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Inspired by [will.octagon.gibson's answer](https://proposals.codidact.com/posts/296122/296143#answer-296143) and [Monica Cellio's answer](https://proposals.codidact.com/posts/296122/296144#answer-296144) (which are 2 very different approaches), I've found another. Spoilers for both of those answers are hidden so that you can try finding a solution yourself.
<details><summary>Spoiler from Monica Cellio's answer</summary>
The rational number 1.1 is a Droppable Square number.
</details>
<details><summary>Spoiler from will.octagon.gibson's answer</summary>
There are irrational Droppable Square numbers.
</details>
<details><summary>My new Droppable Square number (includes spoilers for both of the other answers)</summary>
I used will.octagon.gibson's approach to find a Droppable Square number between $1$ and $1.1$. I chose to drop the $2$ into the second decimal place (since dropping the $2$ into the first decimal place would just give the same $1.1$ solution again). So the solution needs to be of the form:
$$1.0...^2=1.02...$$
Representing the missing part of the left side as $X$, the missing part of the right side will be $\frac{X}{10}$ (since it is shifted $1$ decimal place to the right by the insertion of the $2$):
$$(1+X)^2=1.02+\frac{X}{10}$$
$$X^2+2X+1=1.02+\frac{X}{10}$$
Multiplying both sides by $50$ for the convenience of making the coefficients integer:
$$50X^2+100X+50=51+5X$$
$$50X^2+95X-1=0$$
~~Using the quadratic formula $X=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:~~
$$X=\frac{-95\pm\sqrt{95^2-4(50)(-1)}}{2(50)}$$
$$X=\frac{-95\pm\sqrt{9225}}{100}$$
I'm looking for a number greater than zero, so the $\pm$ can be changed to a $+$:
$$X=\frac{-95+\sqrt{9225}}{100}$$
$$X=\sqrt{\frac{9225}{10000}}-\frac{95}{100}$$
$$X=\sqrt{\frac{369}{400}}-\frac{19}{20}$$
So $1+X=\sqrt{\frac{369}{400}}+\frac{1}{20}$ is a Droppable Square number.
In decimal form this is $1.0104686356149273...$
Squaring this gives $1.0\color{red}{2}\color{black}{104686356149273}...$
</details>
<details><summary>Estimating the total number of Droppable Square numbers (a large number of spoilers)</summary>
Instead of $1.02...$ being chosen as the square of the number, $1.002...$ or $1.0002...$ could have been chosen. Would these also give new Droppable Square numbers? Is there a limit to how many extra zeroes could be inserted this way?
Since the square is greater than $1$, the number must also be greater than $1$. This means that the number is less than its square, leading to the conclusion that the same number of zeroes can be added to the right of the decimal point of the number as were added to the right of the decimal point of its square. This gives starting points such as the following:
$$1.00...^2=1.002...$$
$$1.000...^2=1.0002...$$
$$1.0000...^2=1.00002...$$
Since the infinite string of digits represented here by "$...$" will always be shifted $1$ decimal place to the right upon squaring (due to the insertion of the $2$), the same approach to solving can be used again, generalised to:
$$(1+X)^2=1.0...02+\frac{X}{10}$$
Here the "$...$" represents a string of all zeroes.
A similar process to before gives:
$$X^2+\frac{19X}{10}-0.0...02=0$$
Since the $a$ and the $c$ in the quadratic formula are positive and negative respectively, the $b^2-4ac$ will always be positive so there will always be a real square root. Since in this case $b^2-4ac>b^2$ it follows that $X$ will always have $1$ positive value.
Will this positive value always be small enough to not overwrite any of the zeroes?
$$X^2+\frac{19X}{10}-0.0...02=0$$
$$X^2+\frac{19X}{10}=0.0...02$$
$$X(X+\frac{19}{10})=0.0...02$$
$$X=\frac{0.0...02}{X+\frac{19}{10}}$$
Since $X$ is positive, $X+\frac{19}{10}>1$, so it follows that:
$$X<0.0...02$$
This means $X$ will never overwrite any of the zeroes, so there will be a Droppable Square number for every number of zeroes that can be inserted.
This is a family of Droppable Square numbers with infinitely many members, but it is not all of the Droppable Square numbers. The $2$ linked answers both show numbers outside this family. I suspect there are other families, perhaps other infinite families, so there is still plenty of room for more answers.
