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

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.
