Factors of Sums of Powers of 100,000

Theorem
All integers $n$ of the form:
 * $n = \ds \sum_{k \mathop = 0}^m 10^{5 k}$ for $m \in \Z_{> 0}$

are composite.

Proof
Notice that:
 * $\ds \sum_{k \mathop = 0}^m 10^{5 k} \times R_5 = R_{5 \paren {m + 1} }$

where $R_i$ is the $i$th repunit.

Suppose $p \divides m + 1$, where $p$ is a prime that is not $5$.

By Divisors of Repunit with Composite Index:
 * $R_p \divides R_{5 \paren {m + 1} }$

By Prime not Divisor implies Coprime, $p$ and $5$ are coprime.

By Condition for Repunits to be Coprime, $R_p$ and $R_5$ are coprime.

By Euclid's Lemma:
 * $R_p \divides \dfrac {R_{5 \paren {m + 1} } } {R_5} = n$

Suppose $25 \divides m + 1$.

By Divisors of Repunit with Composite Index:
 * $R_5 \divides R_{25}$

and:
 * $R_{25} \divides R_{5 \paren {m + 1} }$

So we have:
 * $\dfrac {R_{25}} {R_5} \divides \dfrac {R_{5 \paren {m + 1} } } {R_5} = n$

The final case is $m + 1 = 5$.

This is the case $n = 100 \, 001 \, 000 \, 010 \, 000 \, 100 \, 001$.

We have:
 * $n = 21 \, 401 \times 25 \, 601 \times 182 \, 521 \, 213 \, 001$

Thus all cases are covered.