Is a Sum a Prime Number?

Problem

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).

 

|Contact| |Up| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

71471736