation and thus has infinitely many solutions (You can see any number theory book, like Elementary Number Theory by Gareth and Mary Jones). So even if we have $i.j \geq 0$, we can always find $i',j'$ which is not necessarily greater than $0$ (for the proof of this, you can see my solution, where I explained how infinitely many solutions of $d_1x+d_2y=1$ can be derived.)

[Nadim Ul Abrar]

It is stated that given AP's are $a_0,a_1,...$ And $b_0,b_1,...$ .
Then how can we consider negative $i',j'$ negative ??

( I think I'm Asking Question's like a stupid , )

[*Mahi*]

One of $i',j'$ can be negative.

[Nadim Ul Abrar]

Actually i wanted to say that what will $a_{-1}$ be ?

[*Mahi*]

There will not be any $a_{-1}$ because $\gcd(2,3)=1$ and we can find infinitely many solutions of $2x+3y=1$, in many of those $i'$ or $j'$ is negative.