Solving this system of equations without a CAS: $4=a+c$, $5=d+ac+b$, $4=ad+bc$, $4=bd$

Question

Solving this system of equations : $$4=a+c$$ $$5=d+ac+b$$ $$4=ad+bc$$ $$4=bd$$

Anyone knows how to solve this without a CAS?

Answer 1

We obtain $$c=4-a$$ and $$d=\frac{4}{b},$$ which gives $$\frac{4}{b}+b+a(4-a)=5$$ and $$a\cdot\frac{4}{b}+b(4-a)=4.$$ From the last equation we obtain: $$a=\frac{4b^2-4b}{b^2-4},$$ which gives $$\frac{4}{b}+b+\frac{4b^2-4b}{b^2-4}\left(4-\frac{4b^2-4b}{b^2-4}\right)=5$$ or $$\frac{b^2-5b+4}{b}+\frac{16b(b-4)(b-1)}{(b^2-4)^2}=0$$ or $$(b-1)(b-4)\left(\frac{1}{b}+\frac{16b}{(b^2-4)^2}\right)=0$$ or $$(b-1)(b-4)=0$$ because $$(b^2-4)^2+16b^2>0.$$ Can you end it now?

Answer 2

A particular solution is $(a,b,c,d)=(4,4,0,1)$. This becomes from the following: $$\begin{cases}a+c=4\\ac+b+d=5\\ad+bc=4\\bd=4\end{cases}\Rightarrow\begin{cases}c(4-c)+b+d=5\\d(4-c)+bc=4\\bd=4\end{cases}\Rightarrow\begin{cases}4c-c^2+\dfrac 4d+d=5\\4d-cd+\frac{4c}{d}=4\end{cases}$$ The third system with two unknowns give as resultant $$d^6-5d^5+12d^4-40d^3+48d^2-80d+64=(d-1)(d-4)(d^2+4)^2=0$$ which gives the solution above and others whose calculation is left for the OP.