How do you factor this?

$n^{2}-7n+7 = k^{2} / \times 4$

$4n^{2}-28n+28 = 4k^{2}$

$4n^{2}-28n + 49 - (2k)^{2} = 21$

$(2n-7)^{2} - (2k)^{2} = 21$

$(2n-7-2k)(2n-7+2k)=21$

21 factors to ±3 and ±7, and ±21 and ±1, so we can have four cases where we have sets of two linear equations. For example, one of the sets is:

$2n-7-2k=21$

$2n-7+2k=1$

By adding these two equations together, we get $4n - 14 = 22 => n=9$

The two other natural solutions are 1 and 6. Their sum is $\boxed{1+6+9=16}$