Find all ordered tuples of positive integers $(a_1,a_2,a_3,\ldots,a_n)$ such that $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$

The only thing I have been able to think about is using inequalities. I have tried to apply AM-GM, Titu's lemma etc.. Cauchy-Schwarz gives the following thing:

$$(\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n})(a_1+\cdots a_n) \ge (\sqrt{1}+\sqrt{2}+\cdots \sqrt{n})^2$$

$$(a_1+\cdots a_n)^2\ge 2(\sqrt{1}+\cdots \sqrt {n})^2$$

which doesn't really help us at all.

I have also tried considering smaller cases. For $n=2$,

$$a_1a_2(a_1+a_2)=4a_1+2a_2$$ which tells us that $2a_2=ka_1$ and $8a_1=pa_2=ka_1p\implies kp=8$. This should now give us all the solutions by checking all the cases.

So how can we even begin to attack this problem?

  • $\begingroup$ A remark: $a_i\mid 2i$ for each $i$ and note that solutions with $a_1 = 1$ are in bijective correspondence with solutions having $a_1 = 2$ just by taking the $n$-tuple with $2i/a_i$ in place of $a_i,$ so you can assume $a_1 = 1$ without loss of generality. Also, you can reduce it to an integer equation, but I don't know how helpful that is. $\endgroup$
    – cats
    Commented Mar 3, 2015 at 7:07
  • $\begingroup$ It should be $a_1 a_2 (a_1+a_2)=4a_1+2a_2$ $\endgroup$ Commented Mar 3, 2015 at 7:16
  • $\begingroup$ There seems to be no solution for $n<4$, according to WolframAlpha. $\endgroup$ Commented Mar 3, 2015 at 7:34
  • $\begingroup$ @ghosts_in_the_code Consider $a_n = \sqrt{2n}$ and it's easy to prove that this is solutions for all $n$ using induction. $\endgroup$
    – Stefan4024
    Commented Mar 3, 2015 at 12:26
  • $\begingroup$ @Stefan4024, why would $\sqrt{2n}$ be an integer for all $n$? $\endgroup$
    – rah4927
    Commented Mar 3, 2015 at 14:07

1 Answer 1


Partial solution: If $n+1$ is a square, then set $a_1=a_2=\cdots=a_n=\sqrt{n+1}$.

Then the LHS is $$(1+2+\cdots+n)\frac{1}{\sqrt{n+1}}=\frac{n(n+1)}{2\sqrt{n+1}}=\frac{1}{2}n\sqrt{n+1}$$

which agrees with the RHS.

More of a partial solution, following @Stefan4024's induction idea. Start with a solution of the preceding form, then add two more terms to the LHS, namely $\frac{n+1}{b}+\frac{n+2}{b}$. The equation will still balance if $$\frac{1}{b}(n+1+n+2)=\frac{b+b}{2}$$ or $b^2=2n+3$. Hence, we seek $n$ satisfying (1) $n+1=a^2$; and (2) $2n+3=b^2$. We can eliminate the $n$ to get Pell's equation: $$b^2-2a^2=1$$ This has infinitely many solutions. The smallest is $b=3, a=2$ ($n=3$, 5 terms). The next smallest is $b=17, a=12$ ($n=143$, 145 terms). The next one is $b=99, a=70$ ($n=4899$, 4901 terms).

  • $\begingroup$ I had found it but failed to prove it is the only solution besides $(1,2,3)$. Do you think this is the case? $\endgroup$ Commented Mar 3, 2015 at 18:45
  • $\begingroup$ Here's another: $(2,2,2,3,3)$. $\endgroup$
    – vadim123
    Commented Mar 3, 2015 at 20:02
  • $\begingroup$ Yeah. I liked how you improved your answer, +1. $\endgroup$ Commented Mar 3, 2015 at 20:30
  • $\begingroup$ @vadim123, someone edited the original problem to omit the condition that the $a_i$'s are distinct. Just thought I should let you know about the original problem. $\endgroup$
    – rah4927
    Commented Mar 4, 2015 at 6:57
  • $\begingroup$ Ack! I saw the original version, too. I suspect the solution is just $(1,2,3)$, since it's a contest problem. The modified version may be too hard. $\endgroup$
    – vadim123
    Commented Mar 4, 2015 at 14:29

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