67th NMO Selection Tests for BMO and IMO

a) Clearly, $K=(a_{2n-1}+a_{2n})/a_n$ is a positive rational number. In fact, $K$ must be integral. To prove this, write $K=p/q$ in lowest terms to deduce that the $a_n$ are all divisible by $q$. Divide them all by $q$ to obtain a new sequence whose corresponding ratios are again $K$. Repetition of the process to the new sequence and its successors shows that the $a_n$ are all divisible by arbitrarily large powers of $q$, so $q=1$ and $K$ is indeed integral. Since the $a_n$ form a strictly increasing sequence, it follows that $K>2$, and since the latter is integral, it is at least $3$. To rule out the case $K=3$, we consider the positive integers $b_n=a_{n+1}-a_n$, show that for every index $m$ there exists an index $n>m$ such that $b_n<b_m$ and reach thereby a contradiction. Indeed, if $K=3$, then $3b_n=b_{2n-1}+2b_{2n}+b_{2n+1}$, so at least one of the three $b$'s in the right-hand member must be less than $b_n$. Consequently, $K\ge 4$.

b) If $N=4$, let $a_n=2n-1$; in this case, the verifications are obvious. If $N\ge 5$, set $a_1=1$ and let $a_{2n-1}=\lfloor (N{a}_n-1)/2\rfloor$ and $a_{2n}=\lfloor N{a}_n/2\rfloor +1$. This sequence satisfies the required ratio condition, $a_{2n-1}$ is obviously less than $a_{2n}$, and it is sufficient to prove that $a_{2n}<a_{2n+1}$. This can be done by noticing that $a_2<a_3$, and showing that if $a_n<a_{n+1}$, then $a_{2n}<a_{2n+1}$. Indeed, $a_{2n+1}-a_{2n}\ge (N{a}_{n+1}-2)/2-(N{a}_n/2+1)=N(a_{n+1}-a_n)/2-2\ge N/2-2\ge 1/2$.