Let $a$ be the common difference and $an+b$ the sequence. If $ak+b=m^2$ for some $k,m$, then $\infty$ many squares are given by $(na+m)^2=n^2a^2+2nam+m^2=(k+an^2+2mn)\cdot a + b$. This immeadiatly generalizes to any powers.

Here is the general of this problem Problem 1 If an arithmetic sequence contain infinite p th power of an integer if and only if contains infinite $ p^{th}$ power of an integer . proof (It is an fundamental of Polynomial congruent) because $ x_n$ is an arithmetic sequence then $ x_{n + 1} = an + b$ Suppose exist an p th power of an integer in this sequence $ a_{n_0 + 1} = k^p$ Imply that $ k^p\equiv - b (\mod a)$ Consider the number $ (k + ta)^p$ $ t > 0$ Because $ (k + ta )^p\equiv k^p\equiv - b (\mod a)$ so $ (k + ta)^p$ is an term of the sequence . So there are many p th power in this sequence. Problem 2 If a arithmetic sequence contain an $ m,n$ power the it contains mn power . Prove by congruence . where $ \gcd(m,n)=1$