Number Theory - Prove $n^2$ and $n-1$ are coprime

Question

In my foundations of math class we have just finished our section on number theory. I am having a really hard time with the questions involving co-primality, gcd's, and Bezout's identity.

The question I'm working on now asks us to prove $gcd(n^2,n-1)=1$ for all $n>=1$

I know that the above implies $1=an^2 + b(n-1)$ for some $a,b \in \mathbb{Z}$.

I seem to be just running into dead ends and gaps in logic/knowledge.

Answer 1

You don't need Bezout's identity.

Let $p$ be a prime dividing $n^2$. Then $p$ divides $n$. If $p$ divided $n-1$ , then $p$ would divide $1$.

Answer 2

$n^2-(n+1)(n-1)=1-n^2+n^2=1$ implies the result.