Let's determine the coefficient values of $A, B, C, D$ (non-zero real numbers) for the following two polynomials such that their product consists of only 3 terms. - Is there a value system where both polynomials are perfect squares?

$$
\begin{aligned}
& x^{6}+4 x^{5}+A x^{4}+B x^{3}+C x^{2}+D x+1 \\
& x^{6}-4 x^{5}+A x^{4}-B x^{3}+C x^{2}-D x+1
\end{aligned}
$$
a) The product of the two polynomials is of the 12th degree, and it only contains even powers of $x$. The coefficient of $x^{12}$ and the constant term are 1. We choose the exponent of the allowed third term to be 6, so the coefficients of $x^{10}$, $x^{8}$, $x^{4}$, and $x^{2}$ must vanish:

$$
\begin{array}{r}
2 A-16=0 \\
A^{2}-8 B+2 C=0 \\
2 A-2 B D+C^{2}=0 \\
2 C-D^{2}=0
\end{array}
$$

From (3), $A=8$, from (6), $C=D^{2} / 2$, from (4) and (6), $B=8+C / 4=8+D^{2} / 8$, and substituting these into (5):

$$
\begin{aligned}
& D^{4}-D^{3}-64 D+64=(D-1)\left(D^{3}-64\right)= \\
& \quad=(D-1)(D-4)\left(D^{2}+4 D+16\right)=0
\end{aligned}
$$

From the vanishing of the first two factors, we get one set of valid values each:

$$
\begin{array}{lll}
D_{1}=1, & B_{1}=65 / 8, & C_{1}=1 / 2 \\
D_{2}=4, & B_{2}=10, & C_{2}=8
\end{array}
$$

In this case, the product of the two polynomials is:

$$
x^{12}+\frac{4097}{64} x^{6}+1, \quad \text { or } \quad x^{12}-2 x^{6}+1=\left(x^{6}-1\right)^{2}
$$

b) It is easy to notice that the second set of values obtained also satisfies the second requirement, in which case the two polynomials are:

$$
\begin{aligned}
& \left(x^{3}+2 x^{2}+2 x+1\right)^{2}=\left[\left(x^{3}+1\right)+2 x(x+1)\right]^{2}=(x+1)^{2}\left(x^{2}+x+1\right)^{2} \\
& \left(x^{3}-2 x^{2}+2 x-1\right)^{2}=\left[\left(x^{3}-1\right)-2 x(x-1)\right]^{2}=(x-1)^{2}\left(x^{2}-x+1\right)^{2}
\end{aligned}
$$

In general, (1) and (2) are the square of the polynomials $x^{3}+2 x^{2}+b x+c$, and $x^{3}-2 x^{2}+b' x+c'$, respectively, if the following conditions are met:

$$
\left.\left.\begin{array}{rl}
4+2 b & =A, \\
2 c+4 b & =B, \\
4 c+b^{2} & =C, \\
2 b c & =D, \\
c^{2} & =1 ;
\end{array}\right\} \quad \text { and } \quad \begin{array}{rl}
4+2 b' & =A \\
2 c'+4 b' & =-B \\
-4 c'+b'^{2} & =C \\
2 b' c' & =-D \\
c'^{2} & =1
\end{array}\right\}
$$

Adding the first four conditions on the left and right, and subtracting the fifth conditions:

$$
\left(c+c'\right)+2\left(b-b'\right)=0, \quad c^{2}-c'^{2}=\left(c+c'\right)\left(c-c'\right)=0
$$

Each of these, as well as the first four pairs of conditions, is satisfied if $c'=-c$, and $b'=b$, and finally from the fifth condition, $c=1$. Thus, for any real parameter $b$, the following is valid:

$$
A=2 b+4, \quad B=4 b+2, \quad C=b^{2}+4, \quad D=2 b
$$

and in this case:

$$
\begin{gathered}
x^{6} \pm 4 x^{5}+(2 b+4) x^{4} \pm(4 b+2) x^{3}+\left(b^{2}+4\right) x^{2} \pm 2 b x+1= \\
=\left(x^{3} \pm 2 x^{2}+b x \pm 1\right)^{2}
\end{gathered}
$$

Our observation above follows from this for $b=2$.