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For some natural numbers $a, b, c, d$ if $a\cdot b = c \cdot d$ is it possible that the sum $a + b + c + d$ is prime number, or in other words it doesn't have divisors rather then $1$ and itself.

What I tried

I think that the answer is not, here is my try

Case 1: All numbers $a, b, c, d$ are even, the sum will be even too, so in this case $a+b+c+d$ cannot be prime number

Case 2: One number of $(a, b)$ is even and one is odd, let's say $a$ is even, and $b$ odd. Let it hold for $(c, d)$ too, $c$ is even, $d$ is odd. Obviously $b + c$ is going to be an even number, because two odd numbers sum up to an even number. When we add $a$ and $c$ to this sum, we will end with an even number, and not prime number.

Case 3: Exactly one of the $4$ numbers $a, b, c, d$ is odd. For simplicity let it be $a$. It is obvious that the sum $a+b+c+d$ is going to be odd number.

I'm stuck on case 3, can you give me some hints how to continue and finish the proof.

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    $\begingroup$ $ab+ac=ac+cd$, and that relates $b+c$ to $a+d$. Maybe that will help. $\endgroup$ – Michael Jan 18 '18 at 15:02
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Michael's comment can be used to give an answer.

$a(b+c)=c(a+d)$, and so $\frac{a+d}{b+c}=\frac{a}{c}$. This means that $\frac{a+d}{b+c}$ is not in lowest terms, i.e. $\gcd(a+d,b+c)=h>1$. But then $h$ is a nontrivial factor of $a+b+c+d$.

(This assumes $a,b,c,d>0$; without this assumption $1\times0=2\times 0$ gives a counterexample.)

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Since $d=\frac{ab}{c}$ is a natural number, we have $a=a_1c_1$, $b=b_1c_2$, where $\{a_1,b_1,c_1,c_2\}\subset\mathbb N$ and $c=c_1c_2.$

Thus, $$a+b+c+d=a+b+c+\frac{ab}{c}=\frac{(c+a)(c+b)}{c}=$$ $$=\frac{(c_1c_2+a_1c_1)(c_1c_2+b_1c_2)}{c_1c_2}=(c_2+a_1)(c_1+b_1).$$

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Note that $\dfrac{ab}{c}$ is an integer and so is $$a+b+c+\frac{ab}{c}=\frac{1}{c}(a+c)(b+c),$$ which implies $c \mid (a+c)(b+c)$. Assume $c=uv$ and $u \mid a+c$ and $v \mid b+c$, then $$\frac{1}{c}(a+c)(b+c)=\frac{a+c}{u} · \frac{b+c}{v},$$ where both factors are greater than $1$. Hence, the number is not a prime.

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    $\begingroup$ Please use MathJax for mathematical equations. $\endgroup$ – jvdhooft Jun 15 '18 at 15:06

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