Let's find a square number (i.e., the square of a natural number) that ends with five consecutive digits in ascending order from left to right.
The last digit of a square number is $0, 1, 4, 5, 6$, or 9. In our task, there is no preceding digit for 0 and 1; for 4, ...34 would be the two-digit ending, but this does not work because it is not divisible by 4—since the base of an even square number is even, and thus its square is divisible by 4. Similarly, the 5 ending is ruled out because squares ending in 5 are ...25, not ...45. For the 9 ending, the three-digit ending of the square is ...789, which, when divided by 8, leaves a remainder of 5, and it is known that the square of an odd number is of the form $8m+1$: $(2k+1)^2 = 4k(k+1) + 1$, and here the $k(k+1)$ factor is even.

Therefore, the last digits of the square number we are looking for can only be ...23456. From the divisibility rules for 2, 4, and 8, we can proceed to see that divisibility by $2^5 = 32$ depends on the last five digits, since $10^5$ is divisible by $2^5$. In our case, 23456 is divisible by $2^5$, so the square we are looking for is also. But then it is also divisible by $2^6$, since dividing by $4^2$ gives an even quotient, and thus the base is divisible by 8.

Now, ...6 is the ending of $(\ldots 4)^2$ and $(\ldots 6)^2$, which can be summarized as $(10k \pm 4)^2$. From this, we find the last digit of the $k$ values that are possible:

$$
(10k \pm 4)^2 = 100k^2 \pm 80k + 16 = \ldots 56
$$

Here, the two-digit ending of $\pm 80k$ is 40 (since the two-digit ending of $100k^2$ is 00), or $(-60)$, i.e., the one-digit ending of $\pm 8k$ is 4 or $(-6)$. Thus, $(10k + 4)$ fits 3 and 8, and no others, since $8 \cdot 3$ ends in 4, i.e.,

$$
10k + 4 = 10(10m + 3) + 4 = 100m + 34
$$

and $100m + 84$. Similarly, for $(10k - 4)$, $k = 10m + 2$ and $10m + 7$ fit, since $(-8k) = (-8) \cdot 2$ ends in $(-6)$, so here we have

$$
10k - 4 = 10(10m + 2) - 4 = 100m + 16
$$

and $100m + 66$.

Of the results obtained, $100m + 34$ and $100m + 66$ are not divisible by 8 for any integer $m$, so these bases do not lead to a result. The remaining $100m + 84$ and $100m + 16$ have the common form $100m \pm 16$.

Similarly, we find the last digits of $m$ that are possible:

$$
(100m \pm 16)^2 = 10^4 m^2 \pm 32 \cdot 10^2 m + 256 = 10^3 M + 456
$$

ignoring multiples of $10^3$:

$\pm 2 \cdot 10^2 m$ ends in 200 or -800,

the $(100m + 16)$ form fits $m = 1$ and 6, and the $(100m - 16)$ form fits $m = 4$ and 9, providing the following possibilities for the last three digits of the square:

$$
1000n + 116, \quad \text{or} \quad 616, \quad \text{or} \quad 384, \quad \text{or} \quad 884
$$

Of these, only $1000n \pm 384$ satisfies divisibility by 8.

Considering the four-digit ...3456 ending of the desired square number, we similarly find that the base of the square number can only be of the following forms:

$$
10^2 r + 2384, \quad +7384, \quad +2616, \quad +7616
$$

Testing the second form with $r = 0$ gives:

$$
7384^2 = \ldots 23456
$$

which provides a valid square number.

Essentially, we found that the form $10^4 r + 7384$ fits $r$ with a one-digit ending of 0 and 5, so for any integer $s$, $(50000s + 7384)^2$ is valid.

Similarly, the other three forms fit the base numbers $5 \cdot 10^4 s + 32384$, $5 \cdot 10^4 s + 42616$, and $5 \cdot 10^4 s + 17616$, and the middle one with $s = 0$ already provides six consecutive digits in ascending order from left to right:

$$
42616^2 = \ldots 6123456
$$