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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.

Bill Dubuque
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CODE
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    Is this homework? What have you tried? – Alex Provost May 06 '13 at 15:38
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    I vote against closing this question. – MJD May 06 '13 at 15:45
  • Let the OP respond before closing the question. The question seems pretty good. – Inceptio May 06 '13 at 15:48
  • I already know the answer but it is kinda hard. SO i posted it in here to see if anyone has a better solution. – CODE May 06 '13 at 15:48
  • @CODE: It will be very good of you if you post the answer you know. We can indeed think of another way out. – Inceptio May 06 '13 at 15:49
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    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 06 '13 at 16:10
  • Here you are! I posted one of my own answers :) – CODE May 06 '13 at 16:23
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    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 06 '13 at 18:49

8 Answers8

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

Boris Novikov
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  • This may be nitpicking, but $2 \cdot 3 = 3 \cdot 2$ does not imply $2 = xy$. – orlp May 06 '13 at 19:08
  • @nightcracker: Sorry, I didn't understand. Do you consider the case $a=d=2, b=c=3$? – Boris Novikov May 06 '13 at 19:16
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    @nightcracker: $a=2$, $b=3$, $c=3$, $d=2$, $x=1$, $y=2$, $z=3$, $t=1$ works. – fgrieu May 06 '13 at 19:59
  • Of course, pesky $1$ :D Nevermind. – orlp May 06 '13 at 20:07
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    @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 06 '13 at 20:53
  • How do we construct x, y, z, t? I don't find it obvious that there are integers like that. – Anay Karnik Jan 05 '17 at 17:59
  • @fgrieu The unjustified claim is sometimes called the Four Number Theorem. It is an easy consequence of the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations of integers). There are many closely related formulations of such factorization refinements, e.g. see the link in my answer. – Bill Dubuque Jan 09 '22 at 13:24
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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.

pritam
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    +1. Beautiful. Though this argument won't work (as it is) for $a,b,c,d \in \mathbb{Z}^+$. –  May 06 '13 at 15:51
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    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 06 '13 at 16:01
  • @user17762 It's not clear to me what is wrong with the proof in your eyes. – Thomas Andrews May 06 '13 at 16:39
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    @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 06 '13 at 21:08
  • Yeah, if $a+b-c-d=0$ then $a+b+c+d=2(a+b)$ is not prime. @jerry – Thomas Andrews May 06 '13 at 21:33
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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.

CODE
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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 $\,m,n \le a < M,N$.

Bill Dubuque
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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.

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

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

which cannot be prime.

Ehsan M. Kermani
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Hint:

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

$ab=p_1p_2p_3\dots p_n=cd$

$a=p_1p_2 \dots p_j$

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

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

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

Inceptio
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$ab=cd$ (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.

Maczinga
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