Problem 836

[neverforget]

Problem 836 (View Problem)

I'd add a small note at the bottom saying that for this problem, $57$ is considered to be a prime number.

[hk]

Does that help with finding the correct answer??? Or in spreading more mist??

[neverforget]

perhaps I made a mistake in my calculations, but if you take Axiom of Choice then the premise ($f$ being well-defined for $p=57$ and odd $m$) seems to require $57$ to be prime. And without Axiom of Choice then $f$ being well-defined is equivalent to the Riemann Hypothesis (which AFAIK has not yet been proven)

Also, I had to consider $0$ to be a number, but that's probably much less controversial.

Another note: it's probably obvious that $f$ does not behave well when $p=2$, but may as well mention that $p\neq 2$, in case someone gets confused.

[cat_good]

read the entire problem!!!!!!!!!!!!!!