Arithmetic progression with common difference 2061 If there are 30 consequent members of an arithmetic progression with CD of 2061, show that among them are at most 20 squares of natural numbers. I wrote out $a_1$ through $a_{30}$ and tried to find some patterns. Maybe it's checking modulos, then which ones? How is it done? Help would be appreciated.

Let $\\{a_k\\}_{k=1}^{30}$ be your arithmetic progression and suppose $a_k=y_k^2,\,\forall k.$
Consider that equation $$y_{k+2}^2-y_k^2=2\times 2061$$ $$(y_{k+2}-y_k)(y_{k+2}+y_k)=2\times 3^2\times 229.$$ If $y_{k+2}, y_k$ are integers, should be same parity. Otherwise $y_{k+2}^2-y_k^2$ become an odd integer.
If both are even $y_{k+2}^2-y_k^2$ is divisible by $4.$
If both are odd $y_{k+2}^2-y_k^2$ is divisible by $8.$
Hence $y_{k+2}^2-y_k^2=2\times 2061$ has no integer solutions.

Therefore, consider how that perfect squares can put in to your progression.

Her I give you a one possible progression containing $16$ perfect squares, $$y_1^2, y_2^2 ,.., ..,y_5^2, y_6^2 ,.., ..,y_9^2, y_{10}^2,.., ..,y_{13}^2, \\\y_{14}^2 ,.., ..,y_{17}^2,y_{18}^2,.., ..,y_{21}^2,y_{22}^2,.., ..,y_{25}^2,y_{26}^2,.., ..,y_{29}^2,y_{30}^2.$$

Consider all such possible progressions. Then you have the answer.