congruence for modular forms coefficients Let $A$ be the set of modular forms $f=\sum a(n)q^n$ of weight $k$ on $\Gamma_0(N)$ with character $\chi$ whose coefficients $a(n)$ are in the ring of integers $\mathcal{O}$ of a fixed number field $F$. Let $\mathfrak{P}$ be an ideal of $F$, Serre obtained the following result :
the set of primes $p\equiv 1\pmod{N\mathfrak{P}}$ such that $f|T_p\equiv 2f\pmod{\mathfrak{P}}$ for all $f$ in $A$ has positive density.
My question is about this congruence, taking two co-prime integers $a$ and $b$, is it possible to find infinitely many primes $p\equiv a\pmod{b}$ for which $f|T_p\equiv 2f\pmod{\mathfrak{P}}$ ? (we may replace 2 by other elements of $\mathcal{O}/\mathfrak{P}$). 
 A: There are two questions, the one where you ask whether it is possible to find a $p$ in an arithmetic sequence s.t. $f|T_p \equiv 2 f$ and the one s.t. $f | T_p \equiv \alpha f$ for some $\alpha$ in the residue field. 

Here I answer the first question, by the negative: no, it is not always possible.
Here is a counter-example: take $f=\Delta$, $\mathcal O=\mathbb Z$, $\mathfrak P = 3\mathbb Z$. Then as you say, by Serre's result, $f | T_p  \equiv 2 f$ for a positive density, and actually for all primes $p$ such that $p \equiv 1 \pmod{3}$. But $f|T_p \equiv 0$ for all primes $p$ such that $p \equiv -1 \pmod{3}$. To see both assertion below, since $f$ is an eigenform, it suffices to check that $\tau(p) \equiv 1 + p \pmod{3}$, which is actually a well-known congruence, where $\tau(n)$ is the $n$-th coefficient of $f=\Delta$.

Now I claim that the answer to the second question is also negative. 
Here is a counter-example: take $f=\Delta^3$, $\mathcal O=\mathbb Z$, $\mathfrak P = 2\mathbb Z$. Then for all odd prime $p$, one has $T_p f \equiv 0$ or $T_p f \equiv \Delta$, but never $T_p f \equiv \alpha f$ for $\alpha \in \mathbb F_2$. Thus any $(a,b)$ gives a counter-example.

To give a positive result, let me say that the answer to the second question is positive when $f$ is an eigenform for almost all Hecke operators (or even infinitely many, that's enough.)
