They're Not All The Same Degree!

Since $a^3 \equiv a$ modulo $3$ for all integers $a$ , and since $x^3 + y^3$ must be a multiple of $3$ , it follows that $x+y \equiv 0$ modulo $3$ . If, for example, $x \equiv 1$ and $y \equiv 2$ modulo $3$ , then $\tfrac13(x^3 + y^3) + xy \equiv 2$ modulo $3$ , and hence cannot be equal to $2007$ . Thus it follows that $x,y$ must both be multiples of $3$ . Writing $x = 3X$ , $y = 3Y$ , we need to solve $X^3 + Y^3 + XY = 223$ for positive integers $X,Y$ . But then $1 \le X,Y < \sqrt[3]{223} < 7$ , so there are only $36$ cases to check. The only (unordered) pair that works is $\{X,Y\} = \{1,6\}$ , so that $\{x,y\} = \{3,18\}$ , and hence the answer is $\boxed{54}$ .

We could be really basic, and note that, in the original problem, we must have $x,y \le \sqrt[3]{3\times2007}$ , so that $1 \le x,y \le 18$ . A simple Excel calculation can test the $324$ cases.

Sorry for the confusing typo, everyone!

---

We use the factorisation $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ .

$\begin{aligned} \dfrac{x^3+y^3}{3} +xy &= 2007\\ x^3+y^3+3xy &= 6021\\ x^3+y^3+(-1)^3 - 3xy(-1) &= 6020\\ (x+y-1)(x^2+y^2+1-xy+x+y) &= 6020\\ (x+y-1)((x+y-1)^2 - 3xy - 3x - 3y) &= 2^2 \times 5 \times 7 \times 43 \end{aligned}$

Since the RHS $\equiv 2 \pmod{3}$ , the LHS must also be $\equiv 2 \pmod{3}$ . Thus, we must have $x+y-1 \equiv 2 \pmod{3}$ . All positive factors of 6020 that are equivalent to $2 \pmod{3}$ are $2$ , $5$ , $14$ , $20$ , $35$ , $86$ , $140$ , $215$ , $602$ , $860$ , $1505$ and $6020$ . By observation, we get that only $x+y-1=20$ satisfies, with $x=18$ , $y=3$ or vice-versa. Thus, $xy=54$ .

Transform the equation first:

$\begin{aligned} \dfrac{x^3 + y^3}{3}\ + xy &= \dfrac{(x + y)(x^2 - xy + y^2)}{3}\ + xy \\ &= \dfrac{(x + y)((x + y)^2 - 3xy)}{3}\ + xy \\ & = \dfrac{(x + y)^3 - 3xy(x + y)}{3}\ + xy = 2007 \end{aligned}$

Let:

$\begin{aligned} m &= x + y \\ n &= xy \end{aligned}$

Now we have:

$\begin{aligned} \dfrac{m^3 - 3mn}{3}\ + n &= 2007 \\ \dfrac{m^3}{3}\ - mn + n &= 2007 \\ \dfrac{m^3}{3}\ + n(1 - m) &= 2007 \\ \dfrac{m^3}{3}\ - 2007 &= n(m -1)\\ \dfrac{m^3 - 6021}{3(m - 1)}\ &= n \end{aligned}$

Since $n$ has to be a positive integer, it must be that:

$\begin{aligned} m^3 > 6021 \Rightarrow m >18\end{aligned}$ .

Also by known inequality:

$\begin{aligned} xy &\leq \dfrac{(x + y)^2}{4}\ \\ n &\leq \dfrac{m^2}{4}\ \\ \dfrac{m^3 - 6021}{3(m - 1)}\ &\leq \dfrac{m^2}{4} \\ \dfrac{m^3 - 6021}{3}\ &\leq \dfrac{m^3 - m}{4} \\ 4m^3 - 24084 &\leq 3m^3 - 3m \\ m^3 + 3m &\leq 24084\end{aligned}$

Knowing that $30^3 = 27000$ and by testing, we find that: $m < 29$ .

Having that: $3|m^3 - 6021$ and $3|6021$ it must be that $3|m^3 \Rightarrow 3|m$ . So, $m$ can be either $21$ , $24$ , or $27$ . By replacing $m$ with each of these numbers, we can conclude that $n$ is integer only if $m = 21 (n = 54)$ , which means there is only one unordered pair of integers $x$ and $y$ that satisfies the starting equation. We were asked to find the sum of products of all such pairs, and knowing that there is only one of them, the solution is $n = xy = \boxed{54}$