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I came across this question that asks weather there are infinitely many primes of the form $an+d$ with $\gcd$($a, d$) $=$ $1$. To show there are infinitely many primes of the form(s) $4k-1$ and $6k-1$ can be proven with Euclid's Theorem:

For $4k-1$:

Suppose there are finitely many primes (in general), they are $2$, $p$, $p_2$,... $p_n$. Now let $q$ be the product of all these primes. $2q$ is still divisible by all the primes $p_n$ on our list. $2q-1$, However is $-1$ $\pmod {p_n}$ for all primes $p_n$ on the list of primes first created. Therefor, $2q-1$ must be divisible by a prime $n$ that was not on our list. Since $2$ was on our list of primes, and the product $2q$ is divisible by $4$, $2q-1$ is $3$ $\pmod 4$, and therefore must have a prime factor $3$ $\pmod 4$ which was not on our list of primes.

The $6k-1$ proof is similar by first starting with $2, 3, p, p_2$, $p_n$. instead. Given the product $q, q-1$ has a prime factor $n$ $=$ $5$ $\pmod 6$, but $n$ is not on our list.

Is there a proof like these for the $+1$ sides or any other progressions? Thanks.

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In general it's true that whenever $(a,d) = 1$ then the arithmetic sequence $ax + d$ has infinitely many primes. This is true due to the Dirichlet's Theorem. I doubt that there's an easy proof of this.

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    $\begingroup$ This is kind of orthogonal to the question. $\endgroup$ – Erick Wong Feb 9 '18 at 6:59

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