Prove that $\gcd(4k, 2k+1)=1$ for $k\in\mathbb Z$

Question

As part of solving a problem in elementary number theory, I got to this condition which I'd like to prove.

I went about it as follows:

If $d=\gcd(4k, 2k+1)$ then $4k \equiv 0 \equiv 2k+1\pmod d$ and therefore $0 \equiv 2(2k+1) - 4k \equiv 2\pmod d$, whence $d \mid 2$. Therefore either $d = 1$ or $d = 2$.

However, clearly d cannot be 2 since $2k+1$ is an odd number, therefore it must be that $d = 1$, QED.

This seems to prove the result, however I am bothered by the fact I cannot find a representation of d ($=1$) in the form $1 = ax + by$ for x, y integers where $a=4k$ and $b=2k+1$. I would expect to be able to find such an identity given Bezout's Identity.. what am I missing?

Answer 1

Let $x=-(k+1), y=(2k+1)$

$$4kx+(2k+1)y=-4k(k+1)+(2k+1)^2=1$$