Greatest common divisor for $x+1$ and $2x+1$?

Question

I was just wondering, what would the gcd of $x+1$ and $2x+1$ be?

I know that $x+1$ is not even or odd, whereas, $2x+1$ is odd. Is the gcd $1$?

Answer 1

We always have $\gcd(x,x+1)=1$. Hence it is also always true that $\gcd(x,x+(x+1))=\gcd(x,2x+1)=1$.

Answer 2

Hint: Note that $2(x+1)-(2x+1)=1$.

Answer 3

If $d(x)|x+1$ and $d(x)|2x+1$ then $d(x)|(2x+1)-(x+1)= x$. And so $d(x)|x$ and $d(x)|x+1$ so $d(x)|(x+1) -x = 1$ so $d(x)|1$. And assuming we are talking $\mathbb Z[x]$ this is as far as we go and $d(x) =1$ is the only (positive) common divisor of $2x+1$ and $x+1$. So the greatest common divisor of $2x + 1$ and $x+1$ is $1$.

In general you have the rule $\gcd(a,b) = \gcd(a \pm k b, b)$ so $\gcd(2x+1, x+1) = \gcd(x, x+1) = \gcd(x, 1) = 1$.

Euclid's algorithm confirm:.

$(2x+1) = (x+1) + (x)$

$(x+ 1) = (x) + (1)$

$(x) = (1)\cdot x + 0$ so $\gcd(2x+1, x) = 1$.

As does Bezout's Lemma:

$2\cdot(x+1) - 1\cdot(2x+1) = 1$