Is $\gcd(2^{2n}+1, 3)=1$?

Question

Can any one prove that $2^{2n}+1$ and $3$ are relatively prime for any integer $n$?
I tried with a Matlab program and computed this gcd upto $n= 25$. I got 1 for all of them. So I suppose that the result is true.

Answer 1

In mod $3$, $$2^{2n}+1\equiv (2^2)^n+1\equiv 4^n+1\equiv 1^n+1\equiv 2.$$

Answer 2

Hint $\,\ 4^n\!+1 = (\color{#c00}{4^n\!-1})+2,\ $ and $\ 3 = 4\!-\!1\mid \color{#c00}{4^n-1}\,$ by the Factor Theorem.

Said in modular language: $\ {\rm mod}\ 3\!:\,\ \color{#c00}{4\equiv 1}\,\Rightarrow\, \color{#c00}4^n\equiv \color{#c00}1^n\equiv 1\ $ by the Congruence Power Rule

Answer 3

Yes it is 1. The only thing you need to check is if 3 divides $2^{2n}+1$. But $2^{2n}+1 \equiv 2 \mod 3$. Thus, the gcd is 1.

Answer 4

$2^{2n}=4^n$. Proof by induction:

Base case: $n=0, 4^n=1$, so $4^n+1\equiv2 \mod 3$.

Assume true for $n$, i.e. $4^n+1\equiv2\mod3$. Multiply through by $4$ to get $4(4^n+1)\equiv8\mod3 \implies 4^{n+1}+4\equiv2\mod3\implies 4^{n+1}+1\equiv 2\mod3$. So it is also true for $n+1$.

As it is true for $n=0$, it is true for $n=1$. As it is true for $n=1$, it is true for $n=2$. And so for all $n\in\mathbb{N}$.