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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.

Hungry Blue Dev
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4 Answers4

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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 $b^2$.

Barry Cipra
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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.

Nate
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    Can you actually show that it gives every remainder mod $p$? I haven't learned modular arithmetic yet. (Not as well I want to at least) – Hungry Blue Dev Feb 22 '17 at 18:49
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    If it doesn't hit all the residue classes mod $p$ then of the first $p$ terms two of them say $A_i$ and $A_j$ must be the same class mod $p$ by the pigeonhole principle. But that means $p$ divides $ (A_j-A_i)= d(j-i)$, but $p$ doesn't divide $d$ or $(j-i)$ as $i,j<p$ which contradicts the fact that $p$ is prime. – Nate Feb 22 '17 at 19:00
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    I don't like this proof, because it is more complicated than the obvious proof given by the OP. (Yes, the OP's proof is longer, but it can be made shorter: the difference between successive squares grows without bound, but the difference between successive elements of an AP is constant.) – TonyK Feb 22 '17 at 22:38
  • I'm a little surprised by the fact that (a) I've received a handful of down votes on this and (b) that you took the time to comment that you don't like this proof. I'll note that I led off by saying "here's another way" not "here is a better way", and that OP specifically asked for other ways to prove this result. – Nate Feb 23 '17 at 17:09
  • @Nate - do you understand why mathematicians like clean, elegant, beautiful proofs? Then you should understand why they also actively dislike proofs which are ugly, excessively complex, or unhelpful in understanding why the theorem is true. – jwg Feb 24 '17 at 08:17
  • @jwg - If my answer and OP's solution had both been posted as answers to the same question I'd probably vote for his over mine, my point was that OP specifically asked for other approaches so I think posting this approach was appropriate. I hardly agree that reducing things modulo a prime is ugly or excessively complex though. Also, you seem to think OP's proof gets at the "real" reason why this result is true while mine does not, to that I'd suggest you try to prove the same result for perfect squares in $\mathbb{Z}[\sqrt{2}]$. – Nate Feb 24 '17 at 15:48
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    That's a lot of bitching for a single downvote. – jwg Feb 24 '17 at 18:44
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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).

Erick Wong
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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.

Fluidized Pigeon Reactor
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