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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.

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    $\begingroup$ The title is quite confusing to me without any quantifiers in it. $\endgroup$ Jan 4, 2011 at 17:43

1 Answer 1

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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.

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    $\begingroup$ Nice. For the property of $K$ and $L$ you mention, don't we need that $L$ is Galois (which is satisfied in your case)? For general $L$ we would need that every rational prime above which there is a degree 1 prime in $L$ splits in $K$, i.e. we would require the property for more rational primes. Compare with Proposition 15 in Section VIII.5 of Weil's Basic number theory. $\endgroup$
    – GH from MO
    Jan 4, 2011 at 22:51
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    $\begingroup$ You're right, thanks. I thought I had been careful to phrase this in a way which avoided needing Galois extensions, but I failed. $\endgroup$ Jan 4, 2011 at 23:52
  • $\begingroup$ Nice. It is interesting to follow step by step your proof, adapted to the square root (instead of the fourth root) of $2$. Here we know that the property is true for $N=8$. At we end, we find that $\mathbb Q(\sqrt2)$ embeds into $\mathbb Q(\zeta_8)$, which is true because $\sqrt2=\zeta_8+(\zeta_8)^7$. $\endgroup$ Jan 5, 2011 at 10:34

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