For how many positive integers $n$ is $4^n − 1$ a prime number?

Question

This was the initial question and one of the methods to work it out include this:

The second method uses the fact when $n$ is a positive integer $x − 1$ is a factor of $x^n − 1$, since, for $n ≥ 2$, $$x^n − 1 = (x − 1)(x^{n−1} + x^{n−2} + \ldots + x + 1).$$

Hence, putting $x = 4$, we deduce that, for each positive integer $n, 3$ is a factor of $4^n − 1$. Therefore $4^n −1$ is not prime except when $n = 1$ and $4^n −1 = 3$. So there is just one positive integer $n$ for which $4^n − 1$ is a prime number.

However, I don't understand the initial method and how they deduced that $x-1$ is a factor of $x^n - 1$

Possible duplicate of Understanding Infinite primes of form $4n+3$ — Shaun, Oct 26 '17 at 12:56
Are you asking about $4n-1$ or $4^n-1$? The latter is the difference of two squares so... — lulu, Oct 26 '17 at 13:00
From OPs calculations it's pretty clear that it should be $4^n$ instead of $4n$. I corrected this now. The duplicate suggested above is therefore not correct. — Winther, Oct 26 '17 at 13:15
It follows from the Bezout theorem (aka polynomial remainder theorem) that $(x-1)$ is a factor of the polynomial $x^n-1$. — kingW3, Oct 26 '17 at 13:17
The fact that $(x-1) \mid (x^n-1)$ can also be proven using induction. Note that we can write $x^{n} - 1 = x(x^{n-1}-1) + (x - 1)$ so if $(x-1) \mid (x^{n-1} - 1)$ then $(x-1)\mid (x^n-1)$. — Winther, Oct 26 '17 at 13:21
Or, $$\begin{align} (x-1)(x^{n-1}+x^{n-2}+...+1) &=x\cdot (x^{n-1}+x^{n-2}+...+1)-1\cdot (x^{n-1}+x^{n-2}+...+1)\ &=(x^{n}+x^{n-1}+...+x)-(x^{n-1}+x^{n-2}+...+1)\ &=x^n-1 \end{align}$$ — r.e.s., Oct 26 '17 at 14:20

Barry Cipra · Answer 1 · 2017-10-26T14:05:30.513

For the question at hand, if $P(x)$ is a polynomial and $P(r)=0$, then $P(x)=(x-r)Q(x)$ for some polynomial $Q$. If the coefficients of $P$ and $r$ are all integers, then the coefficients of $Q$ are also all integers.

The specific case of $x^n-1$ might be best understood by example:

$$\begin{align} (x-1)(x^3+x^2+x+1) &=(x-1)x^3+(x-1)x^2+(x-1)x+(x-1)\\ &=(x^4-x^3)+(x^3-x^2)+(x^2-x)+(x-1)\\ &=x^4+(x^3-x^3)+(x^2-x^2)+(x-x)-1\\ &=x^4-1 \end{align}$$

The key step is the "telescopic" rearrangement of the parentheses. This is a standard technique (aka trick) when working with sums and series; you'll run into it again and again. (The term "telescopic" refers to old-style spyglasses, like you see in pirate movies, which collapse down to compact form.)

Another, possibly easier, way of seeing that $n=1$ is the only exponent for which $4^n-1$ is prime:

$$4^n-1=2^{2n}-1=(2^n-1)(2^n+1)$$

score 2 · Answer 2 · answered Oct 26 '17 at 13:52

Knowing that $x-1$ is a factor of $x^n-1$ is somewhat a matter of experience. It comes up often in various problems or theorems, and it's something you'll get used to. A couple of ways to look at it:

Recall that $x^2-1 = (x-1)(x+1)$. The identity in your question is a generalization of this.
Think of the geometric formula $1+x+\ldots+x^n = \frac{1-x^{n+1}}{1-x}$. Multiplying both sides by $1-x$ gives you the identity.

score 1 · Answer 3 · answered Oct 26 '17 at 13:50

I don't understand the initial method and how they deduced that $x-1$ is a factor of $x^n - 1$

Most likely they didn't deduce it but remembered it. It's a standard and elementary result (the sum of a geometric progression) that $$\sum_{i=0}^n r^i = \frac{r^{n+1}-1}{r-1}$$

The fact that $x-1$ is a factor of $x^n - 1$ also follows immediately from a well-known identity of cyclotomic polynomials, $$x^{n}-1 = \prod _{d\mid n}\Phi _{d}(x)$$ since $\Phi_1(x) = x - 1$.

For how many positive integers $n$ is $4^n − 1$ a prime number?

3 Answers3