Mathematical Reflections 2014, Issue 5 - Problem O313

Problem:
Find all positive integers $n$ for which there are positive integers $a_0, a_1, \ldots, a_n$ such that $a_0 + a_1 + \ldots + a_n = 5(n-1)$ and $$\dfrac{1}{a_0}+\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_n}=2.$$

Proposed by Titu Andreescu

Solution:
Assume without loss of generality that $a_0 \leq a_1 \leq \ldots \leq a_n$. By the AM-GM Inequality, we have $$10(n-1)=\left(a_0 + a_1 + \ldots + a_n\right)\left(\dfrac{1}{a_0}+\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_n}\right) \geq (n+1)^2,$$ so $n \leq 6$. Clearly, $n>1$. We have five cases.

(i) $n=2$. We immediately get $a_0=1,a_1=2,a_2=2$.

(ii) $n=3$. We immediately get $a_0=1,a_1=3,a_2=3,a_3=3$.

(iii) $n=4$. Observe that $\dfrac{5}{a_0}\geq 2$, so $a_0=1,2$. If $a_0=1$, then $a_1+a_2+a_3+a_4=14$ and $\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}=1$, but this would imply $$14=(a_1+a_2+a_3+a_4)\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}\right)\geq 16,$$ contradiction. So, $a_0=2$, which gives $a_1+a_2+a_3+a_4=13$ and $\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}=\dfrac{3}{2}$. Now, $\dfrac{4}{a_1} \geq \dfrac{3}{2}$ implies that $a_1=2$. So, $a_2+a_3+a_4=11$ and $\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}=1$. Since $\dfrac{3}{a_2} \geq 1$, we get $a_2=2$ or $a_2=3$. If $a_2=2$, then $a_3+a_4=9$ and $\dfrac{1}{a_3}+\dfrac{1}{a_4}=\dfrac{1}{2}$, which gives $a_3=3,a_4=6$. If $a_2=3$, then $a_3+a_4=8$ and $\dfrac{1}{a_3}+\dfrac{1}{a_4}=\dfrac{2}{3}$, so no solutions.

(iv) $n=5$. Reasoning as before, we get $a_0=2,a_1=2,a_2=4,a_3=4,a_4=4,a_5=4$.

(v) $n=6$. Observe that $\dfrac{7}{a_0} \geq 2$, so $a_0 \leq 3$. But if $a_0=i$ for $i=1,2$, then $$(25-i)\left(2-\dfrac{1}{i}\right)=\left(a_1+\ldots+a_6\right)\left(\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_6} \right) \geq 36$$ gives a contradiction. So, $a_0=3$, which gives $a_1+\ldots+a_6=22$ and $\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_6}=\dfrac{5}{3}$. Therefore, $a_1=3$, which gives $a_2+\ldots+a_6=19$ and $\dfrac{1}{a_2}+\ldots+\dfrac{1}{a_6}=\dfrac{4}{3}$. Then $a_2=3$, which gives $a_3+a_4+a_5+a_6=16$ and $\dfrac{1}{a_3}+\dfrac{1}{a_4}+\dfrac{1}{a_5}+\dfrac{1}{a_6}=1$, so $a_3=4,a_4=4,a_5=4,a_6=4$.

We conclude that $n=2,3,4,5,6$.