# Pair of consecutive squarefree numbers with $n$ distinct prime factors

Let $(a,a+1)$ be the smallest pair of squarefree numbers with $n$ distinct prime-factors, if such a pair exists.

The solutions for small $n$ are:

\begin{array} nn & (a&a+1) & \text{Factorisation}\\ \hline 2 & 14 & 15 & [2, 1; 7, 1] & [3, 1; 5, 1]\\ 3 & 230 & 231 & [2, 1; 5, 1; 23, 1] & [3, 1; 7, 1; 11, 1]\\ 4 & 7314 & 7315 & [2, 1; 3, 1; 23, 1; 53, 1] & [5, 1; 7, 1; 11, 1; 19, 1]\\ 5 & 378014 & 378015 & [2, 1; 7, 1; 13, 1; 31, 1; 67, 1]& [3, 1; 5, 1; 11, 1; 29, 1; 79, 1]\\ 6 & 11243154 & 11243155 & etc. \end{array}

Does a pair exist for any natural number $n$?

• Just checking: you mean consecutive squarefree numbers with exactly $n$ distinct prime factors each, right? Jan 29, 2014 at 1:27
• Yes, exactly n distinct prime factors. Jan 29, 2014 at 1:28
• This is related to oeis.org/A052215 (which unfortunately doesn't give any results beyond what's figured out in the answers to this question). Jan 28, 2016 at 14:44

To show that it is possible for two consecutive numbers to have $n$ or more prime factors each (revised problem as mentioned in comments) is fairly easy: choose any $n$ different primes you like and let $m_1$ be their product; choose any other $n$ primes you like and let $m_2$ be their product. Then you want to find $a$ satisfying $$a\equiv0\pmod{m_1}\quad\hbox{and}\quad a\equiv-1\pmod{m_2}\ ;$$ but $m_1$ and $m_2$ have no common factor, so the Chinese Remainder Theorem guarantees that this system of congruence equations has a solution.
However $a$ and $a+1$ will very likely have prime factors other than those we have "planted", so this does not ensure that $a$ and $a+1$ are squarefree.