Find the sum of all the integer values of x and y which can be plugged in the equation shown.

In this proof we will use the fact that for $m,n \in \mathbb{Z}$ , either $m^2 = n^2$ or $\bigl \vert m^2 - n^2 \bigr \vert \ge 2n - 1$ ; this is because the closest square to $n^2$ is $(n-1)^2 = n^2 - 2n + 1$ .

We complete the square on $\ y^4 + y^3 + y^2 + y + 1$ ; the simplest way is by equating coefficients. We get

$y^4 + y^3 + y^2 + y + 1 = (y^2 + \frac{1}{2}y + \frac{3}{8}) + \alpha y + \beta$

Since the perfect square has fractional coefficients, we multiply our equation by the square of the LCD to simplify finding $\alpha$ and $\beta$ :

$(8x)^2 = (8y^2 + 4y + 3)^2 + 40y + 55$

Now we apply the fact mentioned at the top. If $\ (8x)^2 = (8y^2 + 4y + 3)^2$ , then $\ 40y + 55 = 0$ , but then clearly $y \not\in \mathbb{Z}$ . Thus it must be true that

$\begin{aligned} \bigl \vert (8x)^2 - (8y^2 + 4y + 3)^2 \bigr \vert &\ge 2 \ (8y^2 + 4y + 3) -1 \\ \\ \bigl \vert 40y + 55 \bigr \vert &\ge 16y^2 + 8y + 5 \\ \\ 16y^2 + 8y + 5 \le 40x + 55 \qquad &\text{OR} \qquad 16y^2 + 8y + 5 \le \text{-}40x - 55 \\ \\ 16y^2 - 32y -50 \le 0 \qquad &\text{OR} \qquad 16y^2 + 48y + 60 \le 0 \end{aligned}$

The second quadratic has a negative discriminant, and since its leading coefficient is positive, the second inequality has no solution in $\mathbb{R}$ . Solving the first, we get

$\begin{aligned} \dfrac{4 - \sqrt{66}}{4} \le \ &y \ \le \dfrac{4 - \sqrt{66}}{4} \\ \\ \text{-}1.031... \le \ &y \ \le 3.031... \end{aligned}$

So $y \in \{\text{-}1,0,1,2,3\}$ . Testing these in the polynomial, the only ones that yield squares are $\text{-}1,0$ and $3$ , and the six solutions $(x,y)$ are

$(\text{-}1,\text{-}1); (1,\text{-}1); (\text{-}1,0); (1,0); (\text{-}11,3); (11,3)$

and only the last one satisfies the condition in the question. Thus our answer is $\ x + y = 11 + 3 = \boxed{14}$

Note: we could have shortened the proof a little by applying the condition that our solutions had to be greater than 2 earlier, but this way we found all solutions over the integers.