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How to find perfect squares such that their base 9 representation is all 1s?

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I am working on solving all of the problems in from the Macalester College Problems of the Week that are available in this archive, and am currently working on MacPOW 1114:

Squares of 1

Find all perfect squares whose base 9 representation consists of only 1s.

Here is my current progress on this:

  • First of all, we can see that 1 is the only trivial solution - there of course have to be other non-trivial solutions.

So really our problem here is finding square numbers of the formn=0k9nhowever I'm not really sure about how I should go about finding square numbers of that form. I guess that if I replace 9 with x, then the problem really is equivalent to finding polynomials p(x)=xn+a1xn1+a2xn2++1 such thatp(x)2=x2n+x2n1+x2n2++1although my question is: How do I solve this problem? I'm not sure what to do from here, so any help is appreciated.

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First I plug in k=1,...,4 and only find k=1 as a solution. Hence I conjecture that there are not solutions apart from k=1

Before we prove this we should recall the geometric series formula i=0k9i=9k191. (prove this via induction or using polynomials)

This means we are solving the equation (9k1)=8n2, for positive integers n and k.

Notice that 9k=(3k)2, so we can factor the left hand side to get (3k+1)(3k1)=8n2.

We compute gcd(3k+1,3k1)=gcd(2,3k1)=2. This means that one of the two factors is 2a2 and the other factor is 4b2, where a is odd.

Looking mod 3 we see that 3k+12a2, so we need to solve 3k1=2a2, 3k+1=4b2.

We can rearrange the second equation to get 3k=(2b+1)(2b1). Since 3 is prime, we see that both factors are powers of 3. Since they differ by 2 we must have 2b+1=3 and 2b1=1. This corresponds to the solution a=b=n=k=1 and this is therefore the only solution.

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