Suppose that $a$,$b$ are two distinct primes. Let $r>0$. I want to show that for every $r>0$ there exists a prime $p$ such that $p\equiv 1\mod a^{r}$ and $p\equiv -1\mod b^{r}$. By CRT we know that there exists an integer $x$ such that $x\equiv 1\mod a^{r}$ and $x\equiv -1 \mod b^{r}$. Now my question is, how does one apply the Dirichlet theorem on primes in arithmetic progressions to complete the proof?
Chinese Remainder Theorem in combination with Dirichlet’s Theorem
elementary-number-theoryfield-theorynumber theory
Best Answer
By Lulu's comment I think that this is the solution:
Since $x$ is a solution we find the arithmetic progression $\{x+na^{r}b^{r}\}_{n}$ consisting of solutions. And we clearly see that since $x = ka^{r} + 1 = mb^{r}-1$ for some $k,l\in\mathbb{Z}$ that $gcd(x,a^{r}b^{r}) = 1$. Then by Dirichlet's theorem we find that there exists infinitely many prime satisfying $p\equiv x\mod a^{r}b^{r}$. So $p = x + sa^{r}b^{r}$, which indeed satisfies $p\equiv 1 \mod a^{r}$ and $p\equiv -1 \mod b^{r}$.
So if im right, we can actually find for every $r>0$ infinitely many primes $p$ satisfying $p\equiv 1 \mod a^{r}$ and $p\equiv -1\mod b^{r}$.