Set of Integers with Sums of Two Distinct Elements Giving Squares

additive-combinatoricsnt.number-theorypythagorean triples

Are there arbitrarily large sets $\mathcal S=\{a_1,\ldots,a_n\}$ of strictly positive integers such that all sums $a_i+a_j$ of two distinct elements in $\mathcal S$ are squares?

Considering subsets in $\mathbb Z$ should essentially give the same answer since such a set can contain at most one negative integer.

An example of size $3$ is given by $\{6,19,30\}$.
(Allowing $0$, one gets $\{0,a^2,b^2\}$ in bijection with
Pythagorean triplets $c^2=a^2+b^2$.)

There is no such example with four integers in $\{1,\ldots,1000\}$.
(Accepting $0$, solutions are given by Euler bricks:
$\{0,44^2,117^2,240^2\}$ is the smallest example. I suspect thus that there are strictly positive solutions in $\mathbb N^4$).

An equivalent reformulation: Consider the infinite graph with vertices $1,2,3,\ldots$ and edges $\{i,j\}$ if $i+j$ is a square.
Does this graph contain arbitrarily large complete subgraphs?
(Trivial observation: Every edge $\{a,b\}$ is only contained in finitely many different complete subgraphs.)

Motivation This is somehow a variation on question Generalisation of this circular arrangement of numbers from $1$ to $32$ with two adjacent numbers being perfect squares

Best Answer

The size of such sets is bounded by some (unknown) constant, assuming a big conjecture in arithmetic geometry.

The Bombieri-Lang conjecture (non-trivially via the Uniformity Conjecture, see Stanley Yao Xiao's comment) implies that for any $f(x)\in \mathbb{Z}[x]$ of degree $5$, with no repeated roots, there are at most $B$ many rational numbers $m$ for which $f(m)$ is a square - here $B$ is some absolute constant (so the conjecture goes), completely independent of $f$.

This implies that the size of a set $A$ such that $a+a'$ is a square for any two distinct elements $a,a'\in A$ is at most $B+5$. Indeed, take any $5$ distinct elements $a_1,\ldots,a_5\in A$, and consider $f(x) = (x+a_1)\cdots(x+a_5)$. For any $m\in A\backslash \{a_1,\ldots,a_5\}$, we know that $f(m)$ is a square, and so $\lvert A\rvert-5\leq B$.

I learnt of this kind of argument (and the Uniformity Conjecture) via this paper of Cilleruelo and Granville, which has many similar arguments and applications: https://arxiv.org/pdf/math/0608109.pdf.

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[Math] Sequences of Squares with all square differences

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.

[Math] Does the set $\{\binom x3+\binom y3+\binom z3:\ x,y,z\in\mathbb Z\}$ contain all integers

As a supplement to @dario2994's result: I have obtained

2613: 27874,-17441,-25379
3337: 60083,-20882,-59229
3362: 20543,19711,-25367
4447: 105313,-35617,-103935

and any solution to $\binom{x}{3}+\binom{y}{3}+\binom{z}{3}=n$ where $n=1523,3603,4482$ must have $|x|,|y|,|z|>50000$.

The searching method is pretty simple: we have $$ \binom{x}{3}+\binom{y}{3}=\frac16(x+y-2)(x^2-x y+y^2-x-y), $$ therefore it is easy to find all solutions to $\binom{x}{3}+\binom{y}{3}=n-\binom{z}{3}$ for fixed $n$ and $z$ by a factorization of $6n-z(z-1)(z-2)$, and we only need to do a one-dimensional search with respect to $z$.

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[Math] Sequences of Squares with all square differences

[Math] Does the set $\{\binom x3+\binom y3+\binom z3:\ x,y,z\in\mathbb Z\}$ contain all integers

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