# There cannot be an infinite AP of perfect squares.

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.

• It looks perfect to me. – Onir Feb 22 '17 at 18:40
• The proof looks fine. You could, if you like, not bother with asserting the existence of a least $n$ such that $T_n\ge d^2$, and instead just note that $T_{d+1}=a+d^2\ge d^2$, which will trap $T_{d+2}$ between $b^2=T_{d+1}$ and $(b+1)^2$. – Barry Cipra Feb 22 '17 at 18:54
• I think it maybe it should say "Let $T_n=b^2$." – Jonathan Allan Feb 22 '17 at 23:17
• – Gerry Myerson Feb 23 '17 at 5:05
• @JonathanAllan No it's okay. – Hungry Blue Dev Feb 23 '17 at 5:10

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 $b^2$.

• Brilliant! This is a very clever apporach indeed. – Hungry Blue Dev Feb 22 '17 at 19:42

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.

• 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
• 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
• 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
• That's a lot of bitching for a single downvote. – jwg Feb 24 '17 at 18:44

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).

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.

• Isn't this the same proof as OP's but with some details abstracted away? – Erick Wong Feb 23 '17 at 9:24
• Yes, pretty much. I feel like it is way easier to understand though. – Fluidized Pigeon Reactor Feb 23 '17 at 21:04