70th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

The required numbers are $n_1=n_2=\cdots =n_k=1$.

For every integer $m>1$, let $p(m)$ denote the least prime divisor of $m$. We show that, if $m$ and $\ell$ are integers greater than $1$, and $m\mid 2^{\ell }-1$, then $p(m)<p(\ell )$. Since $p(m)$ is odd, $p(m)\mid 2^{p(m)-1}-1$, and since $p(m)\mid 2^{\ell }-1$, it follows that $p(m)\mid 2^{\gcd (\ell ,p(m)-1)}-1$. Notice that $\gcd (\ell ,p(m)-1)>1$, since $p(m)>1$, to infer that $\ell$ has a prime divisor not exceeding $p(m)-1$, and conclude thereby that $p(m)<p(\ell )$.

Suppose now, if possible, that $n_1>1$. Then $n_k>1$, so $n_{k-1}>1$, and so on and so forth all the way down to $n_2>1$. Hence $p(n_1)<p(n_2)<\cdots <p(n_k)<p(n_1)$ which is a contradiction. Consequently, $n_1=1$, so $n_2=1$, and so on and so forth all the way up to $n_k=1$.