# Show that if $(b^n-1)/(b-1)$ is a power of prime numbers, where $b,n>1$ are positive integers, then $n$ must be a prime number.

Show that if $$(b^n-1)/(b-1)$$ is the power of a prime number, where $$b,n>1$$ are positive integers, then $$n$$ must be a prime number.

My solution:

If $$n$$ is composite, then let $$n=mk$$, $$m,k>1$$, \begin{align*} \frac{b^n-1}{b-1} &= 1+b+\cdots+b^{n-1} \\ &=(1+b+\cdots+b^{k-1} )+(b^k+b^{k+1}+\cdots+b^{2k-1}) \\ &\quad\,+\cdots+(b^{(m-1)k}+b^{(m-1)k+1}+\cdots+b^{mk-1}) \\ &=(1+b+\cdots+b^{k-1})(1+b^k+\cdots+b^{(m-1)k}) \end{align*} Which is composite and distinct, thus, for $$(b^n-1)/(b-1)$$ to be a power of primes, $$n$$ is not composite, thus it must be prime.

However, $$(1+b+\cdots+b^{k-1})(1+b^k+\cdots+b^{(m-1)k})$$ might be equal to $$p^x \times p^y$$, where $$p$$ is prime.

Is there any better solution?

• Please use MathJax. Mar 22, 2020 at 7:48
• What's "a power of prime numbers"? Do you mean "a power of a prime number"? Mar 22, 2020 at 10:59

Let $$(b^n-1)/(b-1)=p^x$$ where $$p$$ is a prime and $$x> 0$$. If $$n$$ is composite, there are two cases.

1. There exists a prime $$q$$ such that $$n=q^m$$ for some $$m>1$$. Note $$p^x=\frac{b^n-1}{b-1}=\frac{b^{q^m}-1}{b^{q^{m-1}}-1}\cdot \frac{b^{q^{m-1}}-1}{b-1},$$ we can assume $$\frac{b^{q^{m-1}}-1}{b-1}=p^y$$ for some $$0< y< x$$. Then we have \begin{align} 1+q(b-1)p^y+\sum_{i=2}^q\binom{q}{i}\left((b-1)p^y\right)^i&=\left((b-1)p^y+1\right)^q\\ &=\left(b^{q^{m-1}}\right)^q\\ &=b^{q^m}\\ &=(b-1)p^x+1, \end{align} i.e., $$q+\sum_{i=2}^q\binom{q}{i}\left((b-1)p^y\right)^{i-1}=p^{x-y}.$$ Hence, $$p\mid q$$. Recall that $$p$$ and $$q$$ are both primes, so $$p=q$$, we further have $$1+\binom{p}{2}(b-1)p^{y-1}+\sum_{i=3}^p\binom{p}{i}(b-1)^{i-1}p^{y(i-2)}=p^{x-y-1}.$$ Note the left hand side is no less than 2, so both sides are divisible by $$p$$, i.e., the term $$\binom{p}{2}(b-1)p^{y-1}$$ cannot be divisible by $$p$$, thus $$p=2$$ and $$y=1$$. We further have $$b=p^{x-2}$$, i.e., $$p\mid b$$ (recall that $$b>1$$). However, note $$p^x=\frac{b^n-1}{b-1}=1+b+\cdots+b^{n-1},$$ it is impossible that $$p\mid b$$.

2. There exist two co-prime numbers $$s,t>1$$ such that $$n=st$$. In this case, we have $$p^x=\frac{b^n-1}{b-1}=\frac{b^{st}-1}{b^s-1}\cdot\frac{b^s-1}{b-1},$$ which means $$(b^s-1)/(b-1)$$ is divisible by $$p$$. Similarly, $$(b^t-1)/(b-1)$$ is also divisible by $$p$$. Since $$s$$ and $$t$$ are co-prime, there exists integers $$w_s,w_t$$ such that $$w_ss+w_tt=1$$. Without loss of generality, we assume $$w_s>0$$ and $$w_t<0$$. Then we have $$\frac{b^{w_ss}-1}{b^s-1}\cdot\frac{b^s-1}{b-1}-b\cdot\frac{b^{-w_tt}-1}{b^t-1}\cdot\frac{b^t-1}{b-1}$$ is also divisible by $$p$$. Note the expression above is exactly $$\left(1+b+\cdots+b^{w_ss}\right)-b\left(1+b+\cdots+b^{-w_tt}\right)=1,$$ which is impossible.

As a conclusion, $$n$$ must be a prime.

• How did you conclude "thus $b=p=2$"? Mar 22, 2020 at 21:00
• @joriki I calculated the equality wrongly. $b$ may not be 2. I have edited the answer. The conclusion is the same. Thanks for pointing out this mistake. Mar 23, 2020 at 1:04
• OK. Now, in concluding $p\mid b$ from $b=p^{x-2}$, how did you exclude the case $x=2$? Mar 23, 2020 at 1:33
• @joriki Because $b>1$ as OP mentioned in the title. Mar 23, 2020 at 1:34
• I see. I think everything checks out then. Nice proof :-) I especially like how you used Bézout's identity. I do suspect that there should be a simpler proof, though. Mar 23, 2020 at 1:49