Numbers Divisible by All Primes Less or Equal to Itself

Question

Just to preface, I haven't put too much thought into the following lemma/conjecture: The only number divisible by all primes less or equal to itself is two.

Unless I am overlooking something, this conjecture is not as easy to prove rigorously as one might think. So far, I believe that one way to disprove this claim, to break the "only" part, is to show that the difference in consecutive prime numbers is bounded by a constant. In other words, if $p_i$ denotes the $i$th prime number, and if $$max(p_i - p_{i - 1}) = \Omega(1),$$ we can deduce that there must be some finitely large number that is divisible by all primes less than or equal to itself.

However, I am unable to grasp the intuition or reasoning behind this. To summarize, I wish to know if the conjecture stated earlier has been proven or disproven before, and if not, whether my approach for disproving it is accurate.

6 is divisible by all primes less than or equal to 6, so 2 isn't the only possibilty. — MilesB, Mar 15 '20 at 17:42

score 3 · Answer 1 · answered Mar 15 '20 at 18:13

3

If $n>2$, let $p$ be a prime that divides $n-1$. By assumption, $p|n$, so $p|n-(n-1)=1$, contradiction.

answered Mar 15 '20 at 18:13

saulspatz

53,131

score 1 · Accepted Answer · answered Mar 15 '20 at 17:47

1

It follows from Bertrand's postulate.

answered Mar 15 '20 at 17:47

Robert Israel

448,999

Numbers Divisible by All Primes Less or Equal to Itself

2 Answers2