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I came across a problem in Dudley's Elementary Number Theory. Part (a) is to find one $n$ such that both $6n+1$ and $6n-1$ are composite. So I stumbled across $n=50$, which gives $299=13 \cdot 23$, and $301=7 \cdot 43$

The motivation for this was knowing that $9|10^k -1$ and $7|10^{2k}+1$ and trying to find an even power of ten (obviously $300$ is not a power of ten).

Part (b) asks to prove that there are infinitely many $n$ such that both are composite.

How do we show this?

Setting $6n \pm 1 = 10^{2k} \pm 1$ didn't work.

I looked at $(6n+1)(6n-1) = 36n^2-1 = 6(6n^2) -1$, so I can generate composite numbers of the form $6m -1$, but what about the corresponding $6m+1$ ?

Should I be using my result from (a) to generate these pairs?

I looked at the solutions to this related question, but they seemed focused on proving that both are prime.

Other methods that I've considered are using modular arithmetic (though that hasn't been covered at this point in the book) and using the well-ordering principle in some way... (Hints are acceptable, as I want to be an autonomous mathematician, as much as possible)


Added

If we let $n = 6^{2k} = 36^k$, then $6n \pm 1 = 6^{2k+1} \pm 1$, which (for $k \geq 1$) factors into $(6 \pm 1)(6^{2k} \mp ... +1)$

Is this correct? Here I am using a factorization of sums/differences of odd powers, which "hasn't been covered" (which isn't itself a problem).

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6  
Chinese Remainder Theorem, $6x\equiv 1 \pmod{47}$, $6x\equiv -1 \pmod {61}$. –  André Nicolas Jan 26 '12 at 1:35
    
@André, I will try to digest this, thanks! –  The Chaz 2.0 Jan 26 '12 at 1:47
    
@Andre: That's really nice. I like that a lot. –  mixedmath Jan 26 '12 at 1:56
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You don't need to use $47$ and $61$, any pair of primes relatively prime to $6$ will do. @AndréNicolas. For example $6x\equiv 1\pmod 5$ and $6x\equiv -1 \pmod 7$ yields $x\equiv 1\pmod {35}$. –  Thomas Andrews Jan 26 '12 at 2:10
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@Thomas Andrews: I picked two numbers that were obviously not special, though unfortunately I picked two primes. Thought that perhaps the OP could chase this down to get a family of examples. –  André Nicolas Jan 26 '12 at 2:45

3 Answers 3

up vote 4 down vote accepted

Hint: Instead of even powers of $10$, try odd powers of $6$.

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Ah yes! I will show my work, many thanks. –  The Chaz 2.0 Jan 26 '12 at 1:49
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@TheChaz: Your edit is correct :-) $x^{2n+1} + 1$ is divisible by $x+1$ and $x^{2n+1} - 1$ is divisible by $x-1$. –  Aryabhata Jan 26 '12 at 2:06
    
Great. $$$$ $$$$ –  The Chaz 2.0 Jan 26 '12 at 2:09

Here is a way of solving the problem "non-constructively"

You can find infinitely many disjoint sequences of $8$ consecutive composite numbers, and among any 8 consecutive numbers you can find a pair of the type $6n \pm 1$.

Once you figure this "non-constructive" solution, the standard examples lead to simple solutions. For example, it is easy to figure that $k!+5, k!+7$ work for all $k \geq 7$... This would lead to $n= \frac{k!}{6}+1$.

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Try to make the expressions fit forms that you know factor. Sums and differences of cubes factor. So $n=6^2k^3$ will make the expressions take on those forms.

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That's the final step that I was missing. Thanks –  The Chaz 2.0 Jan 26 '12 at 2:03

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