5

Let us suppose that we are presented with a positive integer $k$ and asked to come up with a positive integer $a$ such that $\frac{a^k - 1}{a-1}$ is not a prime power, or just to prove in an elementary way that this can be done. How do we proceed?

Of course, in practice this is essentially trivial. But is there a cannonical choice? For that matter, is there an elementary proof that shows that this is always possible?

This came up in a discussion with another Brown mathie about interesting things to talk about in an elementary number theory class I will be teaching this summer. To be more specific, we were trying to prove the infinitude of primes in arithmetic progressions of the form $1, 1+p^n, 1+2p^n, ...$ using a naive a completely elementary approach (hopefully to be replicated by my students this summer), and this side topic came up.

davidlowryduda
  • 91,687

2 Answers2

4

take instead of $a$ $a^2$ then we want to prove $$\frac{a^{2k}-1}{a^2-1}$$ is not a prime.

but $\frac{a^{2k}-1}{a^2-1}=\frac{(a^k-1)(a^k+1)}{a^2-1}$

So if we suppose $k>3$, then choose and $a$ such that $a^k-1>a^2-1$ and $a^k+1>a^2-1$

Hence there is at least one prime factor from each $a^k-1$ and $a^k+1$ in the expression $$\frac{a^{2k}-1}{a^2-1}$$

clark
  • 15,327
2

Let $a=2k+1$ then $$ \frac{a^k-1}{a-1} = \frac{\sum_{i=1}^k\binom{k}{i}(2k)^i}{2k}\equiv k\pmod{k^2} $$ which is not $k$, divisible by $k$ but not $k^2$, and hence not a prime or a power for $k>1$.

Zander
  • 10,948
  • 1
  • 26
  • 54