Is there a triple of nonzero even integers $(a,b,c)$ that satisfies the following infinite system of congruences?
$$
a+b+c\equiv 0 \pmod{4} \\
a+3b+3c\equiv 0 \pmod{8} \\
3a+5b+9c\equiv 0 \pmod{16} \\
9a+15b+19c\equiv 0 \pmod{32} \\
\vdots \\
s\_na + t\_nb + s\_{n+1}c \equiv 0 \pmod{2^{n+1}} \\
\vdots
$$
where $(s\_n)$ and $(t\_n)$ are weighted tribonacci sequences defined by
$$
s\_1=s\_2=1, \\
s\_3=3, \\
s\_n = s\_{n-1} +2s\_{n-2} + 4s\_{n-3} \text{ for } n>3,
$$
and
$$
t\_1=1, \\
t\_2=3, \\
t\_3=5, \\
t\_n = t\_{n-1} +2t\_{n-2} + 4t\_{n-3} \text{ for } n>3.
$$
I think there are no nonzero solutions, but I haven't been able to prove this. Computationally, I found there are no nonzero solutions for integers $a$, $b$, and $c$ up to $1000$.


Note the $s\_n$ and $t\_n$ are always odd, and that the ratios $\frac{s\_n}{s\_{n-1}}$ and $\frac{t\_n}{t\_{n-1}}$ approach $2.4675...$.



Is there a nonzero solution to this infinite system of congruences?
Let $u\_n = a s\_n + b t\_n + c s\_{n+1}$. The stronger claim is true: for large enough values of $n$,  

the number $u\_n$ will be exactly divisible
by a fixed power of $2$ that doesn't depend on $n$.


Let $u\_n = a s\_n + b t\_n + c s\_{n+1}$ then (by induction)
$$u\_{n} = u\_{n-1} + 2 u\_{n-2} + 4 u\_{n-3}.$$


The polynomial $x^3 - x^2 - 2 x - 4$ is irreducible and has three roots $\alpha\_1$, $\alpha\_2$, and $\alpha\_3$ in $\overline{\mathbf{Q}}$.
By the general theory of recurrence relations,
$$u\_n = A\_1 \alpha^n\_1 + A\_2 \alpha^n\_2 + A\_3 \alpha^n\_3$$
for constants $A\_1$, $A\_2$, $A\_3$. Since $u\_n \in \mathbf{Q}$, we may additionally deduce that $A\_i$ lie in $\mathbf{Q}(\alpha\_1,\alpha\_2,\alpha\_3)$.
That is because we can solve for $A\_i$ using the equation


$$\left( \begin{matrix} \alpha\_1 & \alpha\_2 & \alpha\_3 \\ \alpha^2\_1 & \alpha^2\_2 & \alpha^2\_3 \\ \alpha^3\_1 & \alpha^3\_2 & \alpha^3\_3 \end{matrix} \right)
\left( \begin{matrix} A\_1 \\ A\_2 \\ A\_3 \end{matrix} \right) = \left( \begin{matrix} u\_1 \\ u\_2 \\ u\_3 \end{matrix} \right)$$
and the matrix on the left is invertible (Vandermonde). In fact we deduce the stronger claim that
any Galois automorphism sending $\alpha\_i$ to $\alpha\_j$ sends $A\_i$ to $A\_j$.
(Simply consider the action of the Galois group on both sides of this equation,
noting that the $A\_i$ are determined uniquely from this equation.) In particular, if one of the $A\_i = 0$,
then all of the $A\_i = 0$.


But now fix an embedding of $\overline{\mathbf{Q}}$ into $\overline{\mathbf{Q}}\_2$. From the Newton Polygon,
we see that there is one root (call it $\alpha\_1$) of valuation $0$, and the other two roots have valuation $1$. Hence
$$\|A\_1 \alpha^n\_1 \|\_2 = \|A\_1\|\_2, \quad \|A\_2 \alpha^n\_2 \|\_2 = \|A\_2\|\_2 \cdot 2^{-n}, \quad \|A\_3 \alpha^n\_3 \|\_2 = \|A\_3\|\_2 \cdot 2^{-n}.$$
In particular, if $A\_1 \ne 0$, then (by the ultrametric inequality) $\| u\_n \|\_2 = \|A\_1\|$ for $n$ large enough. Hence we deduce that either the $2$-adic valuation
of $u\_n$ is eventually constant (as claimed) or that $A\_1 = 0$ and so $A\_i = 0$ for all $i$, which implies that $u\_n = 0$ for all $n$. But if $u\_1 = u\_2 = u\_3 = 0$,
then


$$\left( \begin{matrix} 1 &1 & 1 \\ 1 &3 & 3 \\ 3 & 5 & 9 \end{matrix} \right)
\left( \begin{matrix} a \\ b \\ c \end{matrix} \right) = \left( \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right)$$


The matrix on the left is invertible which implies that $a=b=c=0$.


