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Problem:

By considering this sequence of numbers

$$2^1 + 1,\:\: 2^2 + 1,\:\: 2^4 + 1,\:\: 2^8 +1,\:\: 2^{16} +1,\:\: 2^{32}+1,\ldots$$

prove that there are infinitely many prime numbers.


I am thinking that if I can show that every pair of numbers in the sequence are relatively prime then since each has at least one prime factor this would prove the existence of infinitely many primes.

But I am new to discrete mathematics and number theory so I am not sure on how to proceed.

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marked as duplicate by Alex M., Watson, Shahab, vrugtehagel, Claude Leibovici Mar 23 '16 at 11:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Aren't they Fermat Numbers? Numbers of the form $$2^{2^n} +1$$ $\endgroup$ – TheRandomGuy Mar 22 '16 at 14:03
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    $\begingroup$ It is an open problem to decide if there are infinitely many Fermat primes. Evidence suggests that it may be false. $\endgroup$ – Ángel Valencia Mar 22 '16 at 16:38
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    $\begingroup$ @ÁngelValencia: The OP does not ask about Fermat primes, but about prime numbers in general. $\endgroup$ – Alex M. Mar 23 '16 at 10:09
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Your sequence is given by $a_n=2^{2^n}+1$. You can observe that

$(a_n-2)\cdot a_n=(2^{2^n}-1)(2^{2^n}+1)=(2^{2^n})^2-1=(2^{2\cdot 2^n})-1=(2^{2^{n+1}})-1=a_{n+1}-2$.

Hence if a prime number divides $a_n$ and $a_{n+1}$, then it divides $2$. But $a_n$ and $a_{n+1}$ are both odd, so they are coprime.

You could try to go further comparing $a_n$ and $a_m$ in general.

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    $\begingroup$ In the book Prime Number Records ,the 1st chapter gives about 23 different proofs of the infinitude of primes. (Even a topological proof.) It says that Prof. Leo Morse said ,"It is sufficient to exhibit a strictly increasing sequence of pair-wise co-prime natural numbers, for example the Fermat numbers.".... Fermat numbers are $F(n)=1+2^{2^n}$. It is unknown if $F(n)$ is prime for any $n\geq 5$, but that is not needed in this Q. $\endgroup$ – DanielWainfleet Mar 23 '16 at 4:37
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If $2^{2^n}\equiv -1\pmod p$, then show that $2^{2^{m}}\not\equiv-1\pmod p$ for any $m<n$.

This is because if $2^{2^m}\equiv -1\pmod p$ then: $$-1\equiv 2^{2^n}=\left(2^{2^{m+1}}\right)^{2^{n-m-1}}\equiv 1\pmod p$$ So $p=2$. But $p$ can't be $2$.

So $2^{2^n}+1$ and $2^{2^m}+1$ can't have any prime factors in common if $m\neq n$.

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To continue the approach by @JérémyBlanc: Note that $a_n=2^{2^n}+1$ in general. Now, note that $$a_{n}-2=2^{2^{n}}-1=(2^{2^{n-1}}-1)(2^{2^{n-1}}+1)=a_{n-1}(2^{2^{n-1}}-1)=a_{n-1}(a_{n-1}-2)=a_{n-1}a_{n-2}(a_{n-2}-2)=\dots=a_{0}a_{1}a_{2}\dots a_{n}(a_{0}-2)=a_{0}a_{1}a_{2}\dots a_{n-1}$$

$$\therefore a_n-2=a_0a_1a_2\dots a_{n-1}$$ Equivalently, $$ 2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1), $$

Since $a_{n}$ is odd, note that $\gcd(a_n,a_{n}-2)=1$. This implies that any divisor of $a_{n}-2$ is different from any divisor of $a_n$, thus implying for $t<n$, $\gcd(a_n, a_t)=1$.

This implies that all $a_n$ are coprime. As you have noticed, this completes our proof.

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