</details>

Inspired by [will.octagon.gibson's answer](https://proposals.codidact.com/posts/296122/296143#answer-296143) and [Monica Cellio's answer](https://proposals.codidact.com/posts/296122/296144#answer-296144) (which are 2 very different approaches), I've found another. Spoilers for both of those answers are hidden so that you can try finding a solution yourself.
<details><summary>Spoiler from Monica Cellio's answer</summary>
The rational number 1.1 is a Droppable Square number.
</details>
<details><summary>Spoiler from will.octagon.gibson's answer</summary>
There are irrational Droppable Square numbers.
</details>
<details><summary>My new Droppable Square number (includes spoilers for both of the other answers)</summary>
I used will.octagon.gibson's approach to find a Droppable Square number between $1$ and $1.1$. I chose to drop the $2$ into the second decimal place (since dropping the $2$ into the first decimal place would just give the same $1.1$ solution again). So the solution needs to be of the form:
$$1.0...^2=1.02...$$
Representing the missing part of the left side as $X$, the missing part of the right side will be $\frac{X}{10}$ (since it is shifted $1$ decimal place to the right by the insertion of the $2$):
$$(1+X)^2=1.02+\frac{X}{10}$$
$$X^2+2X+1=1.02+\frac{X}{10}$$
Multiplying both sides by $50$ for the convenience of making the coefficients integer:
$$50X^2+100X+50=51+5X$$
$$50X^2+95X-1=0$$
Using the quadratic formula $X= \large \frac{-b\pm\sqrt{b^2-4ac}}{2a}$:
$$X=\frac{-95\pm\sqrt{95^2-4(50)(-1)}}{2(50)}$$
$$X=\frac{-95\pm\sqrt{9225}}{100}$$
I'm looking for a number greater than zero, so the $\pm$ can be changed to a $+$:
$$X=\frac{-95+\sqrt{9225}}{100}$$
$$X=\sqrt{\frac{9225}{10000}}-\frac{95}{100}$$
$$X=\sqrt{\frac{369}{400}}-\frac{19}{20}$$
So $1+X=\sqrt{\frac{369}{400}}+\frac{1}{20}$ is a Droppable Square number.
In decimal form this is $1.0104686356149273...$
Squaring this gives $1.0\color{red}{2}\color{black}{104686356149273}...$
</details>
<details><summary>Estimating the total number of Droppable Square numbers (a large number of spoilers)</summary>
Instead of $1.02...$ being chosen as the square of the number, $1.002...$ or $1.0002...$ could have been chosen. Would these also give new Droppable Square numbers? Is there a limit to how many extra zeroes could be inserted this way?
Since the square is greater than $1$, the number must also be greater than $1$. This means that the number is less than its square, leading to the conclusion that the same number of zeroes can be added to the right of the decimal point of the number as were added to the right of the decimal point of its square. This gives starting points such as the following:
$$1.00...^2=1.002...$$
$$1.000...^2=1.0002...$$
$$1.0000...^2=1.00002...$$
Since the infinite string of digits represented here by "$...$" will always be shifted $1$ decimal place to the right upon squaring (due to the insertion of the $2$), the same approach to solving can be used again, generalised to:
$$(1+X)^2=1.0...02+\frac{X}{10}$$
Here the "$...$" represents a string of all zeroes.
A similar process to before gives:
$$X^2+\frac{19X}{10}-0.0...02=0$$
Since the $a$ and the $c$ in the quadratic formula are positive and negative respectively, the $b^2-4ac$ will always be positive so there will always be a real square root. Since in this case $b^2-4ac>b^2$ it follows that $X$ will always have $1$ positive value.
Will this positive value always be small enough to not overwrite any of the zeroes?
$$X^2+\frac{19X}{10}-0.0...02=0$$
$$X^2+\frac{19X}{10}=0.0...02$$
$$X(X+\frac{19}{10})=0.0...02$$
$$X=\frac{0.0...02}{X+\frac{19}{10}}$$
Since $X$ is positive, $X+\frac{19}{10}>1$, so it follows that:
$$X<0.0...02$$
This means $X$ will never overwrite any of the zeroes, so there will be a Droppable Square number for every number of zeroes that can be inserted.
This is a family of Droppable Square numbers with infinitely many members, but it is not all of the Droppable Square numbers. The $2$ linked answers both show numbers outside this family. I suspect there are other families, perhaps other infinite families, so there is still plenty of room for more answers.
</details>

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