Infinite series squared question.
Is it possible to only pick integer coefficients \(a_n\) on a sum between 0 and 4 inclusive to get a solution to: \[\left(\sum_{n=0}^\infty \frac{a_n}{5^n} \right)^2 = 2\]
yes, just write \(\sqrt{2}\) down in base 5 ?
Haha yep.
I guess I didn't really think this through, cause in my mind I was going to say there were 2 solutions, and then present this other one using the \(a_n\) from the previous solution to make this one, although this one requires infinitely many digits to the right and I think is not entirely "finished" since you can carry... \[\sum_{n=1}^\infty 4*5^n+ \sum_{n=-\infty}^0 (5-a_n)5^n\] I think I'll just leave it and move on since I need to think about this more; this is like playing around with 5-adic numbers.
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