ciceksiz kakarot
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Next privilege 250 Rep.
 Sep 15 comment A Geometric Proof of $\zeta(2)=\frac{\pi^2}6$? (and other integer inputs for the Zeta) I don't think it's dead... Oct 27 awarded Tumbleweed Oct 24 comment Boundedness on k-tuple euclidean space Thanks very much, now I understand it. And the latter part where you point out the fact that I should not single out a special point is mostly satisfying(that was another question on my mind.) @Greg Martin Oct 24 accepted Boundedness on k-tuple euclidean space Oct 24 asked Boundedness on k-tuple euclidean space Oct 13 accepted A function equal to its integration? Oct 13 comment A function equal to its integration? thank you very much Oct 13 comment A function equal to its integration? thank you very much Oct 13 comment A function equal to its integration? Yeah, thank you very much Oct 13 asked A function equal to its integration? Aug 7 accepted Proving that this relation is transitive Aug 7 awarded Curious Aug 6 comment Proving that this relation is transitive I could not get the last bit. Am I not looking for $Y\cup \{a,c \} \in F$ ? You wrote it in the wrong order, and I did not get the logic actually. Aug 6 revised Proving that this relation is transitive added 6 characters in body Aug 6 comment Proving that this relation is transitive I am sorry, I will change it now, thank you. @KyleGannon Aug 6 asked Proving that this relation is transitive Nov 18 comment In a integral domain every prime element is irreducible I don't understand the part, 1=xb hence b=1 (or b is a unit), how does it tie together, can you help? Sep 3 comment Parametric Equation solving over integers Thanks, I found this myself but it seems that I have not made any progress, but I will struggle at least I do not have anything to do until school. Sep 1 comment Parametric Equation solving over integers Yes, just a case I am stuck at, with 24m+1 and 24m+17. Sep 1 comment Parametric Equation solving over integers I am trying to construct three positive integers x,y,z that satisfies the equation $4xyz=(24m+1)(xy+yz+xz)$, and I want the solution for all m, so I thought induction would work, that's how I got to that equation.