showing all gcd $(n^3-n, 2n^2-1)$ I've been stuck on this problem for a long time.
I must show all gcd of $(n^3-n, 2n^2-1)$. I know that $n^3-n=n(n+1)(n-1)$ and so I must find all gcd of $(n(n+1)(n-1), 2n^2-1)$. But I don't know how. Can somebody help me ?

I'd do it the way I did below without factoring.  But factor was a good thought and it will work:
$n^3-n = n(n+1)(n-1)$ so lets see what factors $n,n+1, n-1$ have with $2n^2 -1$.
First of all any (nontrivial) factor of $n$ will be a factor of $2n^2$ and will not be a factor of $2n^2 -1$.  So $n$ and $2n^2 -1$ have no factors in common and $n$ doesn't "contribute" to the gcd.  So $\gcd(n(n+1)(n-1), 2n^2-1)= \gcd((n+1)(n-1),1)$.
Now do $n+1$ and $2n^2 -1$ have any factors in common?  Note:  $2n^2 -1 = 2(n^2 -1) +1= 2(n+1)(n-1) + 1$.  So any non trivial factor that $n+1$ will have will also be a factor of $2(n+1)(n-1)$ so it will not be a factor of $2(n+1)(n-1)-1$.  $n+1$ and $2n^2 -1$ will have not factor in common.
And by the same argument neither will $n-1$ and $2(n+1)(n-1)+1 = 2n^2 -1$.
So $2n^2-1$ has no factors in common with any of the factors of the factorization of $n(n+1)(n-1)$ so the $\gcd(n^3 -n, 2n^2 -1) = 1$.
=====My way=====
use $\gcd(a,b) = \gcd(a \pm kb, b)$.  And use if $\gcd(k,b) =1$ then $\gcd(ka,b)=\gcd(a,b)$
So $\gcd(n^2 -n, 2n^2-1)$;  $2n^2-1$ is odd any common divisor is odd and multiplying $n^2-n$ by $2$ will not effect the $\gcd$.  So:
$\gcd(n^3 -n, 2n^2-1)= \gcd(2n^3-2n,2n^2-1)$.  Whatever common divisor they have divides $2n^2 -1$ and $2n^3-2n$ so it will divide $n(2n^2-1)$ and will divide $(2n^3-2n)-n(2n^2-1)$
So $ \gcd(2n^3-2n,2n^2-1)= \gcd([2n^3-2n]-n[2n^2 -1],2n^2-1) = \gcd(n,2n^2-1)$.
$\gcd(n,2n^2-1) = \gcd(n, [2n^2-1] -[(2n)n]) = \gcd(n, -1)$ and as the only thing that divide $-1$ are $1, -1$ and $1 > -1$ then $\gcd(n,-1) = 1$.
