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Let $$f_n = 37111111...111$$ with n 1's. Prove that $$f_n$$ will never be prime for $$n\ge1.$$

I tried to look $$f_n$$ in mod(p), assuming $$f_n$$ is prime, for the sake of contradiction. I also tried to apply Wilson and Fermat's small theorem.

I'm sure there must be a simple factorization which I'm overseeing.

Martin Sleziak
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Niko
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  • Hint $f_n = \frac{1}{9} (167 \times 2^{n+1} 5^n-1)$ – JohnWO Feb 17 '15 at 10:58
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    @JohnWO I don't quite see how this hint helps to prove what the author wants. – TZakrevskiy Feb 17 '15 at 11:00
  • Try Factor[371], Factor[3711], Factor[37111] etc. (up to $n=10$) with WolframAlpha. – pharmine Feb 17 '15 at 11:02
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    @TZakrevskiy: We have $$f_n = 37\cdot 10^n+\color{blue}{\tfrac{1}{9}(10^n-1)} = \tfrac{1}{9}(167\times 2^{n+1}5^n-1)$$ The blue addend (hence $f_n$) is divisible by $37$ when $n=3m$; and $f_n$ is divisible by $3$ when $n=3m+2$. So what remains is to verify that $f_n$ is divisible by $7$ or $13$ when $n=3m+1$. – Tito Piezas III Feb 17 '15 at 13:08
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    @TitoPiezasIII as you have just shown, this so-called hint looks like a showing off: it does not help directly, it is specifically overcomplicated, and after recovering a good form of this hint, still some explanation is needed. – TZakrevskiy Feb 17 '15 at 15:26
  • @TZakrevskiy See my answer for one simple, general way to view it. – Bill Dubuque Feb 17 '15 at 21:10

2 Answers2

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Hint: the modular bases you are looking for are $3, 7, 13$ and $37$ for different values of $n$. One of these divides every member of the sequence.

Joffan
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    Ahhh! 37|111, so $$f(0(mod 3))=0(mod 37)$$. And it's easy to see that $$f(2(mod 3))=0(mod 3)$$. So all I have to do is find a modulus that works for $$n=1(mod 3)$$ – Niko Feb 17 '15 at 10:59
  • @Niko - in fact two different bases, for $1 \bmod 6$ and for $4 \bmod 6$ – Joffan Feb 17 '15 at 11:18
  • @Niko - For investigating other bases, it's useful to observe that $f(n+1) = 10\cdot f(n)+1$. So, for example, starting from $i=0, (f(i) \bmod 11) \equiv 4, 8, 4, 8, \ldots$ so $11$ never divides any term of the sequence. – Joffan Feb 17 '15 at 14:45
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Hint $\ $ Examining prime factorizations of the first $\,\color{#c00}6\,$ elements of $\,f_N = 334\cdot 10^{N}\!-1$ reveals

$\, 10^{\large\color{#c00}6}\!-1 = 3^3\cdot 7\cdot 11\cdot 13\cdot 37,\,$ and $\,f_N$ is divisible by one of those prime powers $\,p^i$ for all $\,N <\color{#c00} 6,\,$ therefore $\,f_N\,$ is divisible by one of those $\,q = p^i\,$ for all $\,N,\,$ so $\,f_N/9\,$ is composite for all $\,N\ge 0.\,$

Proof $\ $ Dividing $\,N\,$ by $\,6\,$ yields $\,N = K+6J\,$ for $\,K < 6.\ $ By hypothesis some $\,q\mid f_K$

${\rm mod}\ q\!:\ \color{#c00}{10^{\,6}}\equiv\color{#c00} 1\,\Rightarrow\, \color{#0a0}{10^N} \equiv 10^{K+6J}\equiv 10^K(\color{#c00}{10^6})^J\equiv 10^K\color{#c00}1^J\equiv \color{#0a0}{10^K}\ $ by $ $ Congruence Rules

So $\,q\mid f_K\Rightarrow\, 0\equiv f_K\! \equiv 334\cdot \color{#0a0}{10^K}\!-1\equiv 334\cdot \color{#0a0}{10^N}\!-1 \equiv f_N\Rightarrow\, q\mid f_N,\,$ properly by $\,q < f_0 \le f_N$

Remark $\ $ This is a typical application of covering congruences. See also this question, and see this question for a polynomial analog, and a link to a paper of Schinzel.

Bill Dubuque
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