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Prove that each positive integer $n$: $ x = \displaystyle 10^{10^{10^n}}+10^{10^n}+10^n-1 $ is not prime.

This seems like a very difficult problem, any ideas at all?

I would like to use modular arithmetic and algebraic number theory. This is sort of inspired by: Gerry's Myerson's first proof (algebraic number theory).

Now here is my attempt:

If $x$ is NOT prime then:

$$x \equiv 0 \pmod{j}$$ for some $j$ right?

$$10^{10^{10^n}}+10^{10^n}+10^n \equiv 1 \pmod{j}$$

but here I get stuck on the LHS.

How can I use algebraic number theory?

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    Using the method which determines whether or not a number on base $10$ is divisible by $11$, you can see that if $n$ is even then this number is divisible by $11$. Now you're left with proving the above for odd values of $n$. – barak manos Feb 04 '15 at 15:13
  • @barakmanos I don't think that $\bmod 11$ works for even $n$. The two large terms in the expression put a "1" into the representation at digit positions of the same parity. By contrast I think the $\bmod 11$ argument is fine for odd $n$, for the same reason. – Joffan Feb 04 '15 at 15:21
  • @Joffan: Yep, that's correct. I initially wrote "odd", then changed it to "even". Now I can no longer change it, but whoever reads this comment-thread will know... Thanks. – barak manos Feb 04 '15 at 15:24

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OK, so we can show that for $n>0$ $$a_n = 10^{10^{10^n}}+10^{10^n}+10^n-1$$

is divisible by $10^{2^k}+1$ where $2^k$ is the greatest power of $2$ that divides $n$, ie. $n=2^kd$, with $d$ odd

So, for $n$ odd, $k=0, 2^k=1$ and $n$ is divisible by $11$; for $n \equiv 2 \bmod 4$ , $k=1, 2^k=2$ and $n$ is divisible by $101$; for $n \equiv 4 \bmod 8$ , $k=2, 2^k=4$ and $n$ is divisible by $10001$, etc.

as a separate demonstration for the $n$ odd case, examining powers of $10, \bmod 11$: $$\begin{array}{c|c|c} \hline x & 10^x \bmod 11 \\ \hline 1 & 10 \\ \hline 2 & 1 \\ \hline 3 & 10 \\ \hline etc \\ \end{array}$$

Note that the higher power components are $10^\text{even}$
So when $n$ is odd, $a_n \equiv (1+1+10-1)\bmod 11 \equiv 0 \bmod 11$

For the general case, obviously $10^{2^k}-1 \equiv -2 \bmod (10^{2^k}+1)$ and we can see that $(10^{2^k}+1) \mid (10^{2(2^k)}-1)$ and also $(10^{2^k}+1) \mid (10^{2j(2^k)}-1)$ for $j\geq 1$,

So for example, $\overbrace{99999999999999}^{\text{14 9's}}\equiv -2 \bmod 101$

Now the two larger powers both have a high enough power of two in the exponent that they are each $\equiv 1 \bmod 10^{2^k}$ as above. So these add back $1$ each to the modular sum and the end result is that a suitable divisor can be found depending on the factors of $2$ in $n$.

So all $a_n$ are composite.

Joffan
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  • $10^3\equiv 12 \mod13$, $10^4\equiv 3 \mod13$, $10^5\equiv 4 \mod13$, $10^6\equiv 1 \mod13$ – Julián Aguirre Feb 05 '15 at 10:23
  • @JuliánAguirre - thanks - that's what I get for doing modular arithmetic in my head while driving... modified justification, proof for even numbers to pick up later... – Joffan Feb 05 '15 at 15:49