A (probably trivial) consequence of Dirichlet's theorem

Question

Dirichlet's theorem says that for any two positive coprime integers $a$ and $d$, the arithmetic progression $a,a+d,a+2d,a+3d,\ldots$ contains infinitely many prime numbers. In other word, there are infinitely many prime numbers of the form $a+nd$.

What can be said if $a$ is negative (and $d$ is still positive)?

In particular, if $a=-1$ (for which it seems to be known that the answer is positive, namely, there exist infinitely many prime numbers of the form $-1+nd$).

$a$ and $d$ are still assumed to be coprime, since otherwise there is no chance to obtain primes of the form $a+nd$, because we have: $a+nd= \gcd(a,d)\tilde{a}+n\gcd(a,d)\tilde{d}=\gcd(a,d)(\tilde{a}+n\tilde{d})$; notice that this shows that there are infinitely many primes $P$ such that $a+nd$ equals $\gcd(a,d)P$.

Thank you very much!

Edit: Robin Chapman mentions the case $a=-1$ in this question, but there is no explanation there, only a reference.

Answer 1

If $a$ is negative, you can write $a+nd = (a+kd)+(n-k)d$ where $kd>-a$, which gives an equivalent statement with positive integers. In other words, the sign of $a$ is unimportant. $d$ needs to be positive so that there are infinitely many positive terms in the sequence in the first place.

Answer 2

Consider the arithmetic progression $$x_n = -100 + 3n$$ Note $n \leq 33 \implies x_n < 0 \implies x_n \not\in \mathbb{P}$

But at $n = 34, x_n = 2$. A subsequence $y$ of $x$ can be defined as:

$$\forall n\in \mathbb{Z}^*, y_n = 2 + 3n = x_{n+34}$$

So $y_n$ contains an infinite number of primes (by Dirichlet), therefore so does $x_n$

Now lets consider the general case:

$$\forall a \in \mathbb{Z}^-, d \in \mathbb{N}, n \in \mathbb{Z}^*,$$ \begin{align} \text{Let } s_n &= a + nd\\ \exists a' \in \mathbb{N} \text{ s.t. } a &= -a'\text{ (existence of additive inverse)}\\ \exists i \in \mathbb{N} \text{ s.t. } i &> a' \text{ (infinitude of the natural numbers)}\\ id &\geq i > a' \text{ (looser than necessary for simplicity)}\\ t_n &= s_{n+i} = (a+id) + nd\\ &\text{with }t_n\text{ of the required form }\square\\ \end{align}