How can one find the factorization $a^4 + 2a^3 + 3a^2 + 2a + 1 = (a^2 + a + 1)^2$ from scratch? I want to prove $a^4 + 2a^3 + 3a^2 + 2a + 1$ always be perfect square for $a \in \mathbb{N}$. Using GeoGebra I found $a^4 + 2a^3 + 3a ² + 2a + 1 = (a^2 + a + 1)^2$ which is trivial to verify.
Since this equation has no real roots, it's not so easy to decompose it through division by $a-a_0$ where $a_0$ is a root. I don't want to use complex numbers if possible. How do find this factorization from scratch / without using a computer?

One may notice that $p(x)=x^4+2x^3+3x^2+2x+1$ is a palyndromic polynomial, hence $\frac{p(x)}{x^2}$ is a polynomial in $z=x+\frac{1}{x}$:
$$ \frac{p(x)}{x^2} = \left(x^2+\frac{1}{x^2}\right)+2z+3 = (z^2-2)+2z+3 = z^2+2z+1 = \color{red}{(z+1)^2} $$
so:
$$ p(x) = x^2\left(x+\frac{1}{x}+1\right)^2 = \color{red}{(1+x+x^2)^2}. $$
