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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 (posted below). I want to see if there is any other good proofs.

Bill Dubuque
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CODE
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7 Answers7

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By the Four Number Theorem (a consequence of unique prime factorization), it follows that $ab=cd$ implies $a=xy, b=zt, c=xz, d=yt$ for some integers $x,y,z,t$. Hence $$ a+b+c+d=(x+t)(y+z). $$

Bill Dubuque
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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
  • @fgrieu To make it a little less sketchy: Let $x$ be $a$ and $c$'s highest common factor, $y=a/x$, $z=c/x$, $t=b/z$. $y$ and $z$ are coprime integers by construction, and $by=ab/x=cd/x=dz$. Therefore $t=b/z=d/y$ is an integer. – Rosie F Oct 29 '24 at 08:47
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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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2

Call your sum $\:\!S.\ $ Then $\ a\color{#c00}{S} = a^2 + \!\overbrace{ab}^{\Large cd}\!+ac+ad = \overbrace{(a+c)}^{\large \color{#0a0}M}\,\overbrace{(a+d)}^{\large\color{c00} N}$.

By unique factorization: $\, a\color{#c00}S \!=\! \color{#0a0}MN\,\Rightarrow\,a = \color{#0a0}mn,\begin{align} \color{#0a0}m\:\!&|\:\!\color{#0a0}M\\ n\:\!&|\:\!N\end{align}$ so $\ { \color{#c00}S =\! \dfrac{\color{#0a0}MN}{\color{c00}a}} \!=\! \color{#0a0}{\dfrac{M}m}\dfrac{N}n$

Bill Dubuque
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  • Note: both factors $,\frac{M}m,,\frac{N}n>1,$ by $,m,n \le a < M,N.\ \ $ – Bill Dubuque Aug 08 '24 at 23:13
  • This answer is imported from an older deleted question where $,S=a^2+b^2+c^2+d^2,$ therefore - by exactly the same method that we employed above - we deduce that $,a^2 S = (a^2+c^2)(a^2+d^2),,$ etc. $\ \ $ – Bill Dubuque Oct 29 '24 at 06:20
  • Duplicate of answer posted 2 years prior. Please check other answerS before posting. – nonuser Nov 01 '24 at 18:45
  • @nonuser (aka aqua) I don't see it. Where is your proposed dupe? fyi: the invocation of the Four Number Theorem in Boris's answer was recently added by me (before it said simply "implies" without any hint of what method was intended). $\ \ $ – Bill Dubuque Nov 01 '24 at 19:25
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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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