|
3 | added 2 characters in body | ||
|
Q1. $p$ is $\star$-prime iff equation $xy+x+y=2p$ has no solution and $xy+x+y=2p-1$ has exactly one solution, i.e. $(x+1)(y+1)=2p+1$ has no solution (which is iff 2p+1 $2p+1$ is prime) and $(x+1)(y+1)=2p$ has only one solution $\{x,y\}=\{1,p-1\}$. This last holds iff $p$ is prime. Q2. Why partitions?! The number of $\star$-divisors of $n$ equals $\tau(2n-1)+\tau(2n)-4$.\tau(2n+1)+\tau(2n)-4$. |
||||
|
|
2 | fixed a typo | ||
|
Q1. $p$ is $\star$-prime iff equation $xy+x+y=2p$ has no solution and $xy+x+y=2p-1$ has exactly one solution, i.e. $(x+1)(y+1)=2p+1$ has no solution (which is iff 2p+1 is prime) and $(x+1)(y+1)=2p$ has only one solution $\{x,y\}=\{1,p\}$. \{x,y\}=\{1,p-1\}$. This last holds iff $p$ is prime. Q2. Why partitions?! The number of $\star$-divisors of $n$ equals $\tau(2n-1)+\tau(2n)-4$. |
||||
|
|
1 |
|
||
|
Q1. $p$ is $\star$-prime iff equation $xy+x+y=2p$ has no solution and $xy+x+y=2p-1$ has exactly one solution, i.e. $(x+1)(y+1)=2p+1$ has no solution (which is iff 2p+1 is prime) and $(x+1)(y+1)=2p$ has only one solution $\{x,y\}=\{1,p\}$. This last holds iff $p$ is prime. Q2. Why partitions?! The number of $\star$-divisors of $n$ equals $\tau(2n-1)+\tau(2n)-4$. |
||||
