I can solve this, but I need to feed the cat

The expression $f(n) = 80n^4 + 40n^3 - 600n^2 + 4000n - 10000 = k^2$ for some integer $k$ can also be written as $(3n)^4 - (n - 10)^4 = k^2$

Letting $a = 3n$ , $b = n - 10$ , $c = k$ gives the form $a^4 - b^4 = c^2$ which has no integral solutions for $abc \neq 0$ .

Case 1: $a = 3n = 0$ or $n = 0$

$80(0)^4 + 40(0)^3 - 600(0)^2 + 4000(0) - 10000 = -10000$ -10000 is clearly not a perfect square so $n = 0$ is not a solution

Case 2: $n - 10 = 0$ or $n = 10$

$80(10)^4 + 40(10)^3 - 600(10)^2 + 4000(10) - 10000 = 900$ $n = 10$ is a solution since $900 = 30^2$

Case 3: $3n = n - 10$ or $n = -5$

$n = -5$ is another solution because this is the case when $k^2 = 0$ which is a perfect square.

$10 + (-5) = \boxed{5}$

Note:

The proof for the form $a^4 - b^4 = c^2$ is the only proof given by Pierre de Fermat which consequently lead to the $n = 4$ case of the Fermat's Last Theorem . The non-existence of the solution can be proven using Fermat's Method of Infinite Descent .

We have for a non-negative integer $m$ : $40(n+5)(2n-5)(n^2-2n+10)=m^2.$ Observe that $40 \mid m^2$ . Thus $20 \mid m$ . Write $m=20k$ to get $(n+5)(2n-5)(n^2-2n+10)=10k^2,$ Where $k$ is a non-negative integer. Look mod $5$ . We get $n^3(n-2) \equiv 0 \pmod{5}$ . So we have either $n \equiv 0 \pmod{5}$ or $n \equiv 2 \pmod{5}$ .

If $n \equiv 0 \pmod{5}$ then let $n=5N$ to get $125(N+1)(2N-1)(5N^2-2N+2)=10k^2$ It follows that $25 \mid k^2$ and $k=5K$ . Now we have the following equation $(N+1)(2N-1)(5N^2-2N+2)=2K^2$

I'll complete this later.