Famous Riemann hypothesis state that
There are zeta solutions(s = x + i y) on critical strip.
Which functional equations are used to find zeta solutions it? Range (0<Re(s)<1)
and guessed that all solutions may exist on critical line x= 1/2 .and to his evidence millions of solutions also discovered.
@LalitTolani Personally, I think that the question should be let off the hook---there is context (the link to what Jyrki calls "the famous integral thread" (though "infamous" might be more appropriate)), and the thread relates to something which might be considered historically relevant to the site.
If this thread were to be deleted, THEN I would have no qualms at all. However, as that seems unlikely, I am not convinced that action needs to be taken with respect to the newer thread.
That said, I honestly don't really care, either way---I could make the opposite case, as well.
The following is a solution-verification question and, as such, is about a particular proof. I think it should be reopened as the proposed duplicate has nothing to do with the OP's proof.
I am trying to prove the following result.
Show that any finite group $G$ of even order contains an element $x \in G$ such that $x \neq e$ but $x^2 = e$.
I believe I understand the idea of the proof, but I've tried my best to use some formal notation to make it as rigorous as possible. How does...
@Shaun Not true, the OP's proof is essentially the same as the (standard elementary) proofs in the linked dupe, so not only should it remain closed, but it should also be deleted (as should most such SV dupes with nothing novel).
@ParamanandSingh So would you delete those two bogus answers given that we have all voted on it already? Each has gotten 5 downvotes excluding mine since I brought them up in this room, so it is definitely not unilateral. Thank you.
I've been recently reading Yuan Wang's paper on an approximation to Goldbach's problem, in which he showed that
Proposition 1: For all large even integer $x$, there exists $1<n<x-1$ such that $n(x-n)$ has no prime divisors $\le x^{1/8}$, at most three prime divisors $\le x^{1/2}$.
From this, he c...