15. Given the set
$$
P=\left\{x \mid x=7^{3}+a \times 7^{2}+b \times 7+c, a 、 b 、 c\right. \text { are positive integers not }
$$

exceeding 6 $\}$.
If $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ elements in set $P$ that form an arithmetic sequence, find the maximum value of $n$.
15. (1) Clearly,
$$
\begin{array}{l}
7^{3}+7^{2}+7+1, 7^{3}+7^{2}+7+2, \\
7^{3}+7^{2}+7+3, 7^{3}+7^{2}+7+4, \\
7^{3}+7^{2}+7+5, 7^{3}+7^{2}+7+6
\end{array}
$$

These six numbers are in the set $P$ and form an arithmetic sequence.
(2) Prove by contradiction: Any seven different numbers in the set $P$ cannot form an arithmetic sequence.

Let $x_{1}, x_{2}, \cdots, x_{7}$ be seven different elements in the set $P$ that form an arithmetic sequence, with a common difference of $d (d>0)$.

By the properties of the elements in the set $P$, we know that no element in the set $P$ is a multiple of 7.

Thus, by the pigeonhole principle, among the seven numbers $x_{1}, x_{2}, \cdots, x_{7}$, there exist two numbers that have the same remainder when divided by 7, and their difference is divisible by 7.

Without loss of generality, assume $x_{i}-x_{j} (i, j \in \{1,2, \cdots, 7\}, i<j)$ is divisible by 7. Then
$$
7 \mid (j-i) d \Rightarrow 7 \mid d \text{. }
$$

Let $d=7 m (m \in \mathbf{Z}_{+})$, and suppose
$$
x_{1}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3},
$$

where $a_{1}, a_{2}, a_{3}$ are positive integers not exceeding 6.
Then $x_{i}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3}+7(i-1) m$, where $i=2,3, \cdots, 7$.
$$
\begin{array}{l}
\text{By } x_{7} \leqslant 7^{3}+6 \times 7^{2}+6 \times 7+6, \\
x_{7} \geqslant 7^{3}+1 \times 7^{2}+1 \times 7+1+7 \times(7-1) m,
\end{array}
$$

we know $1 \leqslant m \leqslant 6$, i.e., the common difference $d$ can only be
$$
7 \times 1, 7 \times 2, \cdots, 7 \times 6 \text{. }
$$

Since $1 \leqslant m \leqslant 6$, and $(7, m)=1$, the remainders of $m, 2 m, \cdots, 6 m$ when divided by 7 are all different, and are one of 1, 2, $\cdots, 6$.

Therefore, there exists $k \in \{1,2, \cdots, 6\}$ such that $a_{2}+k m$ is divisible by 7.
Let $a_{2}+k m=7 t (t \in \mathbf{Z}_{+})$. Then
$$
\begin{array}{l}
x_{k+1}=7^{3}+a_{1} \times 7^{2}+a_{2} \times 7+a_{3}+7 k m \\
=7^{3}+a_{1} \times 7^{2}+(a_{2}+k m) \times 7+a_{3} \\
=7^{3}+(a_{1}+t) \times 7^{2}+a_{3} .
\end{array}
$$

Thus, the coefficient of 7 (i.e., the second digit from the left) in the base-7 representation of $x_{k+1}$ is 0, which contradicts $x_{k+1} \in P$.

Therefore, any seven different numbers in the set $P$ cannot form an arithmetic sequence.
Hence, the maximum value of $n$ is 6.