Skip to main content
3 events
when toggle format what by license comment
2 hours ago comment added user1728960 A more direct way, compute table $(i^3+j^3)_{i,j}$ in $\mathbb{Z}/8$, count occurences of each number $N(k)$, then calculate $(\sum_{\text{even}\;k}N(k)^2)/(\sum_{\text{all}\;k}N(k)^2)$. But I still wonder why $8$ is chosen? Choosing other numbers gives different results (however, bias remains).
yesterday audit First answers
yesterday
2 days ago history answered French Man CC BY-SA 4.0