$\sum_i \frac{1}{a_i}=1$ implies $\prod_i a_i$ is a square Let $a_1<\cdots<a_k$ be positive integers such that $\sum_i \frac{1}{a_i}=1$. Prove that if $a_k/2$ is prime then $\prod_i a_i$ is a square.
 A: Let $a_k=2p$ for some prime $p$. $p>2$, otherwise $a_k=4$ and it is easy to verify no such sequence exist. Hence $p^2\not\mid a_i\forall i$.
If $a_k$ is the only term in the sequence divisible by $p$, we can multiply $\sum_{i=1}^k\frac{1}{a_i}$ by $\frac{lcm(a_1\dots a_k)}{p}$. The result has to be an integer. We now do the multiplication term-by-term. Every term except $\frac{1}{a_k}$ would become an integer, as $p^2\not\mid lcm(a_1\dots a_k)$.
Hence there is more than 1 term divisible by $p$. As all terms are positive and distinct, there exist both $p$ and $2p$ in the sequence.
We add them to form $\frac{1}{p}+\frac{1}{2p}=\frac{3}{2p}$.
If we multiply all the terms in the summation by $\frac{lcm(a_1\dots a_k)}{p}$, we get every other term except $\frac{1}{p}$ and $\frac{1}{2p}$ becomes an integer. Hence $\frac{3}{2p}\times\frac{lcm(a_1\dots a_k)}{p}$ must be an integer.
Since $p^2\not\mid lcm(a_1\dots a_k)$, it follows that $p\mid3$, hence $p=3$. It is easy to verify that the only such sequence of $a_i$ is $2, 3, 6$, where the product of all terms in the sequence is $36$, a square number, hence we are done.
