Are there infinitely many square numbers with increasing digits? [duplicate]

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.

• – lulu
Mar 6, 2022 at 15:27

Hint : Show that in base $$10$$, all the numbers $$37^2, \ 337^2, \ 3337^2, \ 33337^2, \ 333337^2, ...$$
• It also works with a $5$ at the end, look at math.stackexchange.com/questions/3580745/…