Can we find a reals$λ,μ$ satisfying this equation Can we find reals $λ,μ$ such that
$$n^{-d-aα}-n^{-b-aα}+n^{ b+αa-1}-n^{d+αa-1}=n^{λ}-n^{μ}$$ for all integer $n>1$.
such that $λ,μ$ are inependent from $n$. Here $a,b,d,α$ are real numbers.
 A: Let's call any expression of the form $\sum_{r \in S} c_r n^r$ where $S$ is a finite set of reals and $c_r \in \mathbb R^\times$ a "generalized polynomial".  You are asking whether it is possible for two generalized polynomials to take the same values for all $n>1$.  It turns out that, just as in the case of regular polynomials, this can only happen if they look identical.  More specifically:

If $\sum_{r \in S_1} c_r n^r = \sum_{r \in S_2} d_r n^r$ for all
  $n>1$, then $S_1 = S_2$ and $c_r = d_r$ for every $r \in S_1$.

It's not too hard to prove this by induction on $|S_1| + |S_2|$ and taking limits as $n\to\infty$ to show that the largest exponents of $S_1$ and $S_2$ must agree.
What it means in your case is that the four terms on the left cannot match the two terms on the right, unless two of four cancel out, which could happen in a few possible ways (for instance if $b+a\alpha = \tfrac12$ and $\lambda = -d-a\alpha$, $\mu= d+a\alpha-1$).  But this type of coincidence will not happen for general values of $a,d,b,\alpha$.
