OliForum Matematico

(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.

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.

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

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 $

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!)

in my opinion, this is Number Theory.