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.