Squares and squares

Let $x = a^2$ where $a \geq 0$ .

We want $a^6 + 2a^4 + 2a^2 + 4$ to be a perfect square. Since this is polynomial of even degree, it suggests that we should attempt to bound it by perfect square polynomials. Notice that $a^6 + 2a^4 + 2a^2 + 4$ is close to $(a^3 + a + 1)^2 = a^6 + 2a^4 + a^2 + 2a + 1$ . Let's find the range that

$(a^3 + a)^2 < a^6 + 2a^4 + 2a^2 + 4 < (a^3 + a + 1) ^ 2 .$

Expanding and simplifying (some work required), we see that this is satisfied when $a > 1$ . Thus, when $a > 1$ , $a^6 + 2a^4 + 2a^2 + 4$ is bounded between two consecutive perfect squares, hence it cannot be a perfect square.

It then remains to check what happens when $a = 0, 1$ .
$a = 0, x = 0, x^3 + 2x^2 + 2x + 4 = 4$ is a perfect square.
$a = 1, x = 1, x^3 + 2x^2 + 2x + 4 = 9$ is a perfect square.
Hence, the sum of solutions is $0 + 1 = 1$ .

I used analysis of the function by taking the first and second en order deravative to find that $x = \frac{-2}{3}$ is a point of inflection. From there on I looked at the first square 1. This yields a perfect square as outcome. Looking at the first deravative tells us that the slope is getting great enough to ensure there is no next perfect square.