Prime factorization of $\frac{100^{69}-1}{99}$?

Question

I'm certain that I'm not the first person to ask this question, but I'm wondering what techniques can be used to attempt to find the prime factorization of $$m=\underbrace{696969\cdots 69}_{69\text{ times}}$$

I know that $$m=69\cdot\underbrace{101010\cdots 101}_{68\text{ times} }=3\cdot 23\cdot \sum\limits_{k=0}^{68}100^k=3\cdot 23\cdot\frac{100^{69}-1}{99}$$ From there, I'm not aware of any good way to find the prime factors $$\frac{100^{69}-1}{99}$$ Are there any methods that might lend themselves to factoring that number other than simply using a computer and trial and error?

Answer 1

Most useful information comes from factorizing the polynomial $F(x) = \frac{x^{138} - 1}{x^2 - 1}$, which can be easily expressed as a product of cyclotomic polynomials:

$$F(x) = \phi_3(x)\phi_3(-x)\phi_{23}(x)\phi_{23}(-x)\phi_{69}(x)\phi_{69}(-x),$$ where $\phi_n(x)$ is the $n$-th cyclotomic polynomial.

Thus it suffices to factorize the numbers $\phi_3(\pm 10)$, $\phi_{23}(\pm 10)$, $\phi_{69}(\pm 10)$. I don't think there are any smart methods to do that, other than calculating the numbers and passing them to a factorization algorithm. This perhaps can be seen from the results: \begin{eqnarray} \phi_3(10) &=& 3 \times 37\\ \phi_3(-10) &=& 7 \times 13\\ \phi_{23}(10) &=& 11111111111111111111111\\ \phi_{23}(-10) &=& 47 \times 139 \times 2531 \times 549797184491917\\ \phi_{69}(10) &=& 277 \times 203864078068831 \times 1595352086329224644348978893\\ \phi_{69}(-10) &=& 31051 \times 143574021480139 \times 24649445347649059192745899.\\ \end{eqnarray}

Answer 2

From $x-1\mid x^n-1$, we conclude that $10^n-1$ divides $100^{69}-1$ for all divisors $n$ of $138=2\cdot 3\cdot 23$. Of these $10^1-1$ and $10^2-1$ may cancel against the denominator, but $10^3-1=999=3^3\cdot 37$ certainly gives you an extra $3$ and $37$, etc.

Answer 3

In general we have

$$x^n - 1 = \prod_{d \mid n} \Phi_d(x)$$

where $\Phi_d(x)$ are the cyclotomic polynomials. This is the complete irreducible factorization of $x^n - 1$. Since $100^{69} = 10^{138}$ and $138 = 2 \cdot 3 \cdot 23$ this gives

$$10^{138} - 1 = \Phi_1(10) \Phi_2(10) \Phi_3(10) \Phi_6(10) \Phi_{23}(10) \Phi_{46}(10) \Phi_{69}(10) \Phi_{138}(10)$$

We have $\Phi_1(10) = 9$ and $\Phi_2(10) = 11$ which corresponds to the factor of $99$, so removing those factors gives

$$\frac{10^{138} - 1}{99} = \Phi_3(10) \Phi_6(10) \Phi_{23}(10) \Phi_{46}(10) \Phi_{69}(10) \Phi_{138}(10).$$

The next few factors are

• $\Phi_3(10) = \frac{10^3 - 1}{10 - 1} = 111 = 3 \cdot 37$
• $\Phi_6(10) = \frac{10^3 + 1}{10 + 1} = 91 = 7 \cdot 17$

and from here things get big. The next one is $\Phi_{23}(10) = \frac{10^{23} - 1}{10 - 1} = \underbrace{111 \cdots 1}_{23 \text{ times}}$ which has no more "obvious" factors. From here if you really want to do this by hand you can use the following fact:

Proposition: A prime $p$ divides $\Phi_n(x)$ if and only if $x$ has multiplicative order $n \bmod p$, and in particular $p \equiv 1 \bmod n$.

So to search for factors of $\frac{10^{23} - 1}{9}$ you can restrict your attention to primes congruent to $1 \bmod 23$, and so forth. But this isn't a big help considering how large it is. In fact it turns out to be prime but I don't know how you'd prove that by hand.