How to find perfect squares such that their base 9 representation is all 1s?
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 form$$\sum_{n=0}^k9^n$$however 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 $p(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+1$ such that$$p(x)^2=x^{2n}+x^{2n-1}+x^{2n-2}+\cdots+1$$although my question is: How do I solve this problem? I'm not sure what to do from here, so any help is appreciated.
1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| CrSb0001 | (no comment) | May 7, 2024 at 13:45 |
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 $$\sum_{i = 0}^k 9^i = \frac{9^k - 1}{9 - 1}.$$ (prove this via induction or using polynomials)
This means we are solving the equation $$(9^k - 1) = 8n^2,$$ for positive integers $n$ and $k$.
Notice that $9^k = (3^k)^2$, so we can factor the left hand side to get $$(3^k + 1)(3^k - 1) = 8n^2.$$
We compute $$\gcd(3^k + 1, 3^k - 1) = \gcd(2, 3^k - 1) = 2.$$ This means that one of the two factors is $2a^2$ and the other factor is $4b^2$, where $a$ is odd.
Looking mod $3$ we see that $3^k + 1 \neq 2a^2$, so we need to solve $$3^k - 1 = 2a^2,$$ $$3^k + 1 = 4b^2.$$
We can rearrange the second equation to get $$3^k = (2b + 1)(2b - 1).$$ 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 $2b - 1 = 1$. This corresponds to the solution $a = b = n = k = 1$ and this is therefore the only solution.
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