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Background

The following question was first asked by Alex Rice, who was thinking about small subsets $A\subset [1,\ldots , N]$ with lots of square differences. Certainly for any set $A$ the maximum number of square differences is going to be $\binom{|A|}{2}$. From the point of view of someone working in additive combinatorics, an infinite set of positive integers can't get much less substantial than the squares, and so it's natural to wonder if there are arbitrarily large sets $A$ inside the squares, all of whose differences are squares [edit: I apparently misunderstood the original motivation, see Alex's answer/comment below]. This question was asked of a few others, including Adrian Brunyate, Jacob Hicks and Nathan Walters before it was asked of me by Adrian in this form:

Definition: We say that a sequence $(a_1, \ldots, a_n) \in \mathbf{Z}^n_{\ge 1}$ is a Super-$n$ if for all $1 \le i \le n$, $a_i$ is an integer square and for all $1 \le i < j \le n$, $a_j - a_i > 0$ is also an integer square.

Clearly a Super-2 defines a Pythagorean triple.

Perhaps less clearly, a Super-3 defines an Euler Brick, and is strongly related the the question of whether there is a perfect rational cuboid.

Question 1: For which positive integers $n$ does there exist a Super-$n$ ?

If the answer is yes to the above question, we may also ask the following:

Question 2: For which positive integers $n$ do there exist infinitely many Super-$n$'s?

One may note that the following problems are related to some problems already asked on MO about rational polytopes and sequences of squares

How many sequences of rational squares are there, all of whose differences are also rational squares?

Totally rational polytopes

What seems to be known already

It has been known for millenia that there are infinitely many Pythagorean Triples.

Euler discovered in 1772 that there are infinitely many Super-$3$'s, and in fact he gave a parametrized family of them.

None of us have been able to find a Super 4 (although I haven't been searching myself).

The connection to algebraic geometry

Definition: The Super-$n$-variety is the intersection of the following $\binom{n}{2}$ quadratic polynomials in projective space over $\mathbf{Q}$. $$d_1^2 = c_2^2 - c_1^2$$ $$\vdots$$ $$d_{\binom{n}{2}}^2 = c_{n}^2 - c_{n-1}^2$$

Clearly the Super-2 variety is a copy of $\mathbb{P}^1_{\mathbf{Q}}$.

In Section 8 of the link given above for Euler's family of "Euler Bricks" we see that the Super-3 variety is birational to a singular K3 surface of Mordell-Weil rank 2. In this setting, one could say that Euler found a rational curve on this variety. It is also noted in the article that Narumiya and Shiga found a different rational curve on this variety.

Question 2': Could there be rational curves on the Super-$n$ variety for all $n$?

But perhaps (probably) this is way too much to ask. More generally, I'd like to know:

Question 3: Is there any interesting geometry to the Super-$n$ variety for $n\ge 4$?

In general this seems like an interesting problem, and one that people may have studied before, but perhaps in some guise that I'm not familiar with, so any input is appreciated.

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Observation: The posiitive differences of a Super-3 also lead to a Pythagorean triple. Thus so does any 3-set of a Super-n. Whereas the constructions in question 72040 (about rational square sequences, thanks for the plug) will create some Pythagorean triples from differences of certain subtriples (and eventually terminate), there may be a way to "invert" the process to generate infinitely many triples from a Super-n. Put another way, maybe the known constructions can be reorganized to provide answers to your and my questions. Gerhard "Ask Me About System Design" Paseman, 2011.08.21 –  Gerhard Paseman Aug 21 '11 at 19:23
    
There's basically no way I could not plug your question in asking this question! I, too feel like perhaps there's some way to exploit the constructions given as answers to your question, but I'm not sure how far that will go. –  stankewicz Aug 21 '11 at 21:14
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Correction/clarification re "K3 surface of Mordell-Weil rank 2": in general the M-W rank depends not only on the surface but also on the choice of elliptic fibration, and typically K3 surfaces have more than one such fibration (but still finitely many up to isomorphism). So more properly what that paper finds is a choice of fibration for which the M-W rank is 2. Once the rank is positive there are infinitely many rational curves on the surface, one for each element of the Mordell-Weil group of the fibration (and usually also other rational curves not of this form). –  Noam D. Elkies Aug 22 '11 at 5:01
    
I am tempted to ask a new question about non trivial rational points on the surfaces in my answer related to perfect cuboids. Is this question too elementary/wildly known and will be closed ? (Haven't seen such simple models) –  joro Aug 22 '11 at 12:34
    
Is the relaxation of allowing one difference to not be a square clear? Suppose such relaxed super-4 exist. –  joro Aug 23 '11 at 13:10

3 Answers 3

up vote 10 down vote accepted

The "Super-$n$" variety, call it $V_n$, seems to be of general type once $n \geq 4$. It probably has no nontrivial rational curves (where "trivial" means that it lies on a hyperplane $c_1=0$ or $c_j=c_k$ some distinct $j,k$; over ${\bf C}$ one must also exclude $c_j=0$ for $j>1$). For $n$ large enough this should follow from the Bombieri-Lang conjectures for the "Super-4" variety $V_4$.

