An infinite square arithmetic progression?

Question

How to prove that there does not exist and infinite arithmetic sequence that all of it's terms are distinct squares of integers?

Answer 1

If $a, b \in \mathbb{N}$ then $$f(m)=a+mb$$ is an increasing function of $m$. Moreover, $f(m+1)-f(m) = b$.

Suppose that $f(m) = n_m^2$ for a sequence of terms $n_m \in \mathbb{N}$. Necessarily $n_m$ is an increasing sequence of integers. We know that the gap between two consecutive squares is given by $$(n+1)^2-n^2 = 2n+1.$$

For sufficiently large $N$, $2n+1 > b$ for all $n > N$. Now let $m$ be large enough so that $n_m > N$. Thus we have $$f(m+1)-f(m) = (n_{m+1})^2 - (n_m)^2 \ge (n_m+1)^2 - (n_m)^2 = 2n_m + 1 > b$$ and this is a contradiction.

Answer 2

Define the difference in the AP to be $d$. Then the interval between $d^2$ and $(d+1)^2$ is $2d+1>d$, so the AP cannot extend beyond that, and of course there are no squares less than zero, so an AP sequence of squares cannot be infinite.