Very interesting question. Let's see... Let $P = x_{1}* x_{2}* \ldots * x_{n}$ $X_{i}= \frac{P}{x_{i}}= \prod_{k \neq i}^{n}x_{k}$ Now: $\frac{1}{2}= \sum_{i=1}^{n}\frac{1}{x_{i}^{2}}= \frac{\sum_{i=1}^{n}X_{i}^{2}}{P^{2}}$ That means: $P^{2}= 2 \sum_{i=1}^{n}X_{i}^{2}$ So $P$ must be even. Because $2 \sum_{i=1}^{n}X_{i}^{2}$ is an even square we must have: $S = \sum_{i=1}^{n}X_{i}^{2}\equiv 2 \mod 4$ or $S = 2^{2k-1}* G^{2}$, where $G$ is odd. I. $S = \sum_{i=1}^{n}X_{i}^{2}\equiv 2 \mod 4$ If all $X_{i}$ are even, then $S$ will be divisible by 4. So there exist $X_{t}$, which is odd. But we have: $X_{t}= \prod_{k \neq t}^{n}x_{k}$, so all $x_{i}$ except $x_{t}$ must be odd. We have 2 cases: 1. $x_{t}$ is even. Then, because $x_{t}| X_{i}$ for $i \neq t$ $X_{i}, i \neq t$ will be even and $X_{t}$ will be odd, which means: $S \equiv 1 \mod 4$. Contradiction. 2. $x_{t}$ is odd. Now all $x_{i}$ will be odd, so as their product $P$. But $P$ must be even. Contradiction. II. A) $S = 2^{2k-1}$ Now $P = 2^{2k}= \prod x_{i}$ So $x_{i}= 2^{y_{i}}$ But $x_{i}\neq 1$, because otherwise $\sum_{i=1}^{n}\frac{1}{x_{i}^{2}}>1$ This means that $\{x_{i}\}_{i=1}^{n}\subset \{2^{i}\}_{i=2}^{\infty}$ So... $\sum_{i=1}^{n}\frac{1}{x_{i}^{2}}< \sum_{i=2}^{\infty}\frac{1}{2^{i}}= \frac{1}{2}$ [EDITED] B) Now, if $S = 2^{2k-1}*G^{2}$, then $x_{i}= z_{i}2^{y_{i}}$ where $\sum y_{i}= 2^{k}$ $\prod z_{i}= G^{2}$ I don't know how to proceed.....

krassi_holmz wrote: Because $2 \sum_{i=1}^{n}X_{i}^{2}$ is an even square we must have: $S = \sum_{i=1}^{n}X_{i}^{2}\equiv 2 \mod 4$ or $S = 2^{2k-1}$ I not think so. Are you sure? Example $S=8\cdot 9$.

Observe that $\frac{\pi^{2}}{6}-1-\frac{1}{2^{2}}> \frac{1}{2}$

What is your solution? Please post concrete!

wojto111 wrote: Observe that $\frac{\pi^{2}}{6}-1-\frac{1}{2^{2}}> \frac{1}{2}$ Nice hint But I think that it's <, not > (RHS it's about 0,39)

wojto, I had the same observation, but this leads only to the fact, that $\frac{1}{4}$ must be in the sequence. (otherwise the sum will be less than $\frac{1}{9}+\frac{1}{16}+...<\frac{1}{2}$)

This seems to me to be an exceptionally difficult problem. At least I can find some solutions. Consider $S= a_1a_2...a_k.$ Then $\sum_ {x\mid S} \frac{1}{x^2} = (1+\frac{1}{a_1^2})(1+\frac{1}{a_2^2})...(1+\frac{1}{a_k^2}).$ So, for example, if $S=2\cdot3\cdot4\cdot5 = 120$, then $\sum_ {x\mid S} \frac{1}{x^2} =22100/14400 = 1.5 + 500/14400 = 1.5 + \frac{(10^2)+(20^2)}{120^2}= 1.5 + \frac{1}{6^2}+\frac{1}{12^2}$. Thus if T is the set of all divisors of S, except 1, 6 and 12, $\sum_ {x\in T} \frac{1}{x^2} =\frac{1}{2}$. In particular $T=(2,3,4,5,8,10,15,20,24,30,40,60,120)$. Of course we note that it is also true that $500=400+64+36$, which generates solution $(2,3,4,5,8,10,12,24,30,40,60,120)$, and T equal to all the divisors of $2\cdot3\cdot4\cdot7=168$, except 1, 14,42, and 84, works as well. I have no idea how to find all the other solutions.

$\frac{1}{2}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{9^2}+\frac{1}{12^2}+\frac{1}{14^2}+\frac{1}{21^2}+\frac{1}{36^2}+\frac{1}{45^2}+\frac{1}{60^2}$ Problem is very difficult........