A student in the BeNeLux olympiad apparently proved that 56 is not a cube by observing that 56 = 4^3 - 2^3 and referring to Fermat's Last Theorem for the exponent 3.
I like the way to prove that 2^1/n is irrational for n>3 by using FLT. A proffessor could joke the students by saying this and completing with "the proof of FLT is left as homework"
It is well known that $50$ % of the primes are of the form $x^2 + y^2$.
Many variants exists where a rational amount of primes is of some integer polynomial form.
But I wonder ; are there integer symmetric polynomials $P$ such that $Q$ % of the primes are of the form $P$ where $Q$ is an irrati...
@skullpatrol I think the suspension was due to her unfriendliness... poor formatting is bad etiquette, but not suspension-worthy in and of itself. (excluding extreme cases like spamming.)
@mick I'm no algebraic number theorist. ;) this looks more like Balarka's flavour
Maybe it's just me, but I think optimization of Python code should be off-topic. Especially since this is a contest problem with >90% probability, as such problems tend to be. This is not the first time I see this one.
@Rafflesiaarnoldii Definitely something contest-like, SPOJ or so. But since the answer is not code optimisation but changing the algorithm, it might be considered on-topic (you need a bit of number theory to get an efficient algorithm).
I solved the case for the non-homogenous constant coefficients case and I wondered if there is a way to find a general solution to a non-constant coefficient case. I don't know how to approach this at all, the substitution $y(x)=\frac{\log (v(x))}{r(x)}$ gets problematic immediately.
Normal users get moderation privileges like close/delete/reopen votes and whatnot after a certain number of reputation. Diamond moderators are elected by the community and can do all kinds of things like suspend users, close/delete/reopen unilaterally
@AlexanderGruber I think it's incorrect to say "they belong to SE". SE has an irrevocable license to distribute the content, which is not the same as owning it.
@UserX Most likely, not at all. The bounty would be auto-awarded to the highest scoring qualifying answer. But to qualify, an answer must be posted after the bounty started and have score +2. If no answer qualifies there, the bounty simply disappears.