infinitely many primes 1 mod p

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nayel
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infinitely many primes 1 mod p

Unread post by nayel » Sun May 19, 2013 7:42 pm

Here is a sequence of steps through which you can find infinitely many primes congruent to $1\bmod p$ for any prime $p$. (Of course, without having to use Dirichlet's theorem!)

A. Let $a>1$ be an integer.
B. Let $N=a^p-1$.
C. What is the order of $a\bmod N$?
D. This order must divide $\phi(N)$, where $\phi$ is Euler's totient function.
E. Consider the prime factorisation of $N$ to write down $\phi(N)$. Use D to show that there exists a prime $\equiv 1\bmod p$.
F. Can you make some clever choices for $a$ to find infinitely many such primes?
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Masum
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Re: infinitely many primes 1 mod p

Unread post by Masum » Thu Dec 19, 2013 7:19 pm

Nice. In a similar technique, I tried very hard to find some sort condition or special choices to force $\varphi(p)$ is a square. But still no luck. Must be. Otherwise I would already have solved $p=n^2+1$ problem.
One one thing is neutral in the universe, that is $0$.

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asif e elahi
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Re: infinitely many primes 1 mod p

Unread post by asif e elahi » Fri Dec 20, 2013 8:41 pm

nayel wrote:Here is a sequence of steps through which you can find infinitely many primes congruent to $1\bmod p$ for any prime $p$. (Of course, without having to use Dirichlet's theorem!)

A. Let $a>1$ be an integer.
B. Let $N=a^p-1$.
C. What is the order of $a\bmod N$?
D. This order must divide $\phi(N)$, where $\phi$ is Euler's totient function.
E. Consider the prime factorisation of $N$ to write down $\phi(N)$. Use D to show that there exists a prime $\equiv 1\bmod p$.
F. Can you make some clever choices for $a$ to find infinitely many such primes?
We prove that $p^{p^{k}}-1$ always has a prime divisor that is congruent to 1$mod p^{k}$.We assume the contrary .Let q be a prime divisor of it.Then $p^{p^{k}}\equiv1(mod q)$.Again by Fermat's theorem
$p^{q-1}\equiv 1(mod q)$.So $p^{gcd(p^{k},q-1)}\equiv 1(mod q)$.$gcd(p^{k},q-1)$=$p^{m}$ for some m.By our assumption ,$m\leq k-1$ or $q\mid p^{p^{m}}-1\mid p^{p^{k-1}}-1$.So all the prime divisors $p^{p^{k}}-1$ are divisors of $p^{p^{k-1}}-1$.But $p^{p^{k}}-1=(p^{p^{k-1}})^{p}=p^{p^{k}}-1)X$.here $gcd(p^{p^{k-1}}-1,X)=gcd(p^{p^{k-1}}-1,p)=1$.So $p^{p^{k}}-1$ has a prime divisor that dos'nt divide $p^{p^{k}}-1$.So our assumption was wrong.So $p^{p^{k}}-1$ always has a prime divisor that is congruent to 1$(mod p^{k})$.
We proved that,for all k,there are at least one prime q that is congruent to 1 $(mod p^{k})$.So there are infinite primes congruent to 1(mod p). :D

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