# Prove $a+b+c+d$ 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.

-
Is this homework? What have you tried? –  A.P. May 6 '13 at 15:38
I vote against closing this question. –  MJD May 6 '13 at 15:45
Let the OP respond before closing the question. The question seems pretty good. –  Inceptio May 6 '13 at 15:48
I, too, find this question interesting. Nevertheless I downvoted it for the way it was asked: No context was given, no own effort is shown and only in the comments it turns out it was merely asked to see if someone will come up with a better solution and that the questioner already has solutions. So my read on it is that it’s basically a test to see whether other people are as clever as the questioner which I find rude. But maybe I misinterpret this? At the very least, I think you, CODE, should add your intentions to the question. –  k.stm May 6 '13 at 16:10
Note this does not work if you include 0 in the natural numbers. (Obvious counterexample: a = 0, b=3, c=0, d=4, so ab = cd = 0, but a+b+c+d=7 which is prime). –  dr jimbob May 6 '13 at 18:49

$ab=cd$ implies $a=xy, b=zt, c=xz, d=yt$ for some integers $x,y,z,t$. Hence $$a+b+c+d=(x+t)(y+z).$$

-
This may be nitpicking, but $2 \cdot 3 = 3 \cdot 2$ does not imply $2 = xy$. –  orlp May 6 '13 at 19:08
@nightcracker: Sorry, I didn't understand. Do you consider the case $a=d=2, b=c=3$? –  Boris Novikov May 6 '13 at 19:16
@nightcracker: $a=2$, $b=3$, $c=3$, $d=2$, $x=1$, $y=2$, $z=3$, $t=1$ works. –  fgrieu May 6 '13 at 19:59
Of course, pesky $1$ :D Nevermind. –  orlp May 6 '13 at 20:07
@nightcracker: Still this answer is more that a bit sketchy, for the construction of $x$, $y$, $z$, $t$ is left as an exercise to the reader. –  fgrieu May 6 '13 at 20:53

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.

-
+1. Beautiful. Though this argument won't work (as it is) for $a,b,c,d \in \mathbb{Z}^+$. –  user17762 May 6 '13 at 15:51
It's not hard to show a slightly simpler identity: $a(a+b+c+d)=(a+c)(a+d)$, with the result following from there by a similar size argument. –  Thomas Andrews May 6 '13 at 16:01
(+1) Brilliant. –  achille hui May 6 '13 at 16:08
@user17762 It's not clear to me what is wrong with the proof in your eyes. –  Thomas Andrews May 6 '13 at 16:39
@ThomasAndrews I think he was getting at the fact that one (or both) of $a-b+c-d$ or $a-b-c+d$ may be $0$, in which case $a+b+c+d$ would divide it even if it were prime. Of course, in that case $a,b$ is the same as $c,d$ modulo reordering, so the sum is divisible by 2 and the result still holds. –  jerry May 6 '13 at 21:08

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.

-

Hint: Plug $a=\frac{cd}{b}$ into the sum to get

$$\frac{(b+c)(b+d)}{b}$$

which cannot be prime.

-
Why can you assume that? –  Thomas Andrews May 6 '13 at 15:57
It may not be correct if $b+d$ was not divisible by $b$. –  CODE May 6 '13 at 16:00
@CODE: See the edit. –  Ehsan M. Kermani May 6 '13 at 16:04
Dear @Thomas Andrews: I added the word hint into the body. –  Ehsan M. Kermani May 6 '13 at 16:09
@Thomas Andrews: thanks, edited! –  Ehsan M. Kermani May 6 '13 at 17:44

Hint:

$ab$ has to have at least $3$ prime factors.(If $a,b,c,d$ are distinct naturals)

$ab=p_1p_2p_3\dots p_n=cd$

$a=p_1p_2 \dots p_j$

$b=p_{j+1} \dots p_n$

$c=p_kp_{k+1} \dots p_l$

$d=p_1p_2 \dots p_{k-1}p_{l+1}p_{l+2} \dots p_n$

-
Who says they have to be distinct? Take, for example, $a=b=2, c=4, d=1$. –  jerry May 6 '13 at 21:29
None says they have to be distinct. You have a more general case if you distinct numbers. –  Inceptio May 7 '13 at 3:44

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 = MN/a= (M/m)(N/n).\$ Each factor is $>1\,$ by $\,m,n \le a < M,N$

-
Imported from a deleted question. –  Bill Dubuque Apr 23 at 2:52