Let $\displaystyle a<b<c$ be positive, pairwise relatively prime integers, i.e. $\displaystyle (a,b)=(b,c)=(a,c)=1$

Without loss of generality, show that $\displaystyle (a^2bc , ax+1)=1$, where $\displaystyle x=(-a-b-c)(ab+bc+ac)^{-1} (\bmod abc)$ . Note: I have already shown that $\displaystyle gcd(ab+bc+ac , abc)=1$, so $\displaystyle x$ is well-defined. See http://www.mathhelpforum.com/math-he...e-theorem.html

Or give a counterexample. *I know this looks random and complicated, but this result has implications for Carmichael Numbers.