# What values of $n$ is $A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$ an integer?

Suppose $$a_i\in \mathbb{N}$$ such that $$a_i\ne a_j$$ and $$\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$$ for $$i \ne j$$ $$($$take $$a_{n+1}=a_1)$$ and $$\displaystyle A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$$.
Then for what values of $$n\ge 2$$ can $$A$$ be an integer?

I found that if $$a_i=(n-1)^i$$ then $$A$$ is a integer but I realised that this violates the condition $$\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$$. It was easy to prove that $$A$$ is not an integer for $$n=2$$, but I was stuck for the cases where for $$n>2$$. I then I plugged in some values for $$a_i$$ but for $$n>2$$ but I found no solution.

Intuitively I think there is no solution but I am unable to prove it. Any help would be appreciated.

Edit-1: As pointed out by @WillJagy it is possible for $$n=3,4,5$$. So can we actually prove it is true for $$n>6$$ or are there some $$n>6$$ for which it isn't true?

Edit-2: I was able to find a solution for $$n=5$$ from a solution for $$n=4$$. $$~~~~~~~~~~\frac{6}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 6$$ $$\implies \frac{2}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 5$$ $$\implies \frac{6}{2} +\frac{2}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 8$$
But I could not find a solution for $$n=6$$ from a solution for $$n=5$$ because when I rewrite $$\frac{4}{3} =1+\frac{1}{3}$$, the $$1$$ in the numerator of $$\frac{1}{3}$$ is used by another $$a_i$$. I don't know whether we can prove by induction. Any suggestions is welcome.

• artofproblemsolving.com/community/c6h114708p651437 Commented Jun 1, 2021 at 1:12
• @FearfulSymmetry This question has an additional constraint of $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$ Commented Jun 1, 2021 at 1:19
• Sure; try and start there and see if you can eliminate some of the general solutions. Commented Jun 1, 2021 at 1:33

$$\frac{81}{2} + \frac{36}{81} + \frac{2}{36} = 41$$

$$\frac{98}{12} + \frac{63}{98} + \frac{12}{63} = 9$$

$$\frac{6}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 6$$

$$\frac{6}{5} + \frac{5}{3} + \frac{3}{10} + \frac{10}{12} + \frac{12}{6} = 6$$

$$\frac{3}{12} + \frac{12}{9} + \frac{9}{10} + \frac{10}{8} + \frac{8}{5} + \frac{5}{3} = 7$$

$$\frac{2}{8} + \frac{8}{10} + \frac{10}{6} + \frac{6}{5} + \frac{5}{4} + \frac{4}{3} + \frac{3}{2}= 8$$

• What was your approach for finding these numbers? Commented May 31, 2021 at 20:15
• @Asher2211 I just wrote a program. It's worth checking something before trying to prove it; with integers, pretty firm indications can come from a computer. There is probably some sort of order to these, the only integers I've gotten ( with 3 variables) are 9, 41, 66 Commented May 31, 2021 at 20:21
• Have you assumed $a_i$ to be ordered in the program? I got $6$ with $a = (2, 12, 9)$ and it seems to follow the restrictions. Commented May 31, 2021 at 20:31
• @AnilCh good point, I did have them ordered, apparently that was not a valid restriction. Commented May 31, 2021 at 20:32
• @Asher2211 alright. I forgot that condition. I suspect it will force the $a_i$ to be a bit larger, but the job seems to be getting easier with more variables. As far as your wish for a proof, it is a bit early to be asking about that, given that you began in the opposite direction. If it is important, you will write your own programs and look for patterns. Commented May 31, 2021 at 23:28

I guess you do not want any ratio $$a_i/a_{i+1}$$ to be an integer, otherwise trivial solutions can be produced.

For $$n=3$$, a solution can be produced as already seen in the answers.

We will use Egyptian fractions to prove that there always exists a sequence of such $$a_i$$'s for $$n\gt 3$$

$$1=\frac 12+\frac 13+\frac 16$$

We can break the last unit fraction $$\dfrac 16$$ using the fact that $$\dfrac 1m=\dfrac 1{m+1}+\dfrac 1{m(m+1)}$$

In this way, you can create an Egyptian fraction representation (with distinct unit fractions) for $$1$$ of arbitrary length, ie, for any $$n\gt 3$$, there exist a sequence of distinct $$p_i (\neq 1)$$'s ($$i=1...(n-1)$$) such that $$1=\displaystyle\sum_1^{n-1}\frac 1{p_i}$$

All that is left is to define our $$a_i$$'s in a suitable way. We define $$a_1=1$$ and $$a_{i+1}=a_ip_i$$ for all $$i=1...(n-1)$$ so that all the $$a_i$$'s are distinct and $$\dfrac{a_i}{a_{i+1}}=\dfrac 1{p_i}$$ are distinct for all $$i=1...(n-1)$$ and sum up to $$1$$ and $$\dfrac{a_n}{a_1}=a_n$$ is an integer, so the resulting sum $$A=a_n+1$$

Examples:

$$\frac 12+\frac 26+\frac 6{36}+\frac{36}1=37\\ \frac 12+\frac 26+\frac 6{42}+\frac{42}{1764}+\frac{1764}1=1765$$

Here's a Tio.run link if you want to generate a sequence of $$a_i$$'s for any $$n$$, but beware that the $$a_i$$'s grow larger quite quickly due to the construction.

Hmm, then again, perhaps this answer also generates trivial solutions because $$a_n/a_1$$ is an integer. I'm unsure if I can fix that. I'll update if I can.

$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{15}{6}+\frac{24}{15}+\frac{10}{24}+\frac{2}{10}=10$$

$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{30}{7}+\frac{14}{30}+\frac{24}{14}+\frac{2}{24}=13$$

$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{9}{7}+\frac{36}{9}+\frac{15}{36}+\frac{42}{15}+\frac{2}{42}=15$$

$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{9}{7}+\frac{10}{ 9}+\frac {28}{ 10}+\frac {27}{ 28}+\frac {36}{ 27}+\frac {2}{ 36}=14$$