Let $x_0=a,x_1=b$ and $x_{n+1}=2x_n-9x_{n-1}$ for each $n\in\mathbb N$, where $a,b$ are integers. Find the necessary and sufficient condition on $a$ and $b$ for the existence of an $x_n$ which is a multiple of $7$.
1. **Initial Conditions and Recurrence Relation:**
   Given the sequence defined by \( x_0 = a \), \( x_1 = b \), and the recurrence relation \( x_{n+1} = 2x_n - 9x_{n-1} \) for each \( n \in \mathbb{N} \).

2. **Modulo 7 Analysis:**
   We need to find the necessary and sufficient condition on \( a \) and \( b \) for the existence of an \( x_n \) which is a multiple of 7. We will consider the sequence modulo 7.

3. **Sequence Modulo 7:**
   Let's compute the first few terms of the sequence modulo 7:
   \[
   x_0 \equiv a \pmod{7}
   \]
   \[
   x_1 \equiv b \pmod{7}
   \]
   \[
   x_2 \equiv 2x_1 - 9x_0 \equiv 2b - 9a \equiv 2b - 2a \equiv 2(b - a) \pmod{7}
   \]
   \[
   x_3 \equiv 2x_2 - 9x_1 \equiv 2(2(b - a)) - 9b \equiv 4(b - a) - 9b \equiv 4b - 4a - 9b \equiv -5b - 4a \equiv 2b - 4a \pmod{7}
   \]
   \[
   x_4 \equiv 2x_3 - 9x_2 \equiv 2(2b - 4a) - 9(2(b - a)) \equiv 4b - 8a - 18(b - a) \equiv 4b - 8a - 18b + 18a \equiv -14b + 10a \equiv 3a \pmod{7}
   \]
   \[
   x_5 \equiv 2x_4 - 9x_3 \equiv 2(3a) - 9(2b - 4a) \equiv 6a - 18b + 36a \equiv 42a - 18b \equiv 0a - 4b \equiv 3b \pmod{7}
   \]

4. **Periodicity and Conditions:**
   We observe that \( x_{i+4} \equiv 3x_i \pmod{7} \). This implies that the sequence has a periodic behavior modulo 7. For some term \( x_n \) to be divisible by 7, we need:
   \[
   x_n \equiv 0 \pmod{7}
   \]
   From the periodicity, we need:
   \[
   b \equiv a \pmod{7} \quad \text{or} \quad b \equiv 2a \pmod{7}
   \]

5. **Sufficiency:**
   If \( b \equiv a \pmod{7} \) or \( b \equiv 2a \pmod{7} \), then the sequence will contain a term that is a multiple of 7. This is because the periodicity ensures that the sequence will eventually hit a multiple of 7.

6. **Primitive Root Argument:**
   Since 3 is a primitive root modulo 7, the period of the sequence modulo 7 divides 24. This ensures that the sequence will cycle through all residues modulo 7, confirming that the conditions \( b \equiv a \pmod{7} \) or \( b \equiv 2a \pmod{7} \) are both necessary and sufficient.

\(\blacksquare\)

The final answer is \( \boxed{ b \equiv a \pmod{7} } \) or \( b \equiv 2a \pmod{7} \)