A nice number is a natural number such that its square and cube combined contain all digits exactly once, without needing to use any leading zeros. It is already known that 69 is the only nice number in base ten, and so far, no nice number has been found in any other base either.

Currently, the bases up to 48 have been completely searched for nice numbers, with none found, aside from 69 in base 10. You can see https://nicenumbers.net/ for the current status of the serach, as well as @Conflux's last post about it: https://beautifulthorns.wixsite.com/home/post/progress-update-on-the-search-for-nice-numbers

As you can see, it is estimated that there should be about one nice number by the time we get to base 118, but this is only an estimate.

This market resolves once the next nice number is found, provided that all bases smaller than the one it is nice in have been exhaustively checked (or otherwise proven not to have nice numbers). If it is proven that 69 is the only nice number, this will resolve to the last option. If it turns out that somehow the code for the seraches that have been done so far was wrong and there actually is another nice number in a base below 49, then I'll resolve to the "49 - 85" option.

## Related questions

Just saw these markets for the first time. I don't have much time to spend on this now, but I got nerd-sniped, and I might have some observations.

I quickly skimmed the "Progress Update on the Search for Nice Numbers" post, and some parts seem incredible sketchy (wrong?) to me. Are you claiming that for bases 3 mod 4 there are no nice numbers? I don't believe that. Are you sure the argument is not accidentally using the (wrong) claim that digit-sum(x^2)=(digit-sum(x))^2 mod (b-1)?

Is there any exponential improvement over brute force search for nice numbers? Or is the complexity to check all cases for base b still roughly O(b^(b/5))? I have some ideas to do this exponentially faster (probably still not fast enough to get to another nice number)

@FlorisvanDoorn oh ignore my first point. The argument seems valid. I should stop thinking about this and go to sleep.

@FlorisvanDoorn The proof in the blog post goes through some steps quickly, but it is definitely valid. To spell it out fully, if x is nice, then x^2+x^3 ≡ b(b-1)/2 (mod b-1). This is because any number is congruent to the sum of its digits in base b, mod b-1, and we know that the sum of the digits in x^2 and x^3 must be the (b-1)th triangular number, b(b-1)/2. If b ≡ 3 (mod 4), then b must be odd, so (b-1)/2 is an integer. This means that b(b-1)/2 ≡ (b-1)/2 (mod b-1), since b ≡ 1 (mod b-1). Then, writing b as 4k+3, we find that x^2+x^3 ≡ 2k+1 (mod b-1) for some k. But 2k+1 is odd, while b-1 is even, so this tells us that x^2+x^3 is odd, since it's residue modulo an even number is odd. On the other hand, we know that x^2+x^3 is always even for any integer x. This is a contradiction, so there are guaranteed to never be nice numbers for bases that are 3 mod 4.

@FlorisvanDoorn As for the first one, yeah, I'm pretty sure no one has found an exponential improvement over brute force.