Is a Sum a Prime Number?
Problem
Solution 1
Note that $\displaystyle \frac{ab}{c}$ is an integer and so is
$\displaystyle a+b+c+\frac{ab}{c}=\frac{(a+c)(b+c)}{c},$
implying $c|(a+c)(b+c).$ Assume $c=uv$ and $u|(a+c)$ and $v|(b+c).$ Then
$\displaystyle \frac{(a+c)(b+c)}{c}=\frac{a+c}{u}\cdot\frac{b+c}{v},$
where both factors are greater than $1.$ Hence, the number is not a prime.
Solution 2
By considering the prime factorizations of $a,b,c,d$ we can prove the existence of four integers $t,u,v,w$ such that
$a=tu,$ $b=vw,$ $c=tw,$ $d=uv.$
It follows that
$a+b+c+d=tu+vw+tw+uv=(u+w)(t+v),$
where both factors are greater than $1.$
Acknowledgment
This is problem 1995-B4 Moscow Mathematical Olympiads, 1993-1999 by R. Fedorov, A. Belov, A. Kovaldzhi, I. Ivashchenko (MSRI/AMS, 2011).
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