When is $\sum_{i=1}^n a_i^{-2}=1$? 
For which natural numbers $n$ do there exist $n$ natural numbers $a_i\ (1\le i\le n)$ such that $\displaystyle\sum_{i=1}^n a_i^{-2}=1$?

I didn't see an easy way of solving this. There is a solution for $n=1$ since we just take $a_i = 1$ and a solution for $n=4$ if we take $a_i = 2$. How do we find all possible values of $n$?

If $n$ works, then $n+3$ and $n+8$ works.
Proof. Take a term, $a_i$, and replace that with $4$ copies of $2a_i$. Take a term, $a_i$, and replace that with $9$ copies of $3a_i$. Both work.
So, since $1$ works, all numbers $\equiv1\pmod 3$ work.
Since $9$ works ($a_i=3\forall i$), all numbers $\equiv0\pmod 3$ greater than or equal $9$ work.
Since $1$ works, $1+8+8=17$ works, all numbers $\equiv2\pmod 3$ greater than or equal $17$ work.
Now these are the exceptions we must check: $2, 3, 5, 6, 8, 11, 14$.
Clearly $2, 3$ are impossible as $\frac{1}{a_i^2}\leq\frac{1}{4}$ for $a_i>2$, making $\sum\frac{1}{a_i^2}<1$.
For $5$, one of the $\frac{1}{a_i^2}\geq\frac{1}{5}$, so $a_i=2$ for some $i$. We have $4$ numbers adding to $\frac{3}{4}$. One of the $\frac{1}{a_i^2}\geq\frac{3}{16}>\frac{1}{9}$, so $a_i=2$ for another $i$. We have $3$ numbers adding to $\frac{1}{2}$. One of the $\frac{1}{a_i^2}\geq\frac{1}{6}>\frac{1}{9}$, so $a_i=2$ for another $i$. We have $2$ numbers adding to $\frac{1}{4}$. One of the numbers $\geq\frac{1}{8}$, so impossible.
$6$ is actually possible. $1=3\times\frac{1}{4}+2\times\frac{1}{9}+1\times\frac{1}{36}$
$8$ is actually possible. $1=2\times\frac{1}{4}+4\times\frac{1}{9}+2\times\frac{1}{36}$, so $8+3=11$ and $11+3=14$ are possible.
So now, the updated set of possible numbers is all integers, except for $2, 3, 5$.
