Euler's proof that the fifth Fermat number is composite begins with the following argument. If $p$ divides $F\_n$, then the order of $2$ in $(\mathbb Z/p\mathbb Z)^\*$ is exactly $2^{n+1}$. Hence $p\equiv1$ mod $2^{n+1}$. If $n\ge2$, this implies that $2$ is a square mod $p$, $2=\omega^2$. Then the order of $\omega$ is exactly $2^{n+2}$, hence $p\equiv1$ mod $2^{n+2}$. This limits the set of prime divisors to be tested. For $n=5$, we find $p=257, 641, ...$, in which the first one is impossible because it is $F\_3$ and the Fermat numbers are pairwise coprime. Thus the first candidate is $641$, which turns out to divide $F\_5$.


Of course, the situation is not so simple for larger values of $n$. But I wander how far this argument could be pushed forward. Because a prime divisor satisfies $p\equiv1$ mod $2^{n+2}$, $2$ should be a $4$-th power if and only if $2^{\frac{p-1}{4}}\equiv1$ mod $p$. If $2=\theta^4$, then ...*(bla-bla)*... $p\equiv1$ mod $2^{n+3}$. So my question is



> 
> Let $m\ge2$ be given. Assume that $p-1$ is divisible by a large enough power of $2$. Does this imply that $2$ is a $2^m$-th power mod $p$ ?
> 
> 
> 


Of course, I am not so naive as to look forward for an elementary solution to the long-standing problem of the primality of some $F\_n$'s, but such an approach can provide nice exercises in undergraduate classes.



is $2$ a $2^n$-th power mod $p$ ?
There is no $N$ such that $p \equiv 1 \mod N$ implies that $2$ is a fourth power modulo $p$.


**Proof:** Suppose otherwise. Without loss of generality, suppose that $4$ divides $N$.


Let $K$ be the field $\mathbb{Q}(2^{1/4})$. If $p \equiv 1 \mod N$, then by hypothesis $x^4-2$ has a root in $\mathbb{F}\_p$. Moreover, since $p \equiv 1 \mod 4$, there is a primitive $4$th root of unity in $\mathbb{F}\_p$, so $x^4-2$ splits in $\mathbb{F}\_p$.


So the prime $p$ splits in the ring of integers of $K$. Recall that, if $K$ and $L$ are two number fields, and every prime which splits in $L$ splits in $K$, then $L$ embeds into $K$. Now, $p \equiv 1 \mod N$ if and only if it splits in the cyclotomic field $\mathbb{Q}(\zeta\_N)$. So we would have that $K$ embeds in $\mathbb{Q}(\zeta\_N)$. But this contradicts that $\mathbb{Q}(\zeta\_N)/\mathbb{Q}$ is abelian, and $K/\mathbb{Q}$ is not Galois.


