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What is the maximum number of perfect squares that can be in an arithmetic progression of positive integer terms of length $10$?

We can find five perfect squares in such a sequence: $$1,25,49,73,97,121,145,169,193,217.$$ I had a harder time finding $6$ perfect squares. Can we get an upper bound on the number of such perfect squares?

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  • $\begingroup$ Just for the sake of curiosity, can I see the sequence with 3 perfect squares? $\endgroup$ – RGS Nov 20 '16 at 22:30
  • $\begingroup$ @RSerrao One example would be $1,25,49,\ldots$. $\endgroup$ – user19405892 Nov 20 '16 at 22:32
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    $\begingroup$ @user19405892 That sequence has $5$, doesn't it? $\endgroup$ – Matthew Conroy Nov 21 '16 at 1:47
  • $\begingroup$ @MatthewConroy Which sequence? $\endgroup$ – user19405892 Nov 21 '16 at 2:21
  • $\begingroup$ @user19405892 1,25,49,73,97,121,145,169,193,217 has 5 squares. $\endgroup$ – Matthew Conroy Nov 21 '16 at 2:59
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You may as well assume that the first term is a square. Suppose that

$$M, M + aD, M + bD, M + cD, M + dD$$

are equal to $v^2$, $w^2$, $x^2$, $y^2$, and $z^2$. Then one obtains a solution to the equations

$$\frac{w^2 - v^2}{a} = \frac{x^2 - v^2}{b} = \frac{y^2 - v^2}{c} = \frac{z^2 - v^2}{d}.$$

We are asking whether this has any solutions in integers $w,v,x,y,z$. Thinking of $[w;v;x;y;z] \in \mathbf{P}^4$, the equations above cut out a curve $C$ which is the intersection of three quadrics (the pairwise differences of the first term and the second, third, and fourth term). Assuming $a,b,c,d$ are distinct, the curve $C$ is smooth of has genus $5$, and hence has only finitely many rational points by Faltings' theorem. Hence there will only be finitely many arithmetic progressions (up to scaling) with at least $5$ squares.

On the other hand, I claim there are infinitely many arithmetic progressions (up to scaling) for which $M$, $M+D$, $M+2D$, and $M+4D$ are all squares. In this case, we obtain the equations:

$$w^2 - v^2 = \frac{x^2 - v^2}{2} = \frac{y^2 - v^2}{4}.$$

This is the intersection $X$ of two quadrics in $\mathbf{P}^3$, which defines a curve of genus one. The curve corresponding to $M$, $M+D$, $M+2D$, and $M+3D$ all being squares turns out not to have rational points (Fermat) but this one is an elliptic curve of rank one.

In fact, $X$ is (clearly) isogenous to

$$Y^2 = (1+X)(1+2X)(1+4X),$$

which is an elliptic curve of conductor $192$. If $P = [0,1]$, then the odd multiples of $P$ give rise to such sequences, e.g.:

$$P: M = 1, \quad D = 0,$$ $$3P: M = 49, \quad D = 120,$$ $$5P: M = 17497489, \quad D = -4269720,$$ $$7P: M = 4085374755361, \quad D = 82153503191760,$$ etc.

Summary: There will be infinitely many arithmetic progressions (even of length $5$) with $4$ squares, however, for any fixed length (say $10$, or $100$, or $1000$) there will only be finitely many arithmetic progressions up to scaling with at least $5$ squares.

To find the finitely many exceptions with $5$ or more squares could be a little bit annoying. For example, one could find all the rational points on the $\binom{9}{4}$ genus $5$ curves coming from assuming the first term and four more are all squares. Some of these will be duplicates by symmetry. It might first make sense to consider the $\binom{9}{3}$ genus one curves coming from assuming the first term and three more are all squares, see which ones have positive rank (although $P = [0,1]$ is a point on $Y^2 = (1+aX)(1+bX)(1+cX)$ and may well have infinite order unless $[a;b;c] = [1;2;3]$). If Michael Stoll is still around, he could probably come up with a complete answer. As far as I know, there may not be any with $5$ squares, although clearly this is false if one replaces $10$ (say) by $25$, since then $1,2,3,4,\ldots,25$ has $5$ squares. (This example also had me confused as to why you had a hard time finding an arithmetic progression with three squares, since $1,2,3,4,5,6,7,8,9,10$ does the job.)

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The first upper bound (quite a naïve one) is 8.

Why?

Let me call a k-span to some $k $ elements of your arithmetic sequence that are consecutive and perfect squares.

Making use of the proof that there are no 4-spans, the biggest span we can have is 3. If we were to have as many 3-spans as possible, we could have 2 of them occupying positions $1-3$, $4-6$, and then we could have two more perfect squares over positions $9$ and $10$.

If our sequence has 9 perfect squares, it has at least a 5-span which is impossible.

So 8 perfect squares are on the table, provided they are separated on positions 4 and 8 or positions 3 and 7.

Can anyone get lower?

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  • $\begingroup$ 8 is impossible. Consider your 2 cases. If each $a+id$ is square for $i=1\dots 10$ except 4 & 8, the 5 odd $i$ give a 5-span. If each except $i=3$ & 7, the 5 even $i$ give a 5-span. $\endgroup$ – Rosie F Jun 1 '18 at 10:52

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