Prove that the gcd is 1. I need to show that for $x$ odd the $\gcd\bigg(x,\dfrac{x^2-1}{2},\dfrac{x^2+1}{2}\bigg)=1$. Im trying it doing it pairwise. I already show that $\gcd\bigg(\dfrac{x^2-1}{2},\dfrac{x^2+1}{2}\bigg)=1$, but I dont have an idea how to show that $\gcd\bigg(x,\dfrac{x^2-1}{2}\bigg)=1$ and $\gcd\bigg(x,\dfrac{x^2+1}{2}\bigg)=1$. 
I will appreciate the help. 

If $p|x$ then $p|x^2$ and $p\not |x^2 \pm 1$ so $p\not|\frac{x^2 \pm 1}k$.
So $x$ and $\frac{x^2 \pm 1}2$ have no prime factors in common. So $\gcd(x, \frac{x^2 -1}2) = \gcd(x,\frac{x^2 +1}2) = 1$.
And $\gcd(\frac{x^2-1}2,\frac{x^2 + 1}2) = \gcd(\frac{x^2 -1}2, \frac{x^2 +1}2 - \frac{x^2-1}2) = \gcd(\frac{x^2-1}2, 1) = 1$.
