Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

$(2x^2-2x+2)^2 = 4x^4-8x^3+12x^3-8x+4$

Now, $2x^2-2x+2$ is always integer for integer $x$. So, $f(x)$ is always a perfect square.

Squaring a function with degree $n$ gives a function with degree $2n$ and the given is a quartic function. So, searching for a function with degree $2$ as $ax^2+bx+c$ is the best option.

$(ax^2+bx+c)^2 = a^x+2abx^3+(2ac+b^2)x^2+2bcx+c^2$. 

Comparing the coefficients, we get $a=2,c=2$ and $b=-2$