A number theory problem by sudoku subbu

$n^{2} -19n +99-m^{2}=0$

Since $n \in I$ so Discriminant $D$ of the above equation is a perfect squre

$D=b^{2}-4ac=361-4(99-m^{2})=k^{2}$

$\Rightarrow -35+4m^{2}=k^{2}$

$\Rightarrow 4m^{2}-k^{2}=35$

By this $m^{2}$ has only two integral values that are $81,9$

Substitute these values in the given equation to get

$n^{2}-19n+99=9$

$\Rightarrow n^{2}-19n+90=0$

$\Rightarrow (n-9)(n-10)=0$

$\Large And$

$\Rightarrow n^{2}-19n+99=81$

$\Rightarrow n^{2}-19n+18=0$ $\Rightarrow (n-1)(n-18)=0$

By this $\large n=1,18,10,9$

So the answer is $1+18+10+9=\huge38$

Another approach is to complete the square:

$n^{2}-19n+99=x^{2}$

$\Rightarrow \left ( n-\frac{19}{2} \right )^{2}+\frac{35}{4}=x^{2}$

$\Rightarrow 4\left ( n-\frac{19}{2} \right )^{2}+35=4x^{2}$

$\Rightarrow \left ( 2n-19 \right )^{2}+35=\left ( 2x \right )^{2}$

$\Rightarrow \left ( 2x \right )^{2}-\left ( 2n-19 \right )^{2}=35$

$\Rightarrow \left ( 2x-2n+19 \right )\left ( 2x+2n-19 \right )=35.$

Therefore, $(2x-2n+19,2x+2n-19)=(7,5)\: or\: (35,1).$

$\Rightarrow \left\{\begin{matrix} 2x-2n+19=35 & & \\2x+2n-19=1 & & \end{matrix}\right.$ and $\left\{\begin{matrix} 2x-2n+19=7 & & \\ 2x+2n-19=5 & & \end{matrix}\right.$ .

We get $x_{1}=9,x_{2}=3$ .

Substitute into $n^{2}-19n+99=x^{2}$ , we have $n=1,9,10,18$ .

Thus, the sum is $1+9+10+18=38$