Sequence of perfect squares

Question

Let $a,b\in \mathbb{N}$. Prove that, if $a$ is quadratic residue modulo $b$, then sequence $(a+kb)$, $k\in \mathbb{N}$, has infinite amount of perfect squares.

How should I approach this problem?

Answer 1

Hint: Since $a$ is a quadratic residue modulo $b$, there is some integer $c$ such that $c^2\equiv a\bmod b$, or equivalently, $c^2=a+kb$ for some $k$. What can you observe about $(c+b)^2$? Then repeat.

Answer 2

You can use the following lemma:

"If an arithmetic progression contains a square, then in contains infinitely many squares"

Proof: Let $k$ be the common difference, let $t^2$ be an element of the progression, then $$t^2+k(2t+k) = (t+k)^2$$ $$t^2+k(4t+4k) = (t+2k)^2$$ $$...$$ $$t^2+k(2nk+n^2k) = (t+nk)^2$$ are all elements of the progression.

Can you finish the proof now?