Does the system of congruence equations
\begin{eqnarray} 
A\_{17k}&\equiv& 0 \pmod {17^2}, \nonumber \\ 
A\_{17k+1}&\equiv& 0 \pmod {17^2}, \tag{1}
\end{eqnarray}
has solutions other than $k=3$? Here $A\_n$ are Apery numbers:
$$A\_n=\sum\limits\_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}^2.$$
Thanks to the recurrence relation
$$n^3 A\_n = (34n^3-51n^2+27n-5)A\_{n-1}- (n-1)^3 A\_{n-2},$$
if $k$ is a solution of (1), then all seventeen consecutive numbers beginning from $A\_{17k}$ are congruent to zero modulo $17^2$. In particular,
$$A\_{51},A\_{52},A\_{53},\ldots,A\_{66},A\_{67}\equiv 0 \pmod {17^2}.$$ 



Congruence equation for Apery numbers
Results of Gessel can be used to solve this question. Gessel showed, in [Theorem 1 here](https://www.sciencedirect.com/science/article/pii/0022314X82900713), that for any prime $p$, if $n=\sum d\_i p^i$ is the base-$p$ expansion of $n$, then
$$(\*) A\_n \equiv \prod A\_{d\_i} \bmod p.$$
In [Theorem 4 here](https://www.sciencedirect.com/science/article/pii/0022314X82900713) he essentially refined this as follows. He proved that there is an (explicit) sequence $\{b\_n\}\_{n \ge 0}$ such that $$(\*\*) A\_{pn+k} \equiv A\_{n} (A\_{k} + pn b\_{k}) \bmod {p^2},$$
where $p$ is a prime, $n \ge 0$ and $0 \le k < p$. The first values of $b\_n$ are given by $b\_0=0,b\_1=12$.




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Specializing $(\*)$ to $p=17$, we see that $A\_n$ is divisible by $17$ iff one of the digits $d\_i$ in the base-17 expansion of $n$ satisfies $A\_{d\_i} \equiv 0 \bmod 17$. A finite check shows that the solutions to $A\_d \equiv 0 \bmod 17$ in the range $0 \le d < 17$ are $d=3,13$. 


Specializing $(\*\*)$ to $p=17$ and $k \in \{0,1\}$, we find that
$$A\_{17n} \equiv A\_{n} \bmod {17^2},$$
$$A\_{17n+1} \equiv A\_{n} (A\_1 + 17\cdot 12\cdot n) \bmod {17^2}.$$
This immediately shows that the first equation in your system implies the second one, and that your system is equivalent to 
$$(\*\*\*)A\_n \equiv 0 \bmod {17^2}.$$
We now proceed to solve $(\*\*\*)$. Let $S$ be the set of solutions to $(\*\*\*)$. Writing $n$ as $17m+m'$ ($0 \le m' < 17$), we find that $(\*\*\*)$ is equivalent to
$$A\_{m} (A\_{m'}+17mb\_{m'}) \equiv 0 \bmod {17^2}.$$
There are three possible cases.


1. $A\_m \equiv 0 \bmod {17^2}$.
2. $A\_m \equiv 0 \bmod {17}$, $A\_{m'}+17mb\_{m'} \equiv 0 \bmod 17$.
3. $A\_{m'} + 17mb\_{m'} \equiv 0 \bmod {17^2}.$


The first case says: if $n \in S$ then $17n+n' \in S$ ($0 \le n' < 17$).


The second case says: $m$ has $3$ or $13$ as one of its base-17 digits, and $m' \in \{3,13\}$.


The third case says: Either $m'=3$ and $m \equiv 0 \bmod 17$, or $m'=13$ and $m \equiv -1 \bmod 17$. 


(Here I've used the values $A\_3=1445=17^2\cdot 5$, $A\_{13} = 364713572395983725$, $b\_{3}=4438$, $b\_{13}=7475161875743183448469/6006$.)


This characterizes the set $S$.


