Consecutive numbers with prime factorization with powers at least two
问题内容
Its easy to show that there are infinite amount of two consecutive numbers $n, n+1$ such that in their prime factoring all primes are in power at least two. It is because if one have such $n, n+1$ then construct another $(2n+1)^2 - 1, (2n+1)^2$ ; we start with $(288,289)$. But
are there three (or more) such consecutive numbers with above property. Set of such triples is infinite or not ? ; if they exists: How to find them?!
回答 (1)
This is an open problem, posed by Erdos. It is widely believed that there are no such triples, or at least that there are only finitely many, but no proof is known. Finiteness, at least, would follow from the $abc-$conjecture, see, e.g., this
For reference, natural numbers $n$ with the property that, for primes $p$, we have $p\,|\,n\implies p^2\,|\,n$, are called powerful. Equivalently, powerful numbers are natural numbers of the form $a^2b^3$, for some natural numbers $a,b$.
It is easy to see that there can not be four consecutive powerful numbers. Indeed, in any string of four consecutive natural numbers, one is of the form $4k+2$ and hence is not powerful.