Romanian Mathematical Olympiad

Assume, by way of contradiction, that $M$ contains the infinite arithmetic sequence $\{b_1^k,b_2^k,b_3^k,\dots \}$, where $1\le b_1<b_2<b_3<\dots$. If we denote by $r$ its common difference, we have $b_{n+1}^k-b_n^k=r$, for all $n\ge 1$. Observe that $$r=b_{n+1}^k-b_n^k=(b_{n+1}-b_n)(b_{n+1}^{k-1}+b_{n+1}^{k-2}{b}_n+\cdots +b_{n+1}{b}_n^{k-2}+b_n^{k-1}).$$ The expression in the second parenthesis is clearly increasing, hence the sequence $b_{n+1}-b_n$ is decreasing, and since all its terms are positive integers, we reached a contradiction.