In general if a smooth variety is the "complete intersection" in some projective space ${\bf P}^{N-1}$ of hypersurfaces $P_i=0$ of degrees $d_1,\ldots,d_r$ then it is of general type iff $\sum_i d_i > N$. Here we have $N=(n^2+n)/2$ and $r=(n^2-n)/2$, with each $d_i=2$, so we would get general type once $n\geq 4$; our variety is not quite smooth but the singularities look mild enough not to change the result.

A (very plausible but extremely hard) conjecture of Bombieri and Lang asserts that all the rational points on a variety $X$ of general type lie on a finite union $X_0$ of subvarieties of lower dimension. (This would vastly generalize Faltings' theorems on curves of genus $>1$ [Mordell's conjecture] and subvarieties of abelian varieties.) For complete intersections in ${\bf P}^{N-1}$, the following naïve but suggestive heuristic points in the same way: try the $\sim H^n$ points $(x_1:x_2:...:x_N)$ with integers $x_m$ such that $H \leq \max_m |x_m| < 2H$; for any choice of such $x_m$ we have $|P_i(\vec x)| \ll H^{d_i}$ for each $i$, and if we imagine these $r$ numbers $P_i(\vec x)$ are more-or-less randomly and independently distributed among integers of those sizes then the expected number of $\vec x$ where they're all zero is of order $H^{N-\sum_i d_i}$. This means that the general-type case is precisely when the exponent is negative, and thus that (summing over $H=1,2,4,8,16,\ldots$) the total number of rational points if finite. This heuristic cannot account for non-random rational points due to polynomial identities, but those are precisely the subvarieties that the Bombieri-Lang conjecture allows.

It seems reasonable to guess that already for $n=4$ there are no nontrivial rational curves, and that the nontrivial part of $X_0$ is finite or even empty. Unfortunately, even assuming the B-L conjecture there is no known way to determine $X_0$. Nevertheless it may be possible to deduce that some $V_n$ has no nontrivial points under the assumption that B-L holds for $V_4$. The reason is that for $n > 4$ there are many maps from $V_n$ to $V_4$, obtained by choosing any $4$ of the $n$ variables $c_1,\ldots,c_n$ in order, and the corresponding six $d$'s. If all nontrivial points of $V_4$ were known to lie on a union $X_0$ of proper subvarieties, then any nontrivial point of $V_n$ would have to lie on the intersection of preimages of $X_0$ under $n \choose 4$ different maps, and whatever $X_0$ turns out to be, such an intersection ought to be trivial if $n$ is large enough.

NB it would likely require some nontrivial [sic] work to make a proof of this even assuming the B-L conjecture for $V_4$, but such an analysis was carried out in a similar context in the famous paper

L.Caporaso, J.Harris, and B.Mazur: Uniformity of rational points, J. Amer. Math. Soc. 10 #1 (1997), 1-45

and something like that should be possible here.

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What a wonderful answer. I don't think I could have hoped for much more! –  stankewicz Aug 22 '11 at 5:40
    
Well, I'm still holding out for a proof/counterexample of no Super-4. I am thinking of searching for distinct positive integers a,b,c such that each of ab, ac, and bc is 4 times a triangular number. Is this more specific system known to be insoluble? Gerhard "Ask Me About System Design" Paseman, 2011.08.21 –  Gerhard Paseman Aug 22 '11 at 6:37
    
(I'll actually need more relations than those above to hold between a,b, and c to build a Super-4, but let us see if the basic 3 relations holld first.) Gerhard "Ask Me About System Design" Paseman, 2011.08.21 –  Gerhard Paseman Aug 22 '11 at 6:44
    
@stankiewicz: Thanks! (Still note that some steps need to be filled in to actually prove general type.)$$ $$ @G.Paseman: You don't have to search very far... $(2,3,5)$ is the first example. But yes, that's a different problem, and not only because you dropped some squareness conditions but also because you've added a condition of integrality. –  Noam D. Elkies Aug 22 '11 at 12:10
    
@Paseman: I agree that a proof/counterexample of the existence of a Super-4 is perhaps really what we as mathematicians should hold out for, but that for me (grad student going on the market) and for now, "Possibly nontrivial algebro-geometric work + Bombieri-Lang for $V_4$ implies only finitely many $n$ for which there exists a Super-$n$" is a pretty good answer, particularly when coupled with what's basically a wikipedia-ready tutorial on applying Bombieri-Lang. Moreover, I think that once some of the details are filled in, it will be easier to profitably re-examine this question. –  stankewicz Aug 22 '11 at 14:12

To clarify, this question arose when my adviser Neil Lyall and I were attempting to provide simple upper bounds on the size of the LARGEST subset of $ [1,2,...,N] $ with NO square differences, for the purposes of an introduction to a talk about generalizations of the Sarkozy-Fursternburg Theorem, which states that this quantity is $o(N)$.

