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1) Taking $p_1,p_2,\dots,p_k$ to be the primes up to $x^{1/2}$ we have a way to determine, with proof, each prime $N$ between $x^{1/2}$ and $x$ by Finding a representation of $N$ as in part c. Find all the primes between $5$ and $100$ in this way, along with these proofs that they are indeed prime.

2) Use the sequence of Fermat numbers to prove that for each integer $k \ge 1$, there are infinitely many primes are congruent to $1 \pmod {2^k}$.

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closed as off-topic by Care Bear, Ivo Terek, Jonas Meyer, Jack D'Aurizio, J. W. Perry Aug 11 at 4:08

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2  
The usual question: what have you attempted? –  J. M. Oct 6 '11 at 8:02
    
@JM! I want to say in different ways that primes are infinite. I just failed to prove it. –  gandhi Oct 6 '11 at 8:05
5  
What is "part c"? –  Gerry Myerson Oct 6 '11 at 11:46

1 Answer 1

Some hints on 2).

I assume this problem comes from, or is based on, some book. Is there something in the book about factors of $2^{2^k}+1$ all being congruent to 1 modulo some power of 2? Is there something in the book about every number of the form $2^{2^k}+1$ being relatively prime to every other number of that form? And can you put these two things together to get what you want?

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No. But, your hit is working to solve further. Thank you –  gandhi Oct 6 '11 at 18:27

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