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Source. N. H. Anning and P. Erdős, Integral distances, Bull. Amer. Math. Soc. 51 (1945), 598--600; the Theorem on p. 598, unnumbered, the proof of its finite half on p. 598, the proof of its infinite half on pp. 599--600, and the remark on $n$-dimensional space on p. 600. The copy read is identified on the source card.

Read depth. Claims checked: the statement and the closing remark were read clause by clause on the page images. The proofs of both halves were read in full and followed in outline, not checked; the remark on $n$-dimensional space is stated in the paper without proof. Nothing here is independently reviewed.

Statement

Theorem (p. 598, unnumbered). "For any $n$ we can find $n$ points in the plane not all on a line such that their distances are all integral, but it is impossible to find infinitely many points with integral distances (not all on a line)."

In the corpus's words: for every $n$ there is a set of $n$ points in $\mathbb R^2$, not all collinear, all of whose pairwise distances are integers; and every infinite set of points in $\mathbb R^2$ all of whose pairwise distances are integers lies on a line.

The points of the finite half's proof (p. 598) all lie on one circle. The paper's last paragraph (p. 600) states, without proof, that "a similar argument" shows that infinitely many points in $n$-dimensional space, not all on a line, cannot have all their distances integral.

Proof pointer

Finite half (p. 598). On the circle $x^2+y^2=1/4$, for each prime $p_i\equiv1\pmod4$ write $p_i^2=a_i^2+b_i^2$ with $a_i,b_i\ne0$ and take the point $(x_i,y_i)$ of the circle at distance $b_i/p_i$ from $(-1/2,0)$. By induction on $j$, the distance from $(x_j,y_j)$ to each earlier $(x_i,y_i)$ is rational: the four concyclic points $(-1/2,0)$, $(1/2,0)$, $(x_i,y_i)$, $(x_j,y_j)$ have five rational distances, and Ptolemy's theorem makes the sixth rational. Enlarging the radius to clear denominators gives $n$ points with integral distances. A footnote (p. 598) records that Anning had given 24 points on a circle with integral distances (Amer. Math. Monthly 22 (1915), p. 321). The paper gives a second configuration on p. 599, the point $(m,0)$ with the points $(0,y_i)$ where $m^2=x_i^2-y_i^2$ for an odd $m^2$ with $d$ divisors.

Infinite half (pp. 599--600), in two steps. First, no line $L$ contains infinitely many of the points: for $P$ off $L$ and $Q_i,Q_j$ on $L$ far from $P$ and from each other, integrality gives $d(PQ_j)\le d(PQ_i)+d(Q_iQ_j)-1$, which a comparison with the foot $R$ of the perpendicular from $Q_i$ to $PQ_j$ rules out once the distances are large, since $d(Q_iR)$ is less than the distance of $P$ from $L$. Second, take a direction $P_1X$ with infinitely many of the points in every angular neighborhood of it and a point $P_2$ off the line $P_1X$. For a point $Q$ of the set far from $P_1$ at small angle $\epsilon$ to $P_1X$, the law of cosines with integer sides forces $d(P_2,Q)=d(P_1,Q)-d(P_1,P_2)\cos\alpha$, where $\alpha$ is the angