There cannot be an infinite AP of perfect squares.

Question

I could not find any existing questions on this site stating this problem. Therefore I am posting my solution and I ask for other ways to prove this theorem too.

The Question

Prove that there cannot be an infinite integer arithmetic progression of distinct terms all of which are perfect squares.

My attempt

We shall prove it using contradiction. First off, there are a couple of things to notice which greatly simplify our discussion:

1. The AP cannot be decreasing as eventually, the terms will be negative and perfect squares are non-negative.
2. There has to be a non-zero, positive difference between the terms otherwise the terms would not be distinct.

Let us therefore, assume an AP with the first term $a$ - a non-negative integer and the positive difference $d$. The $i$th term of the AP is $T_i=a+(i-1)d$.

The AP is increasing, therefore there is a term $T_n$ for the least value of $n$ such that $T_n\geq d^2$. Now, $T_{n+1}$ is also a perfect square. Let $T_{n+1}=b^2$. Therefore, we have $$ d^2 \leq b^2 \implies d \leq b $$

Therefore we have $$ T_{n+1}=b^2+d<b^2+2b+1=(b+1)^2 $$ or $$b^2 < T_{n+1} < (b+1)^2$$

However, there are no perfect squares between two consecutive perfect squares. This contradicts our supposition that every term is a perfect square and an integer at the same time. Therefore, no such arithmetic progression exists.

Answer 1

You can prove a slightly stronger result: Any arithmetic progression with all terms distinct can have at most a finite number of consecutive terms both of which are squares.

Proof: If $d\not=0$ is the difference between consecutive terms and $a^2$ and $b^2$ are two consecutive terms that are both square, then $d=b^2-a^2=(b+a)(b-a)$. But any given integer $d$ has only finitely many factorizations, $d=rs$ (with $r$ and $s$ of the same parity). Setting $b+a=r$ and $b-a=s$ and solving for $a=(r-s)/2$ and $b=(r+s)/2$, we conclude there are only finitely many possibilities for $a^2$ and $b^2$.

Answer 2

You way looks fine to me, here's another way:

Suppose $d$ is the common difference in some arithmetic progression. Let $p$ be any prime other than $2$ that doesn't divide $d$. It's not hard to show that then the arithmetic progression hits every residue class mod $p$, but only $\frac{p+1}{2}$ of them are quadratic residues so they can't all be squares.

Answer 3

Yes, this is a very good way to prove it. It generalizes readily to higher powers and many other sequences. Other methods can give tighter bounds but may require more number theory, for instance:

• The number of terms in an arithmetic progression that are $\le n$ is $\Theta(n)$ with constants depending on the specific progression, but the number of squares up to $n$ is only $O(\sqrt{n})$.

• Pick $p$ to be the smallest odd prime that doesn't divide $d$ (this is at most $O(\log d)$ in size). Then one of the first $p$ terms will be a quadratic non-residue mod $p$.

• Using an infinite descent argument, it can be shown that there is no AP of length 4 in the set of perfect squares. This was first observed by Fermat and can be shown by more modern methods, but the folklore of the elementary proof seems to have a bit of a history (see http://www.mathpages.com/home/kmath044/kmath044.htm as well as https://arxiv.org/abs/0712.3850).

Answer 4

I believe it can also be proven by noting that the common difference between any consecutive squares is unbounded.

For any common difference $d$, there exist two consecutive squares whose difference is greater than $100d$. There must be at least one term in the arithmetic progression which falls between these two squares, and as such, is not a square.