Does the statement the are an infinite number of primes p, such that for any j, $1 $pj\equiv b$ mod a imply dirichlets theorem? that there are an infinite number of primes p such that $p\equiv b$ mod a, for constants b and a
Primes in arithmetic progression
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elementary-number-theory
prime-numbers
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0If $a>4$, and, say, $b=a-1$, then this would mean that (for $j=2$): $2p\equiv -1\pmod a$ and (for $j=4$) $4p\equiv -1\pmod a$. Subtracting, that would mean $2p\equiv 0\pmod a$, which seems like a contradiction, so there can not be any such $p$. – 2012-12-18
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2Can you make your statement clearer? At the moment I am not entirely sure what you are asking. – 2012-12-18
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0@ThomasAndrews, it's not a simultaneous equation, so subtracting doesn't make sense. – 2012-12-18
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0@Ethan, you definitely need some restrictions on $j$ you consider, like $(j,a) = 1$. Second, infinitude of primes for one such $j$ is enough, since you can divide $j$ to the other side, as in the other thread you posted. – 2012-12-18
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0Now im confused – 2012-12-18
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0@Sanchez, he said, there exists infinitely many $p$ such that for all $j$. If he meant, "for all $j$, there exists infinitely many $p$," that would be another thing. But that is not what OP said. – 2012-12-18
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0@ThomasAndrews, Ah I see your point. Yet, as OP is thinking about Dirichlet's theorem, I would guess that he meant for all $j$, blah blah blah instead. – 2012-12-18
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0@Sanchez yeah, I suggested in comments below that was what he meant, and he insisted otherwise, which led me to post this more detailed reasoning in the comment. – 2012-12-18
1 Answers
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Dirichlet's theorem: For positive integers $a,b$ with $(a,b)=1$ there exist infinitely many primes $p$ with $p\equiv b\mod a$.
Your statement (or what your statement should be): For positive integers $a,b$ with $(a,b)=1$ and $1 Your statement implies Dirichlet's theorem: Let $a,b$ be positive integers with $(a,b)=1$ and $a>2$. Let $1 The case $a=2$ only says that there are infinitely many odd primes. :)
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0Is it common to use plain parens to represent gcd? I'm so used to reading those as tuples. – 2012-12-18