Proof that the GCD of $x^2+x+6$ and $x^2+x+4$ is 2

Question

How would one demonstrate/prove that for polynomials $x^2+x+6$ and $x^2+x+4$, where $x$ is an integer, equals 2?

Answer 1

Hint:

Both numbers are even, but only one is a multiple of four. Forthermore, no other prime can divide both. Can you show it?

Answer 2

Note that

$$\gcd(x^2+x+6,x^2+x+4)=\gcd(x^2+x+6-(x^2+x+4),x^2+x+4)\\=\gcd(2,x^2+x+4)=\gcd(2,x(x+1)+4)=2$$

Answer 3

If $a \mid b$ and $a \mid c$ then $a \mid b \pm c$. So, the greatest common divisor between $x^2 + x + 6$ and $x^2 + x + 4$ must divide each. Let the greatest common divisor be $d$. Then $d$ also divides their difference: $$d \mid (x^2 + x + 6) - (x^2 + x + 4) \rightarrow d \mid 2.$$

So, $d = 1$ or $d = 2$. If we note that $$x^2+x = x(x+1)$$ is a product of two consecutive integers then it follows that $2$ divides $x^2+x$. Since $2$ also divides both $4$ and $6$, we have that $d = 2.$

Answer 4

$x^2+x+6=x(x+1)+6$, $x(x+1)$ is always divisible by $2$, since the product of any two consecutive integer is always is divisible by $2$. Thus $x(x+1)+6$ is divisible by $2$. Also $x^2+x+4=x(x+1)+4$ is divisible by $2$ by same reason. Also their are no common factor of $x^2+x+6$ and $x^2+x+4$ other than 2. Therefore $\gcd(x^2+x+6,x^2+x+4)=2$. Also you find the $gcd$ by divison method as in above.