Why are $x$, $x^3+1$ and $x^2+x+1$ always mutually co-prime, for any natural number $x$? I just read that for a natural number $x$, the three numbers $x$, $x^3 + 1$ and $x^2 + x + 1$ are all mutually co-prime.
I couldn't find a reason why this is true. OK, any of them does not divide either of the other two, but is it enough to conclude that their GCD is 1?
Couldn't they have any common factors?
Also on what basis, are two consecutive integers co-prime? I know they cannot have a common factor but don't know why. 
 A: For your second question, say, $x$ and $x+1$ has some common factor $d$. Then consider
$$(x+1)-x = 1$$
The left hand side is a multiple of $d$, so the right hand side should also have $d$ as a factor. Then $d$ can only be $1$.
A: In general $\gcd(a,b)=\gcd(b,a)$ and $\gcd(a,b)=\gcd(a,b-ca)$ (do you understand why?).
Applying that gives: $$\gcd(x,x^3+1)=\gcd(x,1)=1$$ and: $$\gcd(x,x^2+x+1)=\gcd(x,1)=1$$ Also $$\gcd(x^3+1,x^2+x+1)=\gcd(-x^2-x+1,x^2+x+1)=\gcd(2,x^2+x+1)$$ and $x^2+x+1$ is odd for each integer $x$. 
This allows the conclusion that $\gcd(x^3+1,x^2+x+1)=1$
A: We see that the gcd of $m$ and $n$ always divides any expression of the form $am+bn$ since $m=ck, n=cj\implies c|(am+bn)=c(ak+bj)$. Being coprime means the gcd is $1$, so let's show this.
Case 1: For $m=x$ and $n=x^3+1$ we see with $a=-x^2$ and $b=1$ we have

$$-x^3+x^3+1=1$$

so they are coprime.
Case 2: For $m=x$ and $n=x^2+x+1$ we see $a=-(x+1)$ and $b=1$ gives

$$-x^2-x+x^2+x+1=1$$

which verifies coprimality.
Case 3: Finally with $m=x^2+x+1$ and $n=x^3+1$ we see that $a=(x-1)$ and $b=-1$ gives us

$$-x^3+1+x^3+1=2$$

so that $\gcd(x^2+x+1,x^3+1)\big|2$. But $x^2+x+1$ is always odd, so the gcd must be $1$.
A: If integer $d$ divides both $x^2+x+1, x^3+1;$
$d$ must divide $x(x^2+x+1)-(x^3+1)=x^2+x-1$
$\implies d$ must divide $x^2+x+1-(x^2+x-1)=2$
But $2\mid x(x+1)\implies x^2+x\pm1=x(x+1)\pm1$ are odd $\implies d=1$
