7. Solve the system

$\left\{\begin{array}{l}a+c=4 \\ a d+b c=5 \\ a c+b+d=8 \\ b d=1\end{array}\right.$

In the answer, write the sum of all solutions of the given system.
7. We have

$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=$

$=x^{4}+4 x^{3}+6 x^{2}+5 x+2=(x+1)(x+2)\left(x^{2}+x+1\right)$. Since the polynomial $x^{2}+x+1$ does not have

real roots, the identity $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}+3 x+2\right)\left(x^{2}+x+1\right)$

means that $\left\{\begin{array}{l}x^{2}+a x+b=x^{2}+3 x+2 \\ x^{2}+c x+d=x^{2}+x+1\end{array}\right.$, or $\left\{\begin{array}{l}x^{2}+a x+b=x^{2}+x+1 \\ x^{2}+c x+d=x^{2}+3 x+2\end{array}\right.$.

From this, we find that $a=3, b=2, c=1, d=1$ or $a=1, b=1, c=3, d=2$ (since the equality of polynomials means the equality of coefficients of corresponding powers). Answer: $\{14\}$