Is $GCD(2^n,(x^2+1)^{n-1})=1$

Question

Let $x>1,n>2$ be positive integers. I come across the following problem:

Is $$GCD(2^n,(x^2+1)^{n-1})=1$$

or $2^n$ is relatively prime to $(x^2+1)^{n-1}$ ?

Is $$GCD(x^{2n},(x^2+1)^{n-1})=1$$ ?

NB: $GCD$ is the greatest common divisor.

Answer 1

If $x=3$, then $$ (x^2+1)^{n-1}=10^{n-1} $$ and hence $$ \gcd(2ⁿ,(x²+1)ⁿ⁻¹)=2^{n-1}\neq1.$$ The 2nd one is correct.

Answer 2

Coprimality (the property of being coprime) is invariant under taking powers. That is to say that two integers $a,b$ are co prime if and only if $a^m, b^n$ are co prime for any two natural numbers $m,n$. So to answer both your questions, we can safely disregard all the powers.

For the first part, we need to check if $x^2+1$ is coprime to $2$ for odd $x$ which is clearly not the case.

For the second part we need to check if $x$ is co prime to $x^2+1$ which is true.

Answer 3

You can refer to @xpaul's answer for the first case.

Now, consider the second scenario you doubt, i.e., $\gcd(x^{2n},(x^2 + 1)^{n-1})=1$.

If $x$ is prime, $x^{2n}$ is a power of a prime and $(x^2 + 1)^{n-1}$ is a product of a power of a prime plus $1$. You can never have a common factor in such a case and hence the equality holds when $x$ is prime.

If $x$ is even, i.e, $x = 2y$, $x^{2n}$ has powers of $2$ in it and $(x^2 + 1)^{n-1} = ({(2y)}^2 + 1)^{n-1}= (4y^2 + 1)^{n-1}$ has no power of $2$ in it. The equality holds here as well.(Just putting up as a notice; I am not much sure. The inequality doesn't hold if both have a power of $y$ in common.)

I am stopping here for now and sorry for being of not much help. I was about to go to bed when I typed this all up.

NB :

If $x$ is prime, $x^{2n}$ is a power of a prime and $(x^2 + 1)^{n-1}$ is a product of a power of a prime plus $1$. You can never have a common factor in such a case and hence the equality holds when $x$ is prime.

I doubt this, so please verify it by tomorrow. I say so because I haven' tried any test cases and I used only algebra.