$\frac{7}{n}$ as a sum of three unit fractions could $\frac{7}{n}$ be represented as sum of three positive unit fractions given that $n = 2 \mod 7$ ?
I know $\frac{7}{2}$ is not but for all $n>2$ and $n = 2 \mod 7$ it seems that its representable.
 A: This is only a partial answer, but perhaps it will inspire someone else to give a full solution. We are seeking a general identity of the form
$$\frac{7}{7k+2}=\frac1a+\frac1b+\frac1c,$$
where $a,b,c\in\mathbb Z_{>0}$ depend on $k$. The obvious thing I tried was to take the guess $a=k+1$, which gives
$$\frac1b+\frac1c=\frac{7}{7k+2}-\frac1{k+1}=\frac{5}{(2+7k)(1+k)}.$$
Now, we can write
$$\frac{5}{(2+7k)(1+k)}=\frac{3}{(2+7k)(1+k)}+\frac{2}{(2+7k)(1+k)}.$$
Observe that $(2+7k)$ and $(1+k)$ are always of opposite parity, so the second term can always be simplified to a unit fraction. You can verify that $3\mid(2+7k)(1+k)$ if $3\nmid k$, so the first term is a unit fraction in this case as well. We have therefore reduced the problem to expressing
$\dfrac7{21k+2}$ as a sum of unit fractions.
Unfortunately, trying $a=nk+c$ for any other choices of $n,c$ don't seem to give useful results. For instance we have
$$\frac{7}{21k+2}=\frac1{3k+2}+\frac{12}{(2+3k)(2+21k)},$$
but I don't see an obvious way to decompose the second term into a sum of two unit fractions. The best case scenario is to end up with an expression in the denominator which the numerator is guaranteed to divide (like the case for the second term above where we are guaranteed a unit fraction), but since the denominator is a quadratic polynomial in general this is not possible unless the numerator is a power of $2$. But this is insufficient; writing $12=2^3+2^2$ doesn't give anything useful above, for instance.
