Perfect squares in base 9?

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

#NumberTheory

Note by Sharky Kesa
5 years, 2 months ago

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Comments

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

∑k=0n−19k=x2  ⟹  9n−19−1=x2  ⟹  9n−1=8x2(∗)\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=0,1n=0,1 corresponding to x=0,1x=0,1 and show that there are no integer solutions for n,xn,x when n>1n\gt 1.

When n>1n\gt 1, WLOG, we can assume that xx is positive. Now,

9n−1=8x2  ⟹  9n−9=8x2−8  ⟹  8×9n−1=8(x2−1)  ⟹  9n−1=x2−1  ⟹  (x−1)(x+1)=32n−2\begin{aligned}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{aligned}

Now, note that only one of x−1x-1 and x+1x+1 is divisible by 33 since otherwise 33 divides their difference (x+1)−(x−1)=2(x+1)-(x-1)=2 which is obviously false. Hence, one of the two terms must equal to 32n−23^{2n-2} while the latter term equals to 11. So, we have either x=32n−2+1=0x=3^{2n-2}+1=0 or x=32n−2−1=2x=3^{2n-2}-1=2

The first case gives 32n−2=−13^{2n-2}=-1 which obviously doesn't have any integer solutions for nn and the second case gives 32n−2=3  ⟹  2n−2=1  ⟹  n=3/23^{2n-2}=3\implies 2n-2=1\implies n=3/2 which is not an integer.

Hence, when n>1n\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=0,1n=0,1 with the perfect squares being 0,10,1 respectively.

Prasun Biswas - 4 years, 9 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 - 4 years, 9 months ago
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