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For a prime $p_1$ is there an expression, in terms of $p_1$, for a point at which integers may have a prime factor greater than $p_1$?

Also, is it true that for a prime $p_2$, no integers less than ${p_2}^2$ have a prime factor of $p_2$?

ketchupcoke
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    $p!+1$ has a prime factor larger than $p$ for sure. – barak manos Dec 13 '16 at 19:34
  • The only prime factor for prime $p$ is $p$ itself. $1$ is a factor of every integers. – Ng Chung Tak Dec 13 '16 at 19:35
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    The first such number is of course just the next prime. There is no simple known formula for "the next prime" (otherwise there would not be a world record for "largest prime discovered). – Mees de Vries Dec 13 '16 at 19:36
  • What about excluding the primes themselves? Actually, I just realised it would be 2*the next prime. Also @NgChungTak I don't think that was what I was referring to, I am looking for the highest number that does not have a prime factor greater than $p$. Would anyone be able to check the second part to my question? – ketchupcoke Dec 13 '16 at 19:39
  • By definition, a prime number is a natural number that can only be divisible by one and itself, there's only one prime factor for a prime number. – Ng Chung Tak Dec 13 '16 at 19:42
  • @NgChungTak Yes, but I was talking about every integer. So what is the largest integer that definitely does not have a prime factor greater than a prime $p$? – ketchupcoke Dec 13 '16 at 19:44
  • Avoid ambiguity, let $p_1$, $p_2$, etc to clarify your primes – Ng Chung Tak Dec 13 '16 at 19:44
  • $2^n$ for all $n$ has no prime factor greater than $2$. With sufficiently large $n$, you can get an integer as large as you desire. – Fimpellizzeri Dec 13 '16 at 19:47
  • @Fimpellizieri ok, but what about a range of integers where you are certain (or at least, fairly certain) that every number has a prime factor of $p$ or less? – ketchupcoke Dec 13 '16 at 19:50
  • Did you mean $(1)$ for $N>p_2>p_1$ such that $p_2|N$ and determine $N=N(p_1)$? and $(2)$ for $N>p_2>p_1$ such that $p_2|N \implies N>p_2^2$? – Ng Chung Tak Dec 13 '16 at 19:50
  • Is that you want to search another larger prime number $p_2$ such that it divides an integer $N$ (prime or not) larger than the first given $p_1$. For example, the starting prime number is $7$, $8$ has prime factors of $2$ only; $9$ has prime factors of $3$ only; $10$ has prime factors of $2$ and $5$ only. But $11$ has no prime factors less than $7$. – Ng Chung Tak Dec 13 '16 at 19:54
  • @NgChungTak Yes that is what I am looking for, and also the smallest value $N$ could be. But in the case of your example 11 would not be $N$ since it is prime itself. – ketchupcoke Dec 13 '16 at 20:00
  • I guess for given $p_n$ (it means $n$-th prime here) the smallest composite is $N=2p_{n+1}$, that is twice the next prime. – Ng Chung Tak Dec 13 '16 at 20:17

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For a prime $p_1$ is there an expression, in terms of $p_1$, for a point at which they may have a prime factor greater than $p_1$?

Well, $p_1!+1$ has a prime factor greater than $p_1$ for sure.

Though it is not necessarily the smallest integer which has a prime factor greater than $p_1$.

Is it true that for a prime $p_2$, no integers less than ${p_2}^2$ are prime factors of $p_2$?

You probably meant to ask if it is true that for a prime $p_2$, there are no integers less than ${p_2}^2$ whose one of their prime factors is $p_2$.

Well, it is true, with the exception of $p_2$ itself of course.

barak manos
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