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Find a representation that shows $3$ different natural numbers are (pairwise) relatively prime in pairs.

The approach needed is algebraic.

To give an inkling as to what I mean by that, let us consider the problem of $3$ naturals s.t. the sum of any two is divided by the third.
Let the $3$ natural numbers be: $a,b,c$, so:
$\exists l,m,n, \in \mathbb {N}$, s.t.
$a+b=lc, b+c = ma, a+c = nb$.
So, $c = \frac{a+b}{l}, a = \frac{c+b}{m}, b = \frac{a+c}{n}.$

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    $\begingroup$ Pairwise relatively prime or just $\gcd=1$ ? $\endgroup$ – Peter Feb 4 '18 at 16:31
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    $\begingroup$ @Peter I have edited for the missing info. It is pairwise relatively prime pairs. $\endgroup$ – jitender Feb 4 '18 at 16:36
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    $\begingroup$ Hint : Bezout's theorem $\endgroup$ – Peter Feb 4 '18 at 16:40
  • $\begingroup$ @Peter Please vet :: Say, $\exists l,m,n, \in \mathbb{Z}$ s.t. $la+mb=n$. Similarly for other 2 equalities relating to the other two r.p. pairs. In nutshell, the 3rd number will not appear in the linear combination in such equalities for pairwise r.p. numbers. $\endgroup$ – jitender Feb 4 '18 at 16:51
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    $\begingroup$ If $a,b,c$ are pairwise coprime, then each of the equations $ma+nb=1 , oa+pc=1 , qb+rc=1$ has an integral solution. $\endgroup$ – Peter Feb 4 '18 at 17:06

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