Number Theory or Algebra?

Prove that if $4^m-2^m+1$ is a prime number, then all the prime divisors of $m$ are smaller than $5$

I initially thought about putting $4^m-2^m+1=p$ where $p$ is some prime and after eliminating initial cases of $p<5$ , setting $p=6k \pm 1$. However, since I'm just a beginner in Number Theory, I couldn't figure out anything else. What is more surprising that my teacher gave this problem in an Algebra worksheet, so perhaps there is some good algebraic way to do this.

So, thinking on Algebraic lines, I made a substitution $2^m=x$ and did some manipulations to the quadratic thus made. However, still no luck (despite the fact that I'm pretty confident about my Algebraic skills) I look forward for some help with this one. Any (or both) method will do. Thanks in advance!

Solution 1:

Note that $$ 4^m-2^m+1=\frac{8^m+1}{2^m+1} $$ so that when $m=pq$ is composite, there will be additional factors $4^p-2^p+1$ and $4^q-2^q+1$ to consider.

Solution 2:

I believe that this problem can most easily be addressed in the Eisenstein integers $R = \mathbb{Z}[\omega]$, where $\omega = e^{\frac{2 \pi i}{3}},$ although there may be more elementary approaches. It is well known that this is a principal ideal domain. I claim that $2^{2m}- 2^{m}+1$ can only be prime (in $\mathbb{Z}$) when $m$ has no prime factor greater than $3,$ as the problem asks. For If not write $m = hn$ where $h >1$ is relatively prime to $6,$ and $2$ and $3$ are the only prime factors of $n.$ Then $4^{m} -2^{m} + 1 = (2^{m} + \omega)(2^{m} + \omega^{2})$ in $R.$ If $h \equiv 1$ (mod $3$), notice that $\omega^{h} = \omega$, while if $h \equiv 2$ (mod $3$), notice that $\omega^{h} = \omega^{2}. $ Hence, in the ring $R,$ $2^{m}+ \omega$ is divisible either by $2^{n}+ \omega$ or by $2^{n} + \omega^{2},$ and hence is not a prime element of $R$.Thus $4^{m}-2^{m} +1$ is not a rational prime either.

For the purposes of illustration, the smallest non-trivial case is when $m = 10.$ Then we see by the above method that $2^{10}+\omega$ is divisible by $4+\omega^{2}$ in $R.$ Then we see that $(4+\omega)(4+\omega^{2}) = 13$ divides $2^{20}-2^{10}+1 = (2^{10}+\omega)(2^{10}+ \omega^{2})$ in $\mathbb{Z}.$ We can check this by verifying directly
that $2^{20} - 2^{10} + 1 \equiv 1- (3 \times 2^{8}) \equiv 1-27 \equiv 0$ (mod $13$) .