Zeta function of Abelian variety over finite field Let $A/\mathbf{F}_q$ be an Abelian variety of dimension $g$. Suppose one knows $|A(\mathbf{F}_{q^n})|$ for all $1 \leq n \leq g$. Does one know then $\zeta(A,s)$ (equivalently, $|A(\mathbf{F}_{q^n})|$ for all $n$)?
It is true for $g = 1$ by an elementary computation.
 A: This is not an answer.
Here is a possible strategy, that was too long for a comment. I briefly thought it gave a full answer, but there is a lot of stuff missing.
If $\alpha_1,\ldots,\alpha_{2g}$ are the eigenvalues on $H^1(A,\mathbb Q_\ell)$, then $\bigwedge^* H^1 = H^*$ computes all eigenvalues. Moreover, any $\frac{q}{\alpha_i}$ is another one of the $\alpha_i$ (see for example [Suh, Prop. 2.2.1]). Thus, the sum
$$\#A(\mathbb F_{q^r}) = \sum_{k=0}^{2g} (-1)^k\operatorname{tr}(\operatorname{Frob}_q^r|H^k(A,\mathbb Q_\ell))$$
is entirely determined by suitable algebraic expressions in $\alpha_1,\ldots,\alpha_g$ (w.l.o.g.).
The question then becomes the following: if $f \in \mathbb Q(x_1,\ldots,x_g,y)$ is a rational function symmetric in the first $g$ terms, does knowing $f(\alpha_1^r,\ldots,\alpha_g^r,q^r)$ for $r = 1,\ldots,g$ determine $\alpha_1,\ldots,\alpha_g$ up to permutation?
For $f = x_1+\ldots+x_g$, the answer is yes by classical theory, and it seems plausible that therefore the answer is yes for general $f$. But our specific function may be bad; for example $f = x_1\cdots x_g$ clearly does not have this property.
[Suh] Suh, Junecue, Symmetry and parity in Frobenius action on cohomology, Compos. Math. 148, No. 1, 295-303 (2012). ZBL1258.14023.
A: SORRY, THIS IS AN ANSWER TO THE WRONG QUESTION
(count of points on a curve instead of on the abelian variety itself)
WILL DELETE OR CORRECT BEFORE LONG

Yes, the counts for $q^n$ ($n \leq g$), together with the value of $q$, 
are enough.
Let the eigenvalues of Frobenius be $\lambda_i$ ($1 \leq i \leq 2g$),
and let $P(t) = \prod_{i=1}^{2g} (1-\lambda t)$ which is a scaling of
the characteristic polynomial of Frobenius.  Then the power sums 
$\sigma_n := \sum_{i=1}^{2g} \lambda_i^n$ are the Taylor coefficients of
$$
F(t) 
:= -t \frac{P'(t)}{P(t)} = \sum_{i=1}^{2j} \frac{\lambda_i t}{1-\lambda_i t}
= \sum_{i=1}^{2j} 
  \left((\lambda_i t) + (\lambda_i t)^2 + (\lambda_i t)^3 + \cdots \right)
= \sum_{n=1}^\infty \sigma_n t^n.
$$
We are given $\left| A({\bf F}_{q^n}) \right|$, and thus also
$\sigma_n = q^n + 1 - \left| A({\bf F}_{q^n}) \right|$, for $n \leq g$.Thus we know $F(t)$ up to $O(t^{g+1})$.  Thus we know $-F(t)/t$ up to $O(t^g)$.
But that's the logarithmic derivative of $P(t)$, so we know the power series for $\log P(t)$ up to $O(t^{g+1})$ (the constant term vanishes because $P(0)=1$).
Thus
$$
P(t) = \exp \left(-\!\int_{\tau=0}^t F(\tau) \frac{d\tau}{\tau}\right)
$$
gives $P(t)$ up to $O(t^{g+1})$, i.e.\ up to and including the $t^g$
coefficient.  Now the functional equation $P(t) = (qt^2)^g P(1/qt)$
gives us the rest: for each $j=1,\ldots,g$, the $t^{g+j}$ coefficient is
$q^j$ times the $t^{g-j}$ coefficient.  So we know the full expansion 
of $P$, and thus the full list of eigenvalues and the zeta function, QED.
