Explain $\gcd(F_n, n)=1$ where $F_n$ is Fermat number.

Question

Show that Fermat number $F_n$ and its index $n$ are coprime.

I try to understand this problem, but cannot understand that "and so 2 has multiplicative order $2n+1$ mod $p$. Thus $2n+1\mid p−1$"

Can you prove this another method without using mod? Or explain the link more easier?

Answer 1

NOTE: This is answer is a 'little' bit longer, but I tried to go through details, asumming that you have zero knowledge in modular arithemtics, group theory, multiplicative order...

First of all prove that two numbers are comprime, you need to prove their factors are coprime numbers. So in this case we need to prove that $F_n$ and $n$ are coprime numbers.

Euler and Lucas showed that any prime divisor of $F_n$, where $n>1$ is of the form $k2^{n+2} + 1$. So we can check that for every $n$ the following statement holds:

$$k2^{n+2} + 1 > n$$

So how can a bigger number be a divisor of a smaller number?

If you take the Euler and Lucas statement as granted then it's simple, but if you want to prove it by yourself then you have to use modular arithemtics, group theory, multiplicative order...

First of all you should know that multiplicative order is the smallest integer for such

$$a^{ord_p(a)} \equiv 1 \pmod p$$

where $(a,p) = 1$

So now let $p$ be a divisor of a Fermat Number, we have:

$$2^{2^n} + 1 \equiv 0 \pmod p$$ $$2^{2^n} \equiv -1 \pmod p$$

Square both sides we get:

$$2^{2^{n + 1}} \equiv 1 \pmod p$$

From this we get that $ord_p(2) \mid 2^{n+1}$, actually we get that $ord_p(2) = 2^{n+1}$

Let $G_p$ denote the group of non-zero elements of the integers (mod p) under multiplication, which has order $p-1$.

From Lagrange theorem for group order we know that order of an element in a finite group divides the order of the group.

To make it more simplier use Fermat's Little Theorem:

$$2^{p-1} \equiv 1 \pmod p$$

Because $e = ord_p(2)$ is the smallest integer that satisfy the equation:

$$2^e \equiv q \pmod p$$

We obtain that $ord_p(2) \mid p-1$, i.e $2^{n+1} \mid p-1$. This can be rewritten as:

$$p - 1 = k2^{n+1}$$ $$p = k2^{n+1} + 1 \text{ for some natural number $k$}$$

We'll stop here, because to prove that $p = k2^{n+2} + 1$ we need to use quadratic residue, which will make things even more complicated. It's enough to see:

$$k2^{n+1} + 1 > n$$

holds for every number n. And as we said earlier a bigger number can't divide a smaller number.

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Proof of the lemma $ord_p(2) = 2^{n+1}$

Suppose that $p$ divides $2^{2^n} + 1$, then it also divides $2^{2^n+1} - 1 = (2^{2^n} + 1)(2^{2^n} - 1)$

We know that $ord_p(2) \mid 2^{n+1}$, so we obtain $ord_p(2) \le 2^{n+1}$. Also because $2^{n+1}$ is a primer power $ord_p(2)$ must be a power of that prime, in this case $2$. So we have:

$$ord_p(2) \mid 2^{n+1} \implies 2^{n+1} = k \cdot ord_p(2) = 2^l \cdot 2^j$$

Now we know from multiplicative order property that if $ord_p(2)$

$$p \mid 2^{ord_p(2)m} - 1 \text{, for every natural number m}$$

Assume that $n \ge j$ then $2^{n-j}$ is a natural number.

If we chose $m=2^{n-j}$ we get:

$$p \mid 2^{ord_p(2)2^{n-j}} - 1 = 2^{2^j2^{m-j}} - 1 = 2^{2^m} - 1$$

We have that $p$ divide both $2^{2^m} - 1$ and $2^{2^m} + 1$. So it must divide their difference, which is $2$. But that's impossible, because $p$ is an odd prime. This leads to contradicting our assumtion that $n \ge j$, implying that $j > n$, i.e $2^j = ord_p(2) > 2^n$.

But we know that $ord_p(2) = 2^j \le 2^{n+1}$. So this implies that $j=n+1$, i.e.

$$2^{n+1} = ord_p(2)$$

Q.E.D.