1

I came across this question by accident. The question is,

For how many integer values of $m$; will $m, m+4, m+14$ all be prime numbers.

Using grunt work, I found that between $1-100$, only $m=3$ holds. But still I wanted an algebraic proof which I couldn't find. I would be glad for any help. Thanks!

Bill Dubuque
  • 282,220
inc0gnizant
  • 57

1 Answers1

0

All prime numbers except $2$ and $3$ are of the form $6k+1$ or $6k+5$, where $k$ is an integer. If $m = 6k+5$ for some integer $k$, then $m+4$ is $6k+9$, which is divisible by $3$. If $m = 6k+1$ for some integer $k$, then, $m+14 = 6k+15$, which is again divisible by $3$. So, for any prime number $m$, the triplet has at least one composite number, unless $m = 2$ or $3$. Checking for those separately shows that for $m=3$, we have the said property.

Shubhav Jain
  • 344
  • 1
  • 9
  • It would have to prove to OP that any prime other than $2$ and $3$ $\in \mp 1+6\mathbb Z$ https://math.stackexchange.com/questions/41623/is-that-true-that-all-the-prime-numbers-are-of-the-form-6m-pm-1 – Stéphane Jaouen Apr 11 '24 at 05:44
  • 3
    Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Apr 11 '24 at 09:49
  • 1
    @StéphaneJaouen I have corrected the solution. Thanks for pointing it out. – Shubhav Jain Apr 12 '24 at 03:57
  • @BillDubuque Understood. I didn't realize there was a simpler way to do this question, and thus, didn't think it was a dupe answer. – Shubhav Jain Apr 12 '24 at 03:58
  • 1
    It's best for site health to delete this duplicate answer (saves much time for others organizing the site). – Bill Dubuque Apr 12 '24 at 04:09