Problem 1015

Proof: Let $\{a_n\}$ be an arithmetic sequence with a common difference of $r$. Then there exist positive integers $k$ and $l$ such that $a_k = 1$ and $a_l = \sqrt{2}$. Therefore,
$$(\sqrt{2}-1) = a_l - a_k = (l-k) r$$

Thus, $r = \frac{\sqrt{2}-1}{l-k}$. Assume $a_m, a_n, a_p$ form a geometric sequence, and let $\frac{m-k}{l-k} = M, \frac{n-k}{l-k} = N, \frac{p-k}{l-k} = P$. Then,
$$\begin{array}{l} a_m = a_k + (m-k) r = 1 + M(\sqrt{2}-1) \\ a_n = a_k + (n-k) r = 1 + N(\sqrt{2}-1) \\ a_p = a_k + (p-k) r = 1 + P(\sqrt{2}-1) \end{array}$$

By $a_n^2 = a_m a_p$, we have $(1 - N + N \sqrt{2})^2 = (1 - M + M \sqrt{2})(1 - P + P \sqrt{2})$, which simplifies to
$$(1-N)^2 + 2N^2 + 2N(1-N)\sqrt{2} = (1-M)(1-P) + 2MP + [M(1-P) + P(1-M)]\sqrt{2}$$

Since $M, N, P$ are rational numbers, we have $(1-N)^2 + 2N^2 = (1-M)(1-P) + 2MP$ and $2N(1-N) = M(1-P) + P(1-M)$. This gives us $3N^2 - 2N = 3MP - M - P$ and $2N - 2N^2 = M + P - 2MP$. Adding these two equations, we get $N^2 = MP$. Substituting back, we have $2N = M + P$. Thus, $(M + P)^2 = 4MP$, which implies $(M - P)^2 = 0$, so $M = P = N \Rightarrow m = n = p$, a contradiction. Therefore, no three terms in the sequence $\{a_n\}$ can form a geometric sequence.

Proof: Let the common difference of the arithmetic sequence $\{a_n\}$ be $r$. Then there exist positive integers $k$ and $l$ such that $a_k = 1$ and $a_l = \sqrt{2}$. Thus,
$$(\sqrt{2} - 1) = a_l - a_k = (l - k) r$$

which implies $r = \frac{\sqrt{2} - 1}{l - k}$. Assume $a_m, a_n, a_p$ form a geometric sequence, and let $\frac{m - k}{l - k} = M$, $\frac{n - k}{l - k} = N$, $\frac{p - k}{l - k} = P$. Then,
$$\begin{array}{l} a_m = a_k + (m - k) r = 1 + M(\sqrt{2} - 1) \\ a_n = a_k + (n - k) r = 1 + N(\sqrt{2} - 1) \\ a_p = a_k + (p - k) r = 1 + P(\sqrt{2} - 1) \end{array}$$

By $a_n^2 = a_m a_p$, we have $(1 - N + N \sqrt{2})^2 = (1 - M + M \sqrt{2})(1 - P + P \sqrt{2})$, which simplifies to
$$(1 - N)^2 + 2N^2 + 2N(1 - N) \sqrt{2} = (1 - M)(1 - P) + 2MP + [M(1 - P) + P(1 - M)] \sqrt{2}$$

Since $M, N, P$ are rational numbers, we have $(1 - N)^2 + 2N^2 = (1 - M)(1 - P) + 2MP$ and $2N(1 - N) = M(1 - P) + P(1 - M)$.

This implies $3N^2 - 2N = 3MP - M - P$ and $2N - 2N^2 = M + P - 2MP$. Adding these two equations gives $N^2 = MP$, and substituting back yields $2N = M + P$. Thus, $(M + P)^2 = 4MP$, which simplifies to $(M - P)^2 = 0$, so $M = P = N \Rightarrow m = n = p$, a contradiction. Therefore, no three terms in the sequence $\{a_n\}$ can form a geometric sequence.