# Number theory question from Homotopy groups of spheres

Let $n$ be some integer.

Is it true that there exists odd prime $p$ such that $4n = (p-1) \cdot k$,

where $k$ is an integer coprime with $p$?

This question asked Roman Mikhailov. This is corresponds with Homotopy groups of spheres. Unfortunately I do not know the details.

• A very basic check shows that there are no counterexamples with $n\leq 10^7$. One could probably go much further with a more intelligent algorithm. – Neil Strickland Apr 15 '18 at 15:38
• it seems that there are no counterexamples with $n \le 10^{10^6}}$ – Alexey Milovanov Apr 15 '18 at 16:05
• At least the case for even $n$ follows from the odd $n$ case. Also, for prime $n$ it is trivial. – EFinat-S Apr 15 '18 at 16:29
• Also, one only need to check the case where 5 divides $n$. – EFinat-S Apr 15 '18 at 16:39
• Background of question: there are two different proofs of the fact that homotopy groups of $S^3$ are nontrivial in dimensions $\geq 3$ (interesting case is is $\pi_{8k+1}$), one is due to B. Gray, Unstable families related to the image of J (where it's not stated, but result follows), and other by S. Ivanov, R. Mikhailov and J. Wu, arxiv.org/abs/1506.00952v1. If aforementioned number-theoretic result was true, second proof could be significantly simplified. – Denis T. Apr 15 '18 at 19:53

Probably yes. The question is equivalent to: for any (nonzero) integer $n$, there exists an odd prime $p$ not dividing $n$ such that $p-1$ divides $4n$.

The divisors of $4$ are $1,2,4$; of these, $2,4$ are one less than odd primes $3,5$. So if $3\nmid n$ then choose $p=3$, and if $5\nmid n$ then choose $p=5$.

So we may suppose that $3,5$ both divide $n$. The divisors of $4\cdot3\cdot5=60$ that are one less than odd primes are $2, 4, 6, 10, 12, 30, 60$. So if $7\nmid n$ then choose $p=7$, if $11\nmid n$ then choose $p=11$, if $13\nmid n$ then choose $p=13$, if $31\nmid n$ then choose $p=31$, and if $61\nmid n$ then choose $p=61$.

So we may suppose that $3,5,7,11,13,31,61$ all divide $n$. The divisors of $4\cdot3\cdot5\cdot7\cdot11\cdot13\cdot31\cdot61$ that are one less than odd primes are $2, 4, 6, 10, 12, 22, 28, 30, 42, 52, 60, 66, 70, 78, 130, 156, 210, \ 310, 330, 366, 372, 420, 462, 546, 660, 682, 732, 858, 910, 1092, 1302, 1612, 1708, 1830, 1860, 2002, 2310, 2730, 2860, 4026, 4092, 4270, 4620, 4758, 6006, 8052, 8580, 8866, 13420, 14322, 16926, 18910, 20020, 24180, 25620, 28182, 28210, 41602, 47580, 47740, 53196, 55510, 56730, 79422, 84630, 88660, 93940, 104676, 132370, 186186, 294996, 624030, 794220, 873642, 930930, 1221220, 1248060, 1474980, 1622478, 1831830, 1861860, 2912140, 3244956, 3663660, 11357346$; and so it continues....

All we need to prove that the answer to your question is "yes" is for this iterative procedure to contain infinitely many primes. Although that might be hard to prove, it would be incredible if the procedure halted.

• Doesn't this procedure miss every prime that is $1$ mod $8$? – Zack Wolske Apr 15 '18 at 20:35
• And $19,37,101,$ etc. too. If it were, indeed, "every odd prime", we would be able to prove it. Unfortunately, it is not. The "random model" results in a branching birth-death process that explodes like crazy, indeed, so the statement is surely true if there is enough randomness in the primes, which makes looking for a counterexample hopeless. Alas, I doubt very much that any proof will be found in the lifetime of any of the MO participants. – fedja Apr 15 '18 at 23:35
• @fedja Is it already not known how to prove that there are infinitely many primes such that no prime factors of $p-1$ is congruent to $1$ mod $8$? – Will Sawin Apr 18 '18 at 11:39
• @WillSawin Probably it isn't, so I agree that Zack's remark was formally already sufficient to show that the task of proving the statement is rather hopeless. I just wanted to spell it out :-) – fedja Apr 18 '18 at 11:48
• WillSawin and fedja, I would be very interested to hear what the MO community had to say about the plausibility of proving that there are infinitely many primes $p$ such that $p-1$ had no prime factors congruent to $1$ modulo some fixed $q$ (such as $q=8$); I think that's a very interesting problem. – Greg Martin Apr 18 '18 at 16:21