Find all positive integers $n$ such that $n$,$n + 2$, and $n + 4$ are all primes.

  • Having a tough time with this problem, I feel that brute force is a possibility especially considering that my professor said that there are only a small amount of numbers that fit this category, but there has to be a better way. Any help or suggestions are greatly appreciated. I did see a similar post to this question but it did not help me solve this nor did it look to find the same thing, please do not mark this as a duplicate.
  • $\begingroup$ Have you looked at this for small values of $n$, say $n=1,2,3,4,5$? Do you find any small numbers $n$ where this is true? And when you write the three numbers $n$, $n+2$, and $n+4$ for bigger and bigger $n$, what do you find out when you try factoring the three numbers? $\endgroup$ – Steve Kass Apr 13 '16 at 2:51
  • $\begingroup$ I was feeling like I had already answered this same question a year ago, but it turns out that what I actually answered was a version of this question with an added wrinkle: math.stackexchange.com/questions/1415196/… $\endgroup$ – Robert Soupe Apr 13 '16 at 3:17
  • $\begingroup$ $\pmod 3 {}{}{}{}{}{}$ $\endgroup$ – TheRandomGuy Apr 13 '16 at 13:33

Note that one of the three numbers must be a multiple of $3$. So the only possible choice is that one of them is $3$ itself. So $3,5,7$

  • $\begingroup$ Okay so I ran through a few cases, and so far I have 1 and 3 as two of the numbers. I am unsure as to how I would find the last one though? $\endgroup$ – Nick Powers Apr 13 '16 at 3:06
  • $\begingroup$ @NickPowers In standard notation, $1$ is not counted as a prime number. As others have pointed out, $n$ must not be an even number. Also since one of the three numbers must be a multiple of $3$, we would be composite if the multiple isn't $3$ itself. So we actually only have one case: $3,5,7$ $\endgroup$ – lEm Apr 13 '16 at 3:19

Brute force can nudge you towards the answer, but you still have to look at and think about the results of brute force. Have your computer give you a few triples factorized and analyze the results.

Obviously, $n$ has to be a prime number. So there is no need to test $n = 0$ or $n = 1$.

Then, with $n = 2$, we see that $n + 2$ can't prime because that's obviously divisible by 2. No need to look at $n = 4$ either. In fact, we don't need to consider any other even $n$.

Moving on to $n = 3$, we obtain the primes 3, 5, 7. Ding, ding, ding!

With $n = 5$, we get the prime 7 but also $9 = 3^2$.

And with $n = 7$, we get $n + 4 = 11$, but $n + 2 = 9 = 3^2$.

Look at this problem modulo 6: if $n$ is odd, it has to be 1 or $5 \pmod 6$ to not be a multiple of 3. But if $n \equiv 1 \pmod 6$, then $n + 2 \equiv 3 \pmod 6$, or if $n \equiv 5 \pmod 6$, then $n + 4 \equiv 3 \pmod 6$.

So this problem has exactly one solution, or exactly two if negative integers are allowed.

  • $\begingroup$ Maybe I'm not understanding something but why are we disregarding 1, wouldn't that yield 1, 3, 5 which fits our conditions? $\endgroup$ – Nick Powers Apr 13 '16 at 3:08
  • $\begingroup$ Well, if you consider 1 to be a prime number, then this problem has exactly two solutions, or four if negative integers are allowed. $\endgroup$ – Robert Soupe Apr 13 '16 at 3:10
  • $\begingroup$ Okay that makes sense thank you, unfortunately I am still unclear as to how I would find the remaining applicable numbers, sorry. $\endgroup$ – Nick Powers Apr 13 '16 at 3:13
  • 1
    $\begingroup$ So under that understanding wouldn't this mean the only possible value for n is 3? $\endgroup$ – Nick Powers Apr 13 '16 at 3:18
  • 1
    $\begingroup$ @RobertSoupe If negative numbers are allowed (and if $1$ is considered prime) then every odd $n$ between $-7$ and $3$ is a solution, so there are six. $\endgroup$ – Erick Wong Apr 13 '16 at 5:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.