Any integer of the form ## 8^n+1 ##, where n##\geq##1, is composite?

[Math100]

Homework Statement

Establish the following statement:
Any integer of the form $8^n+1$, where n$\geq$1, is composite.
[Hint: $2^n+1\mid 2^{3n} +1$.]

Relevant Equations

None.

Proof:

Suppose $a=8^n+1$ for some $a \in\mathbb{Z}$ such that n$\geq$1.
Then we have $a=8^n+1$
=$(2^3)^n+1$
=$(2^n+1)(2^{2n} -2^n+1)$.
This means $2^n+1\mid 2^{3n} +1$.
Since $2^n+1>1$ and $2^{2n} -2^n+1>1$ for all n$\geq$1,
it follows that $8^n+1$ is composite.
Therefore, any integer of the form $8^n+1$, where n$\geq$1, is composite.

 

[fresh_42]

Correct. You should begin with the next chapter.

 

[fresh_42]

E.g. you have been told that $(2^n+1)\,|\,(2^{3n}+1)$ and your calculation shows how. Now, can you also divide $(2^{3n}+1) \, : \,(2^{n}+1) =$ by long division?

 

[Math100]

It's $2^{2n} +1-2^n$.

 

[fresh_42]

It's $2^{2n} +1-2^n$.

Sure, but how do you get it through division? Here is an example of how it can be done:
https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083

 

[PeroK]

Let $m = 2^n$, so that $a = m^3 + 1$. By the remainder theorem $m + 1$ is a factor of $m^3 + 1$. Also, for $m > 1$ we have $m^3 +1 \ne m + 1$, so the other factor cannot be $1$. Hence $a$ is composite for $n \ge 1$.