# Adjacent numbers placed in a circle cannot have integer ratios

Is it possible to place the numbers $$1,\ldots, n$$ in a circle, such that the ratio between two adjacent integers is it itself an integer?

I think the answer is no for $$n\geq 3$$, and I would like to prove it without using Bertrand's postulate (because it is to me heavy machinery that I'm not able to prove).

I started out with a simpler exercise where $$n$$ was odd, and the ratio of adjacent numbers was supposed to be a prime number. This is impossible. Let $$a_1r_1=a_2$$, $$a_2r_2=a_3$$, and so on.

Then $$a_1r_1r_2\cdots r_n=a_1\\\Rightarrow r_1r_2\cdots r_n=1$$ This is impossible as the primes do not come in pairs ($$n$$ is not even).

I want to generalize to $$n\geq 3$$ any integer, and also the ratio being any positive integer. It is clear that the case when $$n=p$$, $$p$$ a prime, is impossible, as the adjacent integers must then be $$1$$ and $$p$$. Also the result would follow if there is a prime $$p$$ such that $$p, which is true by Bertrand's Postulate.

Is it possible to prove this result without using Bertrand's postulate?

• Do you allow the ratio between consecutive numbers to be prime in either direction, i.e. is $\ldots 6, 12, 4, \ldots$ okay since $\frac{12}{6} = 2$ and $\frac{12}{4} = 3$ are both prime? Commented Jan 3, 2023 at 19:27
• @SammyBlack Yes Commented Jan 3, 2023 at 19:29
• The title says cannot have integer ratios but the question demands them. Commented Jan 3, 2023 at 20:53
• Bertrand's postulate is not adapted to this exercice. Select the highest prime number, less than n. This highest prime number whould be between 1 and1 on your circle. Commented Jan 4, 2023 at 11:25

If $$n=2m$$ is even and at least $$4$$, then any two of the $$m$$ numbers $$m+1,..2m$$ cannot be adjacent since the (largest) ratio between them is between $$1$$ and $$2$$, hence they must go in $$m$$ alternate spots, but then $$m$$ is adjacent to two of them (here we use $$m \ge 2$$) and the ratio $$(m+k)/m, 1 \le k \le m$$ is integral only for $$k=m$$
if $$n=2m+1$$ then any two of the $$m+1$$ numbers $$m+1,...2m+1$$ cannot be adjacent, but that is not possible since there are only $$m$$ remaining spots and that shows the result in this case
More simple : consider $$p$$, the highest prime number in range [1,n] and try to build 2 neighbors to $$p$$.
• I don’t think it is that easy, what if $n=2p+1$? Then we could have $1,p,2p,2$. The existence of largest prime that only appears once from $1$ to $n$ is not that easy to show. Commented Jan 6, 2023 at 14:09