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Prove that there are infinitely many integers $n$ such that $\frac {2^{6n}-4}{6n-1}$ is equal to an integer

I am struggling with this qustion. Could you walk me through it?

Bill Dubuque
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djdogs422
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    You have a mismatched number of parentheses in your question title and based on how badly it is written and context, it seems likely that you mean for all of $6n-1$ to be in the denominator rather than only $6n$. So... confirm... are you asking to show there are infinitely many integers $n$ such that $\dfrac{2^{6n}-4}{6n-1}$ is an integer? Or something else? – JMoravitz Apr 27 '20 at 17:15
  • Have you tried any examples? If you do, a simple pattern should emerge quite quickly. – lulu Apr 27 '20 at 17:23
  • Yes JMoravitz it is the equation that you have written. – djdogs422 Apr 27 '20 at 17:23
  • Yes JMoravitz it is the equation that you have written. – djdogs422 Apr 27 '20 at 17:26
  • I don't know where to begin – djdogs422 Apr 27 '20 at 17:26
  • Always begin with examples. You can do $n≤3$ by hand, but you'll want a computer to check examples for even fairly small $n$. Be sure to go far enough to get a few fails (cases where the quotient is not an integer). Keep a list of $n, 2^{6n}-4,6n-1$ and record which ones work and which ones fail. You should spot a pattern quite quickly and it's not difficult to prove enough to settle the given question. – lulu Apr 27 '20 at 17:32
  • Is there a way to prove it without a computer? – djdogs422 Apr 27 '20 at 17:38
  • Number theory is based on examples...you should be able to work with simple expressions like these. To be clear, however: the computer won't help you with the proof. All it will do is to suggest a pattern which you must then prove. If you can guess a useful pattern without examples, that would work too. But I recommend looking at examples. – lulu Apr 27 '20 at 17:50
  • Hint $\ 6n!-!1 = p,$ prime $,\Rightarrow, 2^{\large 6n}!\equiv 2^{\large p+1}!\equiv 2^{\large 2},\color{#c00}{2^{\large p-1}}!\equiv 4,$ by little $\rm\color{#c00}{Fermat}$. Thus $,6n!-!1\mid 2^{\large 6n}!-4$. But there are infintely many primes of the form $,6n!-!1,$ by a simple generalization of Euclid's famous one-line proof (we've tens if not hundreds of posts on this, e.g. here; for generalizations see here). – Bill Dubuque Apr 27 '20 at 18:47

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Here is one method. First one can start by noticing the similarity in $6n-1$ and $6n$ in the exponent. Factor out a $2$ from the numerator to get $$2 * \frac{2^{6n-1} - 2}{6n-1}$$ The question at hand is equivalent to asking if $$2^{6n-1} - 2 \equiv 0 \mod 6n-1$$ (this means the numerator is divisible by the denominator. modular arithmetic is useful in number theory problems so it would be helpful to become familiar with it). Essentially, if $a \equiv b \mod n$, then $a - b$ is divisible by $n$. Here, all this is saying is that $(2^{6n-1} - 2) - 0 = 2^{6n-1} - 2$ is divisible by $6n-1$, which is equivalent to what you are trying to prove. There is a famous theorem called Euler's theorem, which states that $a^{\varphi(n)} \equiv 1 \mod n$ if $n$ is coprime to $a$, where $\phi(n)$ is the number of integers less than $n$ coprime to $n$. You can read about this function here. Let $a = 2$ and $n = 6n-1$ (which are trivially coprime, as one is even and the other is odd). According to Dirichlet's theorem on arithmetic progressions, $6n-1$ contains infinitely many primes. Let $n$ be any arbitrary integer such that $6n-1$ is prime, then $$2^{\phi(6n-1)} \equiv 2^{6n-2} \equiv 1 \mod 6n-1$$ or $$2^{6n-1} \equiv 2 \mod 6n-1$$ proving that $2^{6n-1} - 2 \equiv 0 \mod 6n-1$ for infinitely many $n$. A lot of useful results help with problems like this when they are translated into modular arithmetic, so I would really encourage reading about it.

Ryan Shesler
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