Let $x_1,x_2,x_3...x_n$, where n is odd, be positive integers that have a product $K$. Prove the following inequality: $gcd(x_1^n+K,x_2^n+K...x_n^n+K)\le 2$$gcd(x_1,x_2,x_3...x_n)^n$
I'm having trouble understanding why n has to be odd. Why is this parameter necessary for the inequality hold true? Also, how do you deal with the exponents on the LHS?
Divide both sides by $\gcd(x_1,x_2\dots,x_n)^n$ to get the equivalent:
$\gcd(x_1^n+K,x_2^n+K,\dots,x_n^n+K)/\gcd(x_1,x_2\dots,x_n)^n\leq 2$.
If we let $y_i=x_i/\gcd(x_1,x_2,\dots,x_n)$ and $L=y_1y_2\dots y_n$ then the inequality is equal to:
$\gcd(y_1^n+L,y_2^n+L,\dots ,y_n^n+L)\leq 2$.
Suppose that a prime power $p^a$ divides each $y_i^n+L$. Then $y_i^n\equiv -y_1y_2\dots y_n$ for all $i$.
This implies $y_1^ny_2^n\dots y_n^n\equiv (-y_1y_2\dots y_n)^n \equiv -y_1^n \dots y_n^n\bmod p^a$
This implies $p|2y_1^ny_2^n\dots y_n^n$. This implies that $p^a=2$ or $p$ divides $y_1y_2\dots y_n$. If we had that $p$ divides $y_1y_2\dots y_n=L$ this forces for $p$ to divide each $y_i$, because $y_i^n+L$ is a multiple of $p$ for all $i$, this is a contradiction since $\gcd(y_1,y_2,\dots,y_n)=1$.
We conclude that $p^a=2$. This proves that $\gcd(y_1^n+L,y_2^n+L,\dots,y_n^n+L)$ is $1$ or $2$.