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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
1 year, 7 months 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.

Prasun Biswas - 1 year, 2 months ago

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Technically, 0 in base 9 is 0, which has a '0' digit but no '1' digit in base 9, so only 1 satisfies.

Sharky Kesa - 1 year, 2 months ago

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