I'm looking at the questions on this sheet: https://www0.maths.ox.ac.uk/system/files/coursematerial/2014/3116/5/MF_2014_Sheet4.pdf and
The Q is Q2 a).
Now, assuming this is true, the rest of the problem is fairly trivial. However, showing that a) is true is causing me some problems. This mainly because of the term $p^{2k-1}$ in the quadratic.
So, we are given that $f$ is a normalised eigenform for $M_k(SL_2(\mathbb{Z}))$ with expansion $f(q)= \sum_{n \ge 1} a_n(f)q^n$, and that $\alpha$ and $\beta$ are roots of $X^2-a_p(f)X+p^{2k-1}$ where $p$ is prime.
Now, for part a): We are RTS this statement
$$ a_{p^r}(f)=\alpha^r + \alpha^{r-1}\beta +...+\beta^r$$
for all $ r \ge 0$. Obviously as $a_1(f)=1$ we have that it isn't true in the $r =0$ case. It is trivially true when $r=1$ by the quadratic equation.
We proceed by induction, and assume that it is true all $r' < r+1$.
Now, consider the Hecke operator $T(p)$ on $f$. As $a_0(f)=0$, we have that by the definition action of $T(p)$ on $f$, $f$ has eigenvalue $a_p(f)$ under $T(p)$. Hence, we have that the coefficient of $q^{p^r}$ in $T(p)(f)$ is $a_p(f)a_{p^r}(f)$ and that it is also $a_{p^{r+1}}(f)+p^{k-1}a_p^{r-1}(f)$. Hence we have a recursion relation giving $a_{p^{r+1}}(f)$. Now, using the fact that $\alpha \beta = p^{2k-1}$, we get that
$$ a_{p^{r+1}}(f) = \alpha^{r+1} + (2-p^{-k})(\alpha^{r}\beta +...+\beta^r \alpha) + \beta^{r+1}$$
So, it would seem to me that the term $p^{2k-1}$ is causing the problem, and perhaps should read as $p^{k-1}$. Or am I being a total idiot?