First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part)

We shall prove that there are no consecutive kooky primes. Order all the primes in an increasing sequence $p_1=2<p_2=3<p_3<\dots$ and, for the sake of contradiction, suppose that $p_n$ and $p_{n+1}$ are both kooky for some $n>1$. This means that there exist positive integers $a$ and $b$ such that $$p_1+\cdots +p_{n-1}=a{p}_n\text{ and }{p}_1+\cdots +p_n=b{p}_{n+1}.$$ Subtracting one equation from the other gives us $(a+1)p_n=b{p}_{n+1}$. Since a prime $p_n$ divides the product $b{p}_{n+1}$ and doesn't divide $p_{n+1}$, it has to divide $b$, hence we can see that $p_1+\cdots +p_n$ must be divisible by $p_n{p}_{n+1}$. But this is impossible: by noting that $p_{n+1}>n$ for all $n$ and $p_i\le p_n$ for all $i\le n$, we get that $$0<p_1+p_2+\cdots +p_n\le n{p}_n<p_n{p}_{n+1},$$ so it can't be a multiple of $p_n{p}_{n+1}$, which yields the desired contradiction.