I am attempting to prove that:
$gcd(n^2-100n, 2n+1) = gcd(n-100, 201)$
To do so, I'm making use of the Euclidean algorithm, but I'm not getting to any result that I find particularly useful in demonstrating the equality. When I divide the terms on the LHS, I arrive at a remainder of $\frac{201}{4}$, whereas the RHS yields $\frac{n}{201} - \frac{100}{201}$. I have other incomprehensible calculations that I don't think would add much to the post. Any suggestions would be appreciated, I'm sure the solution is something rather trivial that I'm botching up.
Hint $\gcd(ab,c) = \gcd(b,c)$ whenever $\gcd(a,c) = 1$. Now note that $n^2-100n = n(n-100)$ and check $\gcd(n,2n+1)$.