Perfect square root recurrence

Question

Spent some time trying to find some recurrence for determine bigger than current perfect square but unsuccessful. For example: current 121 and next 144. Who is next after 144? Can someone help me to take this recurrence?

Answer 1

Hint:

The well-known formula for $(n+1)^2$ can be written as $(n+1)^2=n^2+n+(n+1)$.

Answer 2

As $(n+1)^2 = n^2 + 2n + 1$, letting $f(n) = n^2$, we get -$$f(n+1) = f(n) + 2n+1$$