Proving $\gcd(4k+2,k^2+k) = 2$ by direct proof

Question

I came across this problem in a textbook (not an assignment) and was interested in how to solve it.

I imagine $\gcd(4k+2,k^2+k)$ can easily be proved to be an even integer because:

$4k+2 = 2(2k+1)$, which would be even because its $2$ times and integer, and

$k^2+k$ can be proved to be even by using $2$ cases, one where $k$ is even and is equal to $2n$ and the other for an odd $k$, where we can set it equal to $2n+1$, where $n \in \mathbb{Z}$.

From here, if we let $\gcd(4k+2,k^2+k)=x$, we can say if $x$ is the $\gcd$ of $(4k+2,k^2+k)$ then it must divide any linear combination of the 2. I'm stuck on this part, because there's so many directions this could go.

I'm thinking $x|(k^2+5k+2)$ or $x|(4k^2-2)$, but I can't find a way to prove the gcd isn't some even integer greater than $2$. Any insight is appreciated!

Answer 1

HINT: If a prime $p$ divides both $2k+1$ and $k(k+1)$, then either $p$ divides $k$ or $p$ divides $k+1$. Show that the first leads to an immediate contradiction and the second leads to the conclusion that $p$ divides $k$.

Answer 2

Let $(4k+2,k^2+k)=d$. Then $d|4k+2-2(4(k^2+k)-k(4k+2))=2$.

$d=2$ since both numbers are even. The motivation is to play around until we get a constant (in this case, it is $2$).

It motivates us to multiply $4k+2$ by $k$ and $k^2+k$ by 4 and to subtract those numbers so we have something easier to handle (lower in degree): $4(k^2+k)-k(4k+2)=2k$. Seeing $4k+2$, we are motivated to multiply $2k$ by two and subtract from $4k+2$ to obtain $2$.