Find all sequences $x_1,x_2,\dots,x_n$ of distinct positive integers such that $$\frac{1}{2}=\frac{1}{x_1^2}+\frac{1}{x_2^2}+\dots+\frac{1}{x_n^2}\tag{1}.$$
This is 3rd problem from 1st day of 16-th Hungary–Israel Binational Mathematical Competition 2005. How to solve this?
I couldn't find official solution, only couple of posts on AoPS with no complete solution, such as aops1 and aops2. Some examples of $(x_1,x_2,\dots,x_n)$ found there are $$ (2,3,4,5,7,12,15,20,28,35),\\ (2,3,4,6,7,9,12,14,21,36,45,60),\\ (2,3,4,5,8,10,15,20,24,30,40,60,120) $$
Since $\sum_{i=3}^{\infty} \frac{1}{i^2}<\frac{1}{2}$, we know $\frac{1}{2^2}$ will be always in the sum. Also multiplying out $(1)$ to get rid of fractions and inspecting modulo $x_i^2$, we see $2(x_1\cdots x_{i-1}x_{i+1}\cdots x_n)^2 \equiv 0 \pmod{x_i^2}$, so divisors of each of $x_i$ must "split" among other $x_j$'s (and $2$), but I don't think that helps much.
Edit: As noted by WhatsUp in comments, there is relevant problem in Project Euler. There is actually also one that is almost identical, but with limited search range, specifically https://projecteuler.net/problem=152. So it might help to think about this problem in similar manner, letting $x_n \leq M$ and then characterize solutions with these constrains. Caution, the following spoils the above mentioned Project Euler problem.
We can convert this to more computationally manageable diophantine equation by multiplying $(1)$ by $2\text{lcm}(x_1,x_2,\dots,x_n)^2$ instead of $2(x_1x_2\cdots x_n)^2$. For example considering $\text{lcm}(2,3,4,5,7,12,15,20,28,35)=420$, first example can be written as $$420^2=2(210^2+140^2+105^2+84^2+60^2+35^2+28^2+21^2+15^2+12^2).$$ Then by the construction, all the squares on the right are divisors of the square on the left. Also, square on the left is just square of one of divisors of $\text{lcm} (2,3,\dots,M)$, if we have $M \geq 35$. This leads to following reformulation of the problem:
Find all sequences $y_1,y_2,\dots,y_n$ of distinct positive integers such that $$m^2=2(y_1^2+y_2^2+\dots+y_n^2)\tag{2},$$ where $m=\text{lcm}(y_1,y_2,\dots,y_n)$.
So in principle, for fixed $M$ and $x_i \leq M$, we can find $\text{lcm}(2,3,\dots,M)$, and then enumerate its divisors effectively using prime factorization (using that its prime factors will be $p \leq M$).
Then additionally, we can notice that not all primes $p$ can be present in these numbers. Indeed, if we have $p> M/2$, then by earlier observation, it would have to divide at least two $x_i$'s. Smallest of them can be $x_i=p$, with next smallest $x_j=2p>M$, impossible. In similar way, lots of other primes can be ruled out for $M=80$, by realizing they would require to be split among at least $3$ of $x_i$'s (try $37$).
While these ideas allow us to significantly speed up the search, it is still hard to say which factorizations will satisfy $(2)$ without actually checking the values...
