Iranian Mathematical Olympiad

a) Let $a_1=2,a_2=4,\dots ,a_k=2k,a_{k+1}=1,a_{k+2}=3,\dots ,a_{2k}=2k-1$. We can easily check that $a_i{a}_{i+1}+1$ is a perfect square for $1\le i\le 2k$ except $i=k$, which can be repaired if $2k+1$ is a perfect square which is possible for infinitely many values of $k$.

b) Let $a_1,a_2,\dots ,a_n$ be a cubic permutation. Let $2^k$ be the largest power of $2$ less than or equal to $n$. By the definition of the cubic permutation we know that $2^{k}u+1=x^3$, where $u$ is an element of the permutation. So we have $2^{k}u=(x-1)(x^2+x+1)$. Hence we conclude that $2^k|(x-1)$. Because of the way that $k$ is chosen, we have $n<2^{k+1}$. So we have $2^k\le x-1\le n^{\frac{2}{3}}<2^{\frac{2}{3}(k+1)}$ which is a contradiction. Hence no such permutation exists.