# An infinite square arithmetic progression? [duplicate]

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

• @mathmax Well, this one is a much easier argument than the bounded-length AP question which seems to be what that link concerns. (Hence the answers there could be said to not meet the needs of this question. However, the question is severely lacking in details so I voted to close.) – Erick Wong May 15 '15 at 16:31
• @Erick Wong what kind of details am I lacking exactly? – Arian Tashakkor May 15 '15 at 16:44
• Oh and by the way that mentioned question proves that there cannot exist such a progression with more than 3 terms.All I need is the proof that it cannot be infinite and also I don't understand a thing from the proofs given there. – Arian Tashakkor May 15 '15 at 16:46
• @ArianTashakkor Those are the details you are lacking. What is your background? What have you tried so far? What are you thoughts on the problem? Without those details it isn't clear that you wouldn't understand the answers in the other question. – Erick Wong May 16 '15 at 4:22

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.
• Sorry but I fail to understand from "For sufficiently large $N$ ... " can you please explain a bit more on that? – Arian Tashakkor May 15 '15 at 16:39
• @ArianTashakkor: Regardless of the size of $b$, there is $N\in\mathbb{N}$ with that property (actually there are infinitely many of them). If you want to choose a particular one of them, $N$ has to be "sufficiently large", i.e. greater than the least one, which has the given propert. – mathmax May 26 '15 at 11:37
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.