Factors of Sums of Powers of 100,000/General Result

Theorem

All integers $n$ of the form:

$n = \ds \sum_{k \mathop = 0}^m 10^{r k}$ for $m \in \Z_{> 0}$

are composite for $r \ge 2$.

The only exceptions are:

$r = 2^k, m = 1$ for some $k \in \N$

$r = m + 1 =$ some odd prime

in which cases $n$ may be prime.

Proof

Case $1$: $m + 1$ is Composite

Suppose $m + 1$ is composite.

Then:

$\exists p, q > 1: m + 1 = p q$

By Division Theorem, for each $k$ with $0 \le z \le m$:

$\exists i, j \in \N: 0 \le i \le q - 1, \, 0 \le j \le p - 1: k = i + q j$

Thus:

\(\ds n\)

\(=\)

\(\ds \sum_{k \mathop = 0}^m 10^{r k}\)

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 0}^{q - 1} \sum_{j \mathop = 0}^{p - 1} 10^{r \paren {i + q j} }\)

\(\ds \)

\(=\)

\(\ds \paren {\sum_{i \mathop = 0}^{q - 1} 10^{r i} } \paren {\sum_{j \mathop = 0}^{p - 1} 10^{r q j} }\)

Both sums are greater than $1$, so $n$ is composite.

$\Box$

Case $2$: $m + 1$ is an Odd Prime and $r \ne m + 1$

Notice that:

$\ds \sum_{k \mathop = 0}^m 10^{r k} \times R_r = R_{r \paren {m + 1} }$

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

Suppose $m + 1$ and $r$ are coprime.

By Condition for Repunits to be Coprime, $R_{m + 1}$ and $R_r$ are coprime.

By Euclid's Lemma:

$R_{m + 1} \divides \dfrac {R_{r \paren {m + 1} } } {R_r} = n$

Suppose $r^2 \divides m + 1$.

By Divisors of Repunit with Composite Index:

$R_r \divides R_{r^2}$

and:

$R_{r^2} \divides R_{r \paren {m + 1} }$

So we have:

$\dfrac {R_{r^2}} {R_r} \divides \dfrac {R_{r \paren {m + 1} } } {R_r} = n$

$\Box$

Case $3$: $m = 1$, $r$ is not a Power of $2$

We write $n = 1 + 10^r$.

Since $r$ is not a power of $2$, $r$ has an odd factor greater than $1$.

Write $r = x \paren {2 y + 1}$.

Then:

\(\ds n\)

\(=\)

\(\ds 1 + 10^r\)

\(\ds \)

\(=\)

\(\ds 1 + 10^{x \paren {2 y + 1} }\)

\(\ds \)

\(=\)

\(\ds 1^{2 y + 1} + \paren {10^x}^{2 y + 1}\)

By Sum of Two Odd Powers:

$1 + 10^x$ is a factor of $1^{2 y + 1} + \paren {10^x}^{2 y + 1} = n$.

Thus all cases are covered.

$\blacksquare$