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This is a question that came up while joking around with my friends, but now I am really intrigued by this question.

For sake of brevity, let's call square numbers with monotone increasing digits peculiar squares. Some examples of peculiar squares are $13^2 = 169$ (since $1 \leq 6 \leq 9$) and $15^2 = 225$. Question is, are there infintely many peculiar squares?

To tackle this question, I came up with a more generalized conjecture:

Peculiar Square Conjecture. For all $n \in \mathbb{N}$, there exists only finitely many peculiar squares in base $n$.

I first tried solving for $n=2$. This was pretty easy, since it is equivalent to proving that there are only finitely many squares of form $11\cdots 1_{(2)}$.

Then I tried solving for $n=3$. Simple number theory shows that $11\cdots 122 \cdots 2_{(3)}$ cannot be a square number. Thus, we only need to show that there are finitely many squares of form $11\cdots 1_{(3)}$. This was much easier said than done, and in the end I had to borrow the power of StackExchange. (Integer solutions of $3^n-1=2m^2$)

So up to this point, I know that the Peculiar Square Conjecture holds for $n = 2$ and $n = 3$, but I don't have clear idea of how to prove it for $n = 4$ or beyond. Any help or ideas would be much appreciated.

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Hint : Show that in base $10$, all the numbers $$37^2, \ 337^2, \ 3337^2, \ 33337^2, \ 333337^2, ...$$

are "peculiar squares".

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  • $\begingroup$ Wow... How do you come up with such intuition? $\endgroup$
    – Dimen
    Mar 6, 2022 at 15:31
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    $\begingroup$ It also works with a $5$ at the end, look at math.stackexchange.com/questions/3580745/… $\endgroup$
    – zwim
    Mar 6, 2022 at 15:34
  • $\begingroup$ @zwim Thank you, I will check it out! $\endgroup$
    – Dimen
    Mar 6, 2022 at 15:35
  • $\begingroup$ @Dimen Not sure where I saw it for the first time, but I "knew" this pattern. The link provided by zwim, however, explains why it works. $\endgroup$ Mar 6, 2022 at 15:50

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