Also (n+1)^ = n^2 + 2n +1 i.e. you can have the previous square and just add 2n+...

periram on Dec 8, 2022 | parent | context | favorite | on: Finding divisors of a number with Python (2019)

Also (n+1)^ = n^2 + 2n +1 i.e. you can have the previous square and just add 2n+1 to it to get the next square (which is 3 additions in total). square += i + i + 1 if square < n; break