If $ab=cd$ then $a+b+c+d$ is not prime, for integers $\,a,b,c,d > 0$ 
if  $ab=cd$ then $a+b+c+d$ is not prime, for integers $\,a,b,c,d > 0$

I tried different ways (e.g. eliminating one variable and other substituions), but I was not able to continue to a solution.
 A: Let $\,s\,$ be the sum. Then $\ as = a^2 + \!\!\overbrace{ab}^{\Large cd}\!+ac+ad = \overbrace{(a+c)}^{\large M}\,\overbrace{(a+d)}^{\large N}\,$  
By unique factorization $\, a\mid MN\,\Rightarrow\,a = mn,\,\ m\mid M,\ n\mid N,\,$ therefore
$\ s= \dfrac{MN}a = \dfrac{M}m\,\dfrac{N}n,\ $ both $>1\,$ by  $\,m,n \le a < \overbrace{a\!+\!\color{#c00}c}^{\large M},\,\overbrace{a\!+\!\color{#c00}d}^{\large N}\,$ by $\,\color{#c00}{c,d>0}$

Remark $ $ For completeness here is proof of said useful consequence of unique factorization.
$a\mid MN\,\Rightarrow\,ab = MN\,$ so $\, a(b/d) = M (N/d)\ $ by  cancelling $\ d=\gcd(b,N).\ $
$b'\! := b/d\,$ is coprime to $N/d$ so it divides $M,\,$ so $\,a = (M/b')\,(N/d),\ $ $\underbrace{M/b'}_{\large m}\!\mid M,\,\ \underbrace{N/d}_{\large n}\mid N$
This property is the basis for a refinement based view of unique factorization (which proves esp. convenient in the noncommutative case). See this answer for more on that, including references.
A: HINT: with $$d=\frac{ab}{c}$$ we get $$a+b+c+d=\frac{(a+c)(b+c)}{c}$$
A: let $a = mn, b = pq, c = mp, b = nq$, thus $a+b+c+d = mn+pq+mp+nq = (m+q)(p+n)$, obviously, $m, n, p, q >= 1$, thus $LHS$ is not a prime.  
