# Primes in arithmetic progression

Does the statement the are an infinite number of primes p, such that for any j, $1<j<a$,

$pj\equiv b$ 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

-
If $a>4$, 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 $p$. – Thomas Andrews Dec 18 '12 at 0:08
Can you make your statement clearer? At the moment I am not entirely sure what you are asking. – Eric Naslund Dec 18 '12 at 0:38
@ThomasAndrews, it's not a simultaneous equation, so subtracting doesn't make sense. – user27126 Dec 18 '12 at 0:38
@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. – user27126 Dec 18 '12 at 0:39
Now im confused – Ethan Dec 18 '12 at 0:44

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 $1<j<a$ 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 $1<j<a$ 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. :)