A function f from the integers to the integers satisfies f(m + n) = f(m) + f(n) + 2mn for all…

(a) $m = n = 0$: $f(0) = 2f(0)$, so $f(0) = 0$.

A formula. With $m = 1$: $f(n + 1) = f(n) + 4 + 2n$. Starting from $f(0) = 0$ and adding, $f(n) = 4n + 2(0 + 1 + \dots + (n - 1)) = 4n + n(n - 1) = n^2 + 3n$ for $n \ge 0$. With $m = -n$: $0 = f(n) + f(-n) - 2n^2$, so $f(-n) = 2n^2 - (n^2 + 3n) = n^2 - 3n$, which is the same formula at $-n$. So $f(n) = n^2 + 3n$ for every integer, and it does satisfy the equation: $(m + n)^2 + 3(m + n) = (m^2 + 3m) + (n^2 + 3n) + 2mn$.

(b) $f(5) = 25 + 15 = 40$. (c) $f(-3) = 9 - 3 \times 3 = 0$.

(d) $\sum_{k=1}^{10}(k^2 + 3k) = 385 + 3 \times 55 = 550$.

(e) $n^2 + 3n = n(n + 3) < 0$ for $-3 < n < 0$: the integers $-2, -1$, so 2.