if $ab=cd$ then $a+b+c+d $ is composite Let $a,b,c,d$ be natural numbers with $ab=cd$.
Prove that $a+b+c+d$ is composite.
I have my own solution for this (As posted) and i want to see if there is any other good proofs.
 A: From $ab=cd$, We may assume $a=\frac{cd}{b}$. So $M=a+b+c+d = \frac{cd}{b}+b+c+d = \frac{(b+c)(b+d)}{b}$ and so $bM=(b+c)(b+d)$ and $M|(b+c)(b+d)$. We assume that $M$ is not composite, so it is prime. Now we may know that either $b+c$ or $b+d$ is divisible by $M$. So $M\leq b+c$ or $M\leq b+d$ which both result in contradiction because $M=a+b+c+d > b+c$ or $b+d$. So our assumption was wrong and $M$ is a composite number.
A: $ab=cd$ implies $a=xy, b=zt, c=xz, d=yt$ for some integers $x,y,z,t$. Hence
$$
a+b+c+d=(x+t)(y+z).
$$
A: Hint: Plug $a=\frac{cd}{b}$ into the sum to get 
$$\frac{(b+c)(b+d)}{b}$$
which cannot be prime. 
A: Call the sum $\,f.\ $ Then $\ af = a^2 + \!\overbrace{ab}^{\Large cd}\!+ac+ad = \overbrace{(a+c)}^{\large M}\,\overbrace{(a+d)}^{\large N}\,$
By unique factorization $\, a\mid MN\,\Rightarrow\,a = mn,\,\ m\mid M,\, n\mid N,\,$ thus
$\ f = \dfrac{MN}a = \dfrac{M}m\dfrac{N}n.\ $ Each factor is $>1\,$ by  $\,m,n \le a < M,N$.
A: Write $p=a+b+c+d$ and say $p$ is prime. Then we have $$ab=c(p-a-b-c)$$ so $$(a+c)(b+c) = cp$$
which means that $$p\mid a+c\;\;\;\;{\rm or}\;\;\;\;p\mid b+c$$
in 1st case we get $p=a+b+c+d\leq a+c$ a contradiction. The same contradiction we get in the second case. So $p$ must be composite.
A: From $ab=cd$ you have $$(a+b)^2-(a-b)^2=(c+d)^2-(c-d)^2\Rightarrow(a+b)^2-(c+d)^2=(a-b)^2-(c-d)^2$$ Hence we have $$(a+b+c+d)(a+b-c-d)=(a-b+c-d)(a-b-c+d)$$ Now note that $|a+b+c+d|>|a-b+c-d|$ and $|a-b-c+d|$. If $(a+b+c+d)$ was prime then it must divide one of $(a-b+c-d)$ or $(a-b-c+d)$, which is not possible.
A: Hint:
$ab$ has to have at least $3$ prime factors.(If $a,b,c,d$ are distinct naturals) 
$ab=p_1p_2p_3\dots p_n=cd$
$a=p_1p_2 \dots p_j$
$b=p_{j+1} \dots p_n$
$c=p_kp_{k+1} \dots p_l$
$d=p_1p_2 \dots p_{k-1}p_{l+1}p_{l+2} \dots p_n$
A: $ab=cd$ (with $a, b, c, d$ different from zero otherwise the statement is trivial) implies that 
$$
\frac{a}{c}=\frac{d}{b}=k
$$
Hence $a=ck$ and $d=bk$. Plugging these two last equations into $a+b+c+d$
one finds:
$$
a+b+c+d=ck+b+c+bk=k(b+c)+(b+c)=(k+1)(b+c)
$$
This seems the simplest solution to me but maybe I missed something.
