It's Hard To Find A Square

Let $m = k^2$ for some integer $k$ , and proceed to rewrite the original equation slightly, $k^2 = n^2 + 23n + 175 \Rightarrow n^2 + 23n + 175 - k^2 = 0$ We treat the equation above as a quadratic equation in $n$ and consider its discriminant $D = 23^2 - 4(175-k^2) = 4k^2 - 171$ We wish $D$ to be equal to some perfect square, say $d^2$ , as we are looking for integral solutions. Hence, $D = d^2 \Rightarrow (2 k - d)(2k +d) = 171 = 3^2 \cdot 19$ Solving the equation above yields the following integral values for $k$ , $k = \pm 43, \pm 15, \pm 7$

Plugging those values for $k$ into our quadratic equation in $n$ and solving it, yields the following positive pairs of integral solutions, $(k, n) = (15, 2), (43, 31)$ The sum we are asked to find the value for can now be computed, $\sum_{i=1}^{2} (-30)^{i - 1} m_i n_i = (-30)^{1 - 1} \cdot 43^2 \cdot 31 + (-30)^{2 - 1} \cdot 15^2\cdot 2 = \boxed{43819}$

[This is not a solution.]

This problem previously stated $\sum ( -30) ^{i-1} m_i n_i$ , for which the correct answer would be 43819. Those who previously entered 43819 have been marked correct.