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?

Standard Solution:
Let $a^4+2a^3+3a^2+2a+1=(ba^2+ca+d)^2$ where $b,c,d \in \mathbb{R}$
Observe that
$$(ba^2+ca+d)^2=b^2a^4+2bca^3+(2bd+c^2)a^2+2cda+d^2$$
Comparing coefficient of like terms, we get 
$$b^2=1 \Rightarrow b=\pm 1$$
$$bc=1 \Rightarrow c=\pm 1$$
$$ \pm2d+1=3 \Rightarrow d=\pm 1$$
Thus, $$a^4+2a^3+3a^2+2a+1=[\pm(a^2+a+1)]^2$$
Note: Although it appears that this method assumes that the quartic must be the square of a polynomials, but if you end up getting no answer, it means you arrived at a contradiction and the equation is not the square of a polynomial. Moreover, the if there are 2 answers, they are additive inverses of each other.

Alternate Solution:
Let $f(a)=a^4+2a^3+3a^2+2a+1$
Observe that $f(0)=1$ and that the coefficients follow this pattern: $1,2,3,2,1$
Dividing $f(a)$ by $a^2$ to get 
$$g(a)=\frac{f(a)}{a^2}=a^2+2a+3+\frac{2}{a}+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2+2\left(a+\frac{1}{a}\right)+1$$
Substitute $t=a+\dfrac{1}{a}$ to get 
$t^2+2t+1=(t+1)^2$
Now, back-substitute to get 
$$f(a)=a^2 \cdot \left(a+\frac{1}{a}+1\right)^2=(a^2+a+1)^2$$
But since $x^2=(-x)^2$, we get 
$$\color{blue}{\boxed{\color{red}{a^4+2a^3+3a^2+2a+1=[\pm (a^2+a+1)]^2}}}$$
