Problem 1761

Prove that for any positive integer $x$, we have $[\sqrt{5} x] \geqslant x+1$.
This is because when $x \in \mathbf{N}_{+}$, $\sqrt{5} x > 2 x \geqslant x+1$.
Therefore, $[\sqrt{5} x] \geqslant [x+1] = x+1$.
From this, we can see that the conclusion to be proven is equivalent to: there exists some positive integer $m$, such that for all $n \geqslant m, n \in \mathbf{N}_{+}, a_{n}$ is not a multiple of 5 (at this time $a_{n+1} = [\sqrt{5} a_{n}] > a_{n}$). Below, we will prove this.

To do this, we first prove that the sequence $\{a_{n}\}$ contains two consecutive terms that are not multiples of 5.
We use proof by contradiction to prove this. Assume that in the sequence $\{a_{n}\}$, any two consecutive terms have at least one that is a multiple of 5.
Consider $a_{1}, a_{2}, a_{3}$.
(1) If $5 \mid a_{1}$, then $a_{2} = \frac{a_{1}}{5}$, and by the recurrence relation of $\{a_{n}\}$, we know $a_{3} \leqslant [\sqrt{5} a_{2}]$, so $a_{3} \leqslant [\frac{\sqrt{5}}{5} a_{1}]$, i.e., $a_{3} \leqslant a_{1} - 1$.
(2) If $5 \nmid a_{1}$, then by the assumption in (1), $5 \mid a_{2}$, so $a_{2} = [\sqrt{5} a_{1}], a_{3} = \frac{a_{2}}{5}$, thus $a_{3} = \frac{[\sqrt{5} a_{1}]}{5} < a_{1}$, i.e., $a_{3} \leqslant a_{1} - 1$. $\square$

Combining (1) and (2), we see that in either case, $a_{3} \leqslant a_{1} - 1$, i.e., starting from $a_{1}$, we can find a term $a_{3}$ in the sequence $\{a_{n}\}$ such that $a_{3} \leqslant a_{1} - 1$. Similarly, starting from $a_{3}$, we can find a term $a_{5}$ in the sequence $\{a_{n}\}$ such that $a_{5} \leqslant a_{3} - 1, \cdots \cdots$, and this process can be repeated indefinitely, each new term being at least 1 less than the previous one. Since all terms in the sequence $\{a_{n}\}$ are positive integers, this process cannot continue indefinitely, leading to a contradiction.

Therefore, the assumption in (1) is false.
Thus, we have proven that the sequence $\{a_{n}\}$ must contain two consecutive terms that are not multiples of 5.
Hence, we can find two consecutive terms $a_{k}$ and $a_{k+1}$ that are not multiples of 5.
We now prove a lemma:
Lemma: If $a_{n}$ and $a_{n+1}$ are not multiples of 5, then $a_{n+2}$ is also not a multiple of 5.
Proof of the lemma: Since $a_{n}$ is not a multiple of 5, then by the problem statement,
$a_{n+1} = [\sqrt{5} a_{n}], a_{n+2} = [\sqrt{5} a_{n+1}]$.
Let $a_{n+1} = \sqrt{5} a_{n} - \alpha$, where $0 < \alpha < 1$.
Then $a_{n+2} = [\sqrt{5} a_{n+1}] = [\sqrt{5} (\sqrt{5} a_{n} - \alpha)] = 5 a_{n} + [-\sqrt{5} \alpha]$.
Since $0 < \alpha < 1$,
we have $-3 < -\sqrt{5} \alpha < 0$, so $[-\sqrt{5} \alpha] = -3$ or -2 or -1.
Thus, $a_{n+2} = [-\sqrt{5} \alpha] \equiv -3$ or -2 or $-1 \pmod{5}$.
Therefore, $a_{n+2}$ is not a multiple of 5, and the lemma is proven.
Using the lemma and the fact that $a_{k}$ and $a_{k+1}$ are not multiples of 5, we can conclude that for $n \geqslant k$, $a_{n}$ is not a multiple of 5. Combining this with the initial derivation, we have proven the original proposition.