Number please?

Let $n^2+45$ = $m^2$

$(m^2-n^2$ ) = $45$

Now let us factorize 45

$45$ = $9 × 5$ ,

$45$ = $15 × 3$

$45$ = $45 × 1$

Case I:- $m^2-n^2$ = $9 × 5$

$(m+n)(m-n)$ $9 × 5$

now since ' $m$ ' and ' $n$ ' are positive integers, naturally $m+n>m-n$

$m+n=9$

$m-n=5$

On solving we get, $m=7,n=2$

Similarly for other two cases, $m=9,n=6$

$m=23, n=22$

Possible values for $n$ are $2$ $,6$ and $22$ . Their sum, $2 + 6 + 22$ $=$ $30$