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Perfect squares in base 9?

Find all perfect squares which have a base 9 representation consisting solely of 1's. Give proof.

Note by Sharky Kesa
11 months, 3 weeks ago

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Let the perfect squares we're looking for have solely $$n$$ 1's in their base 9 representation. Then, our work amounts to solving the following diophantine equation:

$\sum_{k=0}^{n-1} 9^k = x^2\implies\frac{9^n-1}{9-1}=x^2\implies 9^n-1=8x^2\qquad\qquad (\ast)$

We note the trivial solutions $$n=0,1$$ corresponding to $$x=0,1$$ and show that there are no integer solutions for $$n,x$$ when $$n\gt 1$$.

When $$n\gt 1$$, WLOG, we can assume that $$x$$ is positive. Now,

\begin{align}9^n-1=8x^2&\implies 9^n-9=8x^2-8\\&\implies 8\times 9^{n-1}=8(x^2-1)\\&\implies 9^{n-1}=x^2-1\implies (x-1)(x+1)=3^{2n-2}\end{align}

Now, note that only one of $$x-1$$ and $$x+1$$ is divisible by $$3$$ since otherwise $$3$$ divides their difference $$(x+1)-(x-1)=2$$ which is obviously false. Hence, one of the two terms must equal to $$3^{2n-2}$$ while the latter term equals to $$1$$. So, we have either $$x=3^{2n-2}+1=0$$ or $$x=3^{2n-2}-1=2$$

The first case gives $$3^{2n-2}=-1$$ which obviously doesn't have any integer solutions for $$n$$ and the second case gives $$3^{2n-2}=3\implies 2n-2=1\implies n=3/2$$ which is not an integer.

Hence, when $$n\gt 1$$, there are no integer solutions to equation $$(\ast)$$.

Along with the trivial solutions noted at the beginning, we conclude that the only solutions are $$n=0,1$$ with the perfect squares being $$0,1$$ respectively. · 6 months, 4 weeks ago