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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

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    @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

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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 with $(a,j)=1$ there exist infinitely many primes $p$ such that $pj\equiv b\mod a$.

Your statement implies Dirichlet's theorem: Let $a,b$ be positive integers with $(a,b)=1$ and $a>2$. Let $1 with $(a,j)=1$. It is clear that $(jb,a)=1$. Then apply your statement with the pair $(jb,a)$ and conclude that there exist infinitely many primes such that $pj\equiv jb\mod a$. Now $pj\equiv jb\mod a$ implies $p\equiv b\mod a$.

The case $a=2$ only says that there are infinitely many odd primes. :)

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    Is it common to use plain parens to represent gcd? I'm so used to reading those as tuples.2012-12-18