India MO 1996

[alexlor8083]

(i) Given any positive integer n, show that there are distinct positive integers
a, b such that a + k divides b + k for k = 1, 2, ... , n.
(ii) If a, b are positive integers such that a + k divides b + k for all positive integers k, show that a = b.

[Alex89]

a)Ok it's only Chinese Theorem.
$ b+1 \equiv 0 \pmod{a+1} $
$ b+2 \equiv 0 \pmod{a+2} $
$ ... $
$ b+n \equiv 0 \pmod{a+n} $
So we have $ b \equiv -k \pmod{a+k} $ for every k such as $ 1 \le k \le n $. But $ b \equiv -k \equiv a \pmod{a+k} $ for every k. For Chinese Theorem there are infinite solution of b for every a.

[Alex89]

b)For the same reason we have $ b \equiv a \pmod{a+k} $ for every k. So
$ b \equiv a \pmod{x} $ for every x integer such as $ x>a $. But if b is not equal to a, we have b>a because if $ a+k|b+k $ we have $ b+k \ge a+k $ and so $ b \ge a $. But if b>a and $ b \equiv a \pmod{x} $ for every x such as $ x>a $ we have that if b=x we have $ a \equiv 0 \pmod{x} $, that is impossible because a is positive and $ a<x $. So b=a

[albert_K]

i) explicitly: $ $ (1, (n+1)! + 1) $ $is a solution

ii) if $ $ a \not = b, \exists k_0 \forall k>k_0 $ $ $ \displaystyle 1 < \frac{b+k}{a+k} < 2 $

[edriv]

Alex89's solution is mistaken

By the way, you noticed that $ ~ a+k \mid b+k \Leftrightarrow a+k \mid b-a $, so b-a has just to be a multiple of $ ~ \mbox{lcm}((a+1),(a+2),\ldots,(a+n)) $ (this is a necessary and sufficient condition), and from now it'easy.

(if you want to know why your solution is mistaken... read the statament of the chinese remainder theorem!)

[albert_K]

in my opinion, this is Number Theory.