Nice one but I was looking for another one!

a+b+c+d=(1/a)*(a^2+ab+ac+ad)=(1/a)*(a^2+ac+ad+cd) = (1/a)*(a+c)*(a+d) As c,d are positive a+c,a+d are greater than a. So, a+b+c+d is composite.

lemma.if $ m|xy , 0<|x|,|y|<m$ then $ m$ is composite. $ ab=cd\Rightarrow{a(a+b+c+d)=(a+c)(a+d)}\Rightarrow a+b+c+d|(a+c)(a+d)$ and obviously $ 0<a+c , a+d<a+b+c+d$ therefore $ a+b+c+d$ is composite. @praneeth:is this the same as your solution? another solution: $ ab=cd\Rightarrow (a+b+c+d)(a+b-c-d)=(a-b+c-d)(a-b-c+d)$ if $ a-b+c-d=0$ or $ a-b-c+d=0$ then $ a+b+c+d$ is even and greater than $ 2$,so is composite. and if $ a-b+c-d\neq0$ and $ a-b-c+d\neq0$ then: $ a+b+c+d|(a-b+c-d)(a-b-c+d) , 0<|a-b+c-d|,|a-b-c+d|<a+b+c+d$ so $ a+b+c+d$ is composite.

Assume $p=a+b+c+d$ where $p\ge 4$ is a prime number. Then working mod $p$ we have $ab=cd\equiv c(-a-b-c)\implies 0\equiv ab+ac+bc+c^2=(c+a)(b+c)$ So $p|a+c\vee p|b+c$ but this is impossible, since $b+c,a+c<a+b+c+d=p$ Contradiction!

tjhance wrote:

Could you kindly show how to prove the existence of $w,x,y,z$?

It's Very simple just observe $\frac{a}{c}=\frac{d}{b}\implies\frac{a+c}{c}=\frac{d+b}{b}$ so we get $a+b=\frac{(d+b)c}{b}$ hence $a+b+c+d=\frac{(d+b)(b+c)}{b}$ which is always composite. We are done ! @below because $b|cd$

@above, Would you enlighten me how $\frac{(d+b)(b+c)}{b}$ is always composite?? Thank you!