In particular, if $S$ is the set of squares, $A \subset [1,2,...,N] $ with $(A-A) \cap S = \emptyset $, and $a_1 < a_2< ... < a_k$ is any collection of non-negative integers such that $a_i-a_j \in S$ whenever $i > j$, then the sets $A+a_1$, $A+a_2,...A+a_k$ are all pairwise disjoint. Therefore, $|A| \leq (N+a_k)/k$. If such a collection was possible for every $k$, then this would immediately provide a remarkably (perhaps disturbingly) elementary proof of the Sarkozy-Fursterberg Theorem that would not require any of the harmonic analysis or ergodic theory tools utilized in other proofs.

Notice that the statement that such a collection of non-negative integers is possible for every $k$ is equivalent to the statement that such a collection of positive squares is possible for every $k$, as you can just translate the smallest element to $0$ and get a set of positive squares with one fewer element satisfying the desired property.

Given that nobody has found a set of more than 3 positive squares (and hence no set of more than 4 non-negative integers) with this property, the above method can currently only show that the size of the largest subset of $[1,2,...,N]$ with no square differences is (asymptotically) less than $N/4$.

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I realize that this probably should've been a comment as opposed to an "answer". I'm a bit new to this... –  Alex Rice Aug 21 '11 at 21:16
    
Comments are limited to 600 characters here, which would not have sufficed to accommodate your contribution. –  Noam D. Elkies Aug 23 '11 at 22:53

This is more of a comment.

If I understand correctly, a perfect cuboid will give a Super-4 according to On Perfect Cuboids

Are there four squares all pairs of which have square differences? For a perfect cuboid we could take the squares of $y_3 z$, $y_2 y_3$ , $x_1 z$ and $x_1 y_3$.

While wasting my time with perfect cuboids, I found 2 surfaces on which they are nontrivial rational points. The first might be for all perfect cuboids, the second is not for all. The surfaces are:

$$ x^{4} y^{2} z^{4} + x^{2} y^{4} z^{4} - 2 x^{4} y^{2} z^{2} - 2 x^{2} y^{4} z^{2} - 4 x^{2} y^{2} z^{4} + x^{4} y^{2} + x^{2} y^{4} - 8 x^{2} y^{2} z^{2} + x^{2} z^{4} + y^{2} z^{4} - 4 x^{2} y^{2} - 2 x^{2} z^{2} - 2 y^{2} z^{2} + x^{2} + y^{2} = 0$$

and

$$ x^{4} y^{3} z^{3} - x^{4} y^{2} z^{4} + x^{2} y^{4} z^{4} + 2 x^{4} y^{3} z + 2 x^{4} y^{2} z^{2} - 2 x^{2} y^{4} z^{2} - 2 x^{4} y z^{3} + 4 x^{2} y^{3} z^{3} - x^{4} y^{2} + x^{2} y^{4} - 2 x^{4} y z + 4 x^{2} y^{3} z - 4 x^{2} y z^{3} + 2 y^{3} z^{3} + x^{2} z^{4} - y^{2} z^{4} - 4 x^{2} y z + 2 y^{3} z - 2 x^{2} z^{2} + 2 y^{2} z^{2} - 2 y z^{3} + x^{2} - y^{2} - 2 y z =0$$

In machine readable form:

x^4*y^2*z^4 + x^2*y^4*z^4 - 2*x^4*y^2*z^2 - 2*x^2*y^4*z^2 - 4*x^2*y^2*z^4 + x^4*y^2 + x^2*y^4 - 8*x^2*y^2*z^2 + x^2*z^4 + y^2*z^4 - 4*x^2*y^2 - 2*x^2*z^2 - 2*y^2*z^2 + x^2 + y^2 = 0

and

2*x^4*y^3*z^3 - x^4*y^2*z^4 + x^2*y^4*z^4 + 2*x^4*y^3*z + 2*x^4*y^2*z^2 - 2*x^2*y^4*z^2 - 2*x^4*y*z^3 + 4*x^2*y^3*z^3 - x^4*y^2 + x^2*y^4 - 2*x^4*y*z + 4*x^2*y^3*z - 4*x^2*y*z^3 + 2*y^3*z^3 + x^2*z^4 - y^2*z^4 - 4*x^2*y*z + 2*y^3*z - 2*x^2*z^2 + 2*y^2*z^2 - 2*y*z^3 + x^2 - y^2 - 2*y*z = 0
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Actually as I read it (page 7, but page 13 of the pdf) it seems that Leech claims that if there were a perfect cuboid there would be a Super-4 –  stankewicz Aug 22 '11 at 12:05
    
That said, your contribution is much appreciated. van Luijk's document is a nice find. –  stankewicz Aug 22 '11 at 12:16
    
@stankewicz you are right, my mistake, fixed it. Thanks. –  joro Aug 23 '11 at 13:14

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