$5^n+n$ is never prime?

Question

In the comments to the question: If $(a^{n}+n ) \mid (b^{n}+n)$ for all $n$, then $ a=b$, there was a claim that $5^n+n$ is never prime (for integer $n>0$).

It does not look obvious to prove, nor have I found a counterexample.

Is this really true?

Update: $5^{7954} + 7954$ has been found to be prime by a computer: http://www.mersenneforum.org/showpost.php?p=233370&postcount=46

Thanks to Douglas (and lavalamp)!

@Clearly when $n$ is a multiple of 5 then $5^{n} +n$ doesnt remain a prime. But we will have to think of other cases. — , Sep 06 '10 at 07:11
It is also clear that n is even if the number is prime. Considering remainder on division by 6, 5^n will always be congruent to 1, so n must be congruent to 0 or 4 mod 6. — Jonas Meyer, Sep 06 '10 at 07:19
@Asaf: Jonas means the number is prime => n is even, not the other direction — , Sep 06 '10 at 08:31
Note: I have checked that $5^n+n$ is composite for all $n \leq 1000$. — Pete L. Clark, Sep 06 '10 at 09:21

Douglas S. Stones · Accepted Answer · 2010-10-14T21:29:43.933

59

A general rule-of-thumb for "is there a prime of the form f(n)?" questions is, unless there exists a set of small divisors D, called a covering set, that divide every number of the form f(n), then there will eventually be a prime. See, e.g. Sierpinski numbers.

Running WinPFGW (it should be available from the primeform yahoo group http://tech.groups.yahoo.com/group/primeform/), it found that $5^n+n$ is 3-probable prime when n=7954. Moreover, for every n less than 7954, we have $5^n+n$ is composite.

To actually certify that $5^{7954}+7954$ is a prime, you could use Primo (available from http://www.ellipsa.eu/public/misc/downloads.html). I've begun running it (so it's passed a few more pseudo-primality tests), but I doubt I will continue until it's completed -- it could take a long time (e.g. a few months).

EDIT: $5^{7954}+7954$ is officially prime. A proof certificate was given by lavalamp at mersenneforum.org.

edited Oct 14 '10 at 21:29

answered Sep 06 '10 at 10:35

Douglas S. Stones

21,396

9

I disagree with that heuristic. A better heuristic is that if there is no covering set, and $\sum 1/\log f(n)$ diverges, then there will be a prime. For example, it seems possible that there may be no primes of the form $2^{2^n}+1$ with $n \geq 5$. – David E Speyer Sep 06 '10 at 12:20
Yes, that is better. – Douglas S. Stones Sep 06 '10 at 12:59
Nice! Please do update us if you do let Primo finish. – Aryabhata Sep 06 '10 at 14:41
2

In fact, if $\sum 1/log f(n)$ diverges and there's no covering set there should be infinitely many primes of form $f(n)$. – Michael Lugo Sep 06 '10 at 19:03
Hmm... this heuristic is still not very good in some situations, like f(n)=n^2. (although, I guess nobody is going to ask if there's any primes of the form n^2) – Douglas S. Stones Sep 07 '10 at 00:43
3

I posted a link at mersenneforum.org (http://www.mersenneforum.org/showthread.php?p=228745) -- I won't be able to complete the proof myself since I will be travelling. Also, their computers are probably significantly faster than mine. – Douglas S. Stones Sep 07 '10 at 01:04
dumb question: what's a 3-probable prime? (I know what a probable prime is) – Jason S Sep 08 '10 at 01:30
3

Fermat's Little Theorem says if p is a prime then 3^p=3 (mod p). A 3-probable prime is a number q that satisfies 3^q=3 (mod q). q might not be prime (but probably is). [if instead q does not satisfy 3^q=3 (mod q) then it is guaranteed to not be a prime] – Douglas S. Stones Sep 08 '10 at 01:43
oh. hmm. well why only test 3? I guess I'm confused why you cited 3-probable + didn't use Miller-Rabin – Jason S Sep 08 '10 at 13:19
2

Two reasons (a) I don't know in advance which n to test, so it's most likely that I would end up testing many composite numbers for primality (so it's more efficient to run weaker tests first, trial division+Fermat's test) and (b) I already have WinPFGW installed and it tests for 3-PRP. – Douglas S. Stones Sep 08 '10 at 15:11
I am going to accept this, as it seems we are almost done proving that this number is prime (in this thread here: mersenneforum.org/showthread.php?p=228745). I suppose you will update the answer once we are sure, with the certificates and all that. – Aryabhata Oct 11 '10 at 18:21
1

Awesome, now if only I could upvote twice... – J. M. ain't a mathematician Oct 17 '10 at 16:40
Unfortunately, the proof certificate linked by the relevant post on mersenneforum.org is no longer accessible. It's a dead link. It will either need to be regenerated, or found by digging through some web archive. – kdbanman Feb 09 '23 at 18:34

score 13 · Answer 2 · edited Jun 11 '13 at 01:50

13

If $n$ is odd, then $5^n + n$ is always even because LSD of $5^n$ is always $5$ for $n \gt 0$. Hence, for odd $n ( n \gt 0)$, $5 ^n + n$ is composite.

edited Jun 11 '13 at 01:50

Jeel Shah

9,560

answered Sep 06 '10 at 08:17

Crazy

141

As this is not an answer this post should be a comment. – Sep 06 '10 at 09:05
51

@yjj: Crazy has only 1 reputation point. Remember what that was like? It means you can't comment. – TonyK Sep 06 '10 at 12:31
@TonyK Do not you think the rep needed for comment is too high? Examples like this are littered all over the site. I find it a bit strange that you are actually allowed to answer but not to comment. Perhaps, if you agree with me, you could bring it up on Meta. – Sawarnik Feb 15 '14 at 12:34

score 7 · Answer 3 · edited Jun 11 '13 at 01:52

7

After reading Douglas S. Stones comment I asked mathematica to check if $5^{2\times 3977} + 2\times 3977$ is prime and after about $27$ seconds, found that it is indeed prime. So the claim $5^n +n$ is never prime is false.

Edit: It turns out the function I used in mathematica is not a deterministic algorithm. However we can still say the claim $5^n +n$ is never prime is false is most likely true.

edited Jun 11 '13 at 01:52

vadim123

83,937

answered Sep 06 '10 at 10:18

6

To prove primality of numbers of this size can take months... – Douglas S. Stones Sep 06 '10 at 10:32
1

Is this primality test perfect or does it find "probable" primes? And does/can Mathematica give a certificate of primality? – ShreevatsaR Sep 06 '10 at 10:37
ShreevatsaR: Mathematica has a package NumberTheoryPrimeQ`` (I don't know the proper context in the newer versions) which uses either of Atkin-Morain or Pratt to provide a primality certificate. As Douglas said, however, it takes lots of time and memory to generate primality certificates for huge enough numbers. (If you have a gamer friend, you might want to borrow his/her PC for this ;)) – J. M. ain't a mathematician Sep 06 '10 at 11:20
On a somewhat off-topic note, we can see that this is (probably) a stereotype of mathematicians rather than gamers since P(fast computer|gamer) >> P(fast computer|mathematician) ;) – kahen Oct 14 '10 at 18:24

$5^n+n$ is never prime?

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