Exercise 2.1.49* (Prime divisors of repunits.)

Proof.

a)

If $p$ is any prime other than $2$ or $5$, then $p\wedge 10=1$. By Fermat’s theorem, $$p\mid 10^{p-1}-1.$$

Therefore, for every positive integer $k$,

$$p\mid 10^{k(p-1)}-1.$$

Since $10^{k(p-1)}-1=999\cdots 9$, where the number of $9$ is $k(p-1)$, $p$ divides infinitely many of the integers $9,99,999,9999,\cdots$.

b)

Let $p$ be any prime other than $2$ or $5$. Note that $9=10-1$ divides $10^{k(p-1)}-1$.

• If $p\ne 3$, $p\wedge 9=1$, and by part (a), $p\mid 10^{k(p-1)}-1=9\cdot \frac{10^{k(p-1)}-1}{9}$. Therefore

  $$p\mid \frac{10^{k(p-1)}-1}{9}.$$

  Since $\frac{10^{k(p-1)}-1}{9}=111\cdots 1$, where the number of $1$ is $k(p-1)$, $p$ divides infinitely many of the integers $1,11,111,1111,\cdots$.

• If $p=3$, $10\equiv 1(\mathit{mod}3)$, thus $10^k\equiv 1(\mathit{mod}3)$ for all $k$, therefore

  $$10^{n-1}+10^{n-2}+\cdots +10+1\equiv 1+1+\cdots +1=n,$$

  so $111\cdots 1$ is divisible by $3$ if the number $n$ of $1$ in its decimal expression is a multiple of $3$.

  Thus $3$ divides $111,111.111,111.111.111,\cdots$, and we obtain the same conclusion.

  If $p$ is any prime other than $2$ or $5$, $p$ divides infinitely many of the integers $1,11,111,1111,\cdots$ (the repunits).