Welcome to the staging ground for new communities! Each proposal has a description in the "Descriptions" category and a body of questions and answers in "Incubator Q&A". You can ask questions (and get answers, we hope!) right away, and start new proposals.

Are you here to participate in a specific proposal? Click on the proposal tag (with the dark outline) to see only posts about that proposal and not all of the others that are in progress. Tags are at the bottom of each post.

Post History

66%

+2 −0

Incubator Q&A Droppable Square numbers

Key observation! Click here to reveal Note that $9^2 < 92 < 10^2.$ From this observation we can get a Droppable Square Number $(DSN)$ of the form $9+X$ where $0 < X < 1.$ Det...

posted 4mo ago by will.octagon.gibson‭ · edited 4mo ago by will.octagon.gibson‭

Answer

#3: Post edited by

will.octagon.gibson‭ · 2026-05-13T04:45:45Z (4 months ago)
Minor edit

Copy Link

Raw

Markdown

<details>
<summary>Key observation! Click here to reveal</summary>
Note that $9^2 < 92 < 10^2.$
~~From this observation we can get a ***Droppable Square Number*** ($DSN$) of the form $9+X$ where $0 < X < 1.$~~
</details>
<br>
<details>
<summary>Details! Click here to reveal</summary>
We want $(9+X)^2 \ = \ 92+X$
Solving we get:
$(9+X)^2 \ = \ 92+X$
$81+18X+X^2 \ = \ 92+X$
$X^2+17X-11 \ = \ 0$
Using the Quadratic formula, $X \ = \ \Large \frac{-17 \pm \sqrt{17^2+44}}{2}$ which has only one positive solution.
So $DSN \ = \ 9 + \Large \frac{-17 + \sqrt{333}}{2}$ is a ***Droppable Square Number*** if $DSN < 10.$
Evaluating numerically we get:
$DSN \ \ = \ \ \ 9.624143795 \cdots \ $ and
$DSN^2 \ = \ 92.624143795 \cdots$
So $9 + \Large \frac{-17 + \sqrt{333}}{2}$ $\ = \ 9.624143795 \cdots$ is indeed a ***Droppable Square Number***.
</details>

<details>
<summary>Key observation! Click here to reveal</summary>
Note that $9^2 < 92 < 10^2.$
From this observation we can get a ***Droppable Square Number*** $(DSN)$ of the form $9+X$ where $0 < X < 1.$
</details>
<br>
<details>
<summary>Details! Click here to reveal</summary>
We want $(9+X)^2 \ = \ 92+X$
Solving we get:
$(9+X)^2 \ = \ 92+X$
$81+18X+X^2 \ = \ 92+X$
$X^2+17X-11 \ = \ 0$
Using the Quadratic formula, $X \ = \ \Large \frac{-17 \pm \sqrt{17^2+44}}{2}$ which has only one positive solution.
So $DSN \ = \ 9 + \Large \frac{-17 + \sqrt{333}}{2}$ is a ***Droppable Square Number*** if $DSN < 10.$
Evaluating numerically we get:
$DSN \ \ = \ \ \ 9.624143795 \cdots \ $ and
$DSN^2 \ = \ 92.624143795 \cdots$
So $9 + \Large \frac{-17 + \sqrt{333}}{2}$ $\ = \ 9.624143795 \cdots$ is indeed a ***Droppable Square Number***.
</details>

#2: Post edited by

will.octagon.gibson‭ · 2026-05-13T04:35:18Z (4 months ago)
Minor font change

Copy Link

Raw

Markdown

<details>
<summary>Key observation! Click here to reveal</summary>
Note that $9^2 < 92 < 10^2.$
~~From this observation we can get a Droppable Square Number ($DSN$) of the form $9+X$ where $0 < X < 1.$~~
</details>
<br>
<details>
<summary>Details! Click here to reveal</summary>
We want $(9+X)^2 \ = \ 92+X$
Solving we get:
$(9+X)^2 \ = \ 92+X$
$81+18X+X^2 \ = \ 92+X$
$X^2+17X-11 \ = \ 0$
Using the Quadratic formula, $X \ = \ \Large \frac{-17 \pm \sqrt{17^2+44}}{2}$ which has only one positive solution.
~~So $DSN \ = \ 9 + \Large \frac{-17 + \sqrt{333}}{2}$ is a Droppable Square Number if $DSN < 10.$~~
Evaluating numerically we get:
$DSN \ \ = \ \ \ 9.624143795 \cdots \ $ and
$DSN^2 \ = \ 92.624143795 \cdots$
~~So $9 + \Large \frac{-17 + \sqrt{333}}{2}$ $\ = \ 9.624143795 \cdots$ is indeed a Droppable Square Number.~~
</details>

<details>
<summary>Key observation! Click here to reveal</summary>
Note that $9^2 < 92 < 10^2.$
From this observation we can get a ***Droppable Square Number*** ($DSN$) of the form $9+X$ where $0 < X < 1.$
</details>
<br>
<details>
<summary>Details! Click here to reveal</summary>
We want $(9+X)^2 \ = \ 92+X$
Solving we get:
$(9+X)^2 \ = \ 92+X$
$81+18X+X^2 \ = \ 92+X$
$X^2+17X-11 \ = \ 0$
Using the Quadratic formula, $X \ = \ \Large \frac{-17 \pm \sqrt{17^2+44}}{2}$ which has only one positive solution.
So $DSN \ = \ 9 + \Large \frac{-17 + \sqrt{333}}{2}$ is a ***Droppable Square Number*** if $DSN < 10.$
Evaluating numerically we get:
$DSN \ \ = \ \ \ 9.624143795 \cdots \ $ and
$DSN^2 \ = \ 92.624143795 \cdots$
So $9 + \Large \frac{-17 + \sqrt{333}}{2}$ $\ = \ 9.624143795 \cdots$ is indeed a ***Droppable Square Number***.
</details>

#1: Initial revision by

will.octagon.gibson‭ · 2026-05-13T04:26:47Z (4 months ago)

Copy Link

Raw

Markdown

<details>
<summary>Key observation! Click here to reveal</summary>

Note that $9^2 < 92 < 10^2.$

From this observation we can get a Droppable Square Number ($DSN$) of the form $9+X$ where $0 < X < 1.$

</details>

<br>

<details>
<summary>Details! Click here to reveal</summary>

We want $(9+X)^2 \ = \ 92+X$

Solving we get:

$(9+X)^2 \ = \ 92+X$

$81+18X+X^2 \ = \ 92+X$

$X^2+17X-11 \ = \ 0$

Using the Quadratic formula, $X \ = \ \Large \frac{-17 \pm \sqrt{17^2+44}}{2}$ which has only one positive solution.

So $DSN \ = \ 9 + \Large \frac{-17 + \sqrt{333}}{2}$ is a Droppable Square Number if $DSN < 10.$

Evaluating numerically we get:

$DSN \ \ = \ \ \ 9.624143795 \cdots \ $ and

$DSN^2 \ = \ 92.624143795 \cdots$

So $9 + \Large \frac{-17 + \sqrt{333}}{2}$ $\ = \ 9.624143795 \cdots$ is indeed a Droppable Square Number.

</details>

Communities

Post History