59th Ukrainian National Mathematical Olympiad

a) If $s_n=p$ is prime for some $n$ larger than $a$ and $b$, then for some $k$ and $l$ the numbers $a^k-1$ and $b^l-1$ are divisible by $p$. If we set $m=kl$, then both $a^{mt}-1$ and $b^{mt}-1$ are divisible by $p$ for all natural numbers $t$. This implies that the number $$s_{n+mt}=a^{n+mt}+b^{n+1+mt}=a^n+b^n+a^n(a^{mt}-1)+b^{n+1}(b^{mt}-1)$$ is divisible by $p$ for all $t$, which gives an infinite amount of composite members in the sequence of $s_n$.

b) All prime divisors of $s_n$ either divide both $a$ and $b$ or none of them. Let $q$ be a common prime divisor of $a$ and $b$, and $q^k$ and $q^l$ are the largest powers of $q$ that are divisors for $a$ and $b$ respectively. If $k\le l$ then $kn<l(n+1)$, and if $k>l$ then $kn>l(n+1)$ for all large enough $n$. Therefore, for sufficiently large $n$, the prime divisor $q$ raised to one of the powers $kn$ or $l(n+1)$ divides $s_n$, implying that the greatest common divisor of $a^n$ and $b^{n+1}$ cannot be larger than $d^{n+1}$, where $d=(a,b)$ – GCD of $a$ and $b$. Let $p$ be a prime number dividing neither $a$ nor $b$. Let $p^k$ be the largest power of $p$ that is divisor of $b+1$ (maybe, $k=0$). For some natural number $m$ both $a^m-1$ and $b^m-1$ are divisible by $p^{k+1}$. Then, for some $n$ divisible by $m$, the number $s_n=(b+1)+(a^n-1)+b(b^n-1)$ is divisible by $p$ raised to the same power as $b+1$. Now we can finally get back to the original problem. For the sake of contradiction, we assume that there are only finitely many primes dividing some $s_n$. In particular, this means that there are only finitely many prime numbers $p_1,p_2,\dots ,p_j$ which do not divide $a$ or $b$, but divide $s_n$ for some $n$. We have just shown that for all such $p_i$ there is $m_i$ such that for all $n$ divisible by $m_i$, the largest number, such that $p_i$ raised to the power of this number is a divisor of, $s_n$ is smaller than the largest number, such that $p_i$ raised to the power of this number, is a divisor of $b+1$. However, if we now choose $n$ divisible by all $m_i$, then we will get that $s_n=a^n+b^{n+1}$ is not greater than $d^{n+1}(b+1)$. However, this cannot be the case for all $n$ because of the fact that one of $a$ or $b$ is bigger than $d$.