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. 

Hint: Since you already proved that $\gcd\left(\frac{x^2-1}{2},\frac{x^2+1}{2}\right)=1$, assume that $\gcd\left(x,\frac{x^2-1}{2},\frac{x^2+1}{2}\right)=d>1$ which means $d \mid \frac{x^2+1}{2}$ and $d \mid \frac{x^2-1}{2}$ or $\gcd\left(\frac{x^2-1}{2},\frac{x^2+1}{2}\right)\geq d>1$ a contradiction.
