Perfect square quadratic

By completing the square $f(n) =(n+6)^{2} - 871$

Let $n+6 = x$ and let the value of $(n+6)^{2} - 871$ be $y^{2}$ .

$x^{2} - 871 = y^{2} => x^{2} - y^{2} = 871 => (x+y)(x-y) = 871$

To find the largest integer value of $x$ , $x+y = 871$ and $x-y = 1$

Therefore, $2x = 872 => x = 436$

since $x = n+6, n =430$ and this is the largest integer value of n such that $f(n)$ is a perfect square.

For every positive integer value of $f(n)$ , there is a negative integer value such that $f(n)$ is a perfect square. This is because $(n+6)^{2} - 871 = (-n-12+6)^{2} - 871$ . Therefore, if $n$ is a positive integer solution, $(-n-12)$ is also an integer solution. As a result, for each pair of positive and negative solutions, their sum is $n-n-12 = -12$ . Since we have 2 pairs of positive and negative integer solutions, their sum is $-24$ .

Finally, the product of the largest value with the sum of the 4 values is $-24*430 = -10320$