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

Question

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.

Answer 1

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 on the Four Number Theorem for more on that, including references.

Aternatively we could assume $\,s\,$ is prime and apply Euclid's Lemma, but that obscures the key role that refinement plays in this and related probelsm.

Answer 2

HINT: with $$d=\frac{ab}{c}$$ we get $$a+b+c+d=\frac{(a+c)(b+c)}{c}$$

Answer 3

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.