If GCD(a,b)=d,   then can we say that GCD(a^n,b^n)=d^n? for all positive integers n.  This doubt is inspired from GCD(a,b)=1 implies GCD(a^n,b^n)=1 for all positive integers n.
Let $d=(a^n,b^n), d_1=(a,b)^n$; need show $d\mid d_1$ &amp; $d_1\mid d$ to prove.  For the general case of $(a,b)\ne 1, (a,b)= (d_1)^\frac{1}{n}$, else $(a,b)=d_1$.  Let, $d_2= (a,b)=d_1^\frac{1}{n}\implies a=a'd_2, b = b'd_2\implies a^n= (a')^nd_2^n, b^n= (b')^nd_2^n$  Unable to progress further.
How to redirect output to screen as well as a file? The number $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$. Prove that for every $n\in \mathbb N$, $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$.   I was able to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$ using induction. First, I check whether the statement is true for $n=1$. The statement holds since: $$2^{3^1}+1=9$$ $$3^{1+1}=9$$ Second, I assume that the statement is true for every $n\in \mathbb N$. Finally, I take that $n=n+1$. From that, the original expression becomes $(2^{3^n})^3+1$, which is a sum of cubes, and the same as: $$(2^{3^n}+1)((2^{3^n})^2-2^{3^n}+1)$$ As I assumed that $2^{3^n}+1$ is divisible by $3^{n+1}$, and one of the factors is $2^{3^n}+1$, the entire expression must be divisible by $3^{n+1}$, and so the statement is true for every $n\in \mathbb N$.  Now, I don't know how to prove that $2^{3^n}+1$ isn't divisible by $3^{n+2}$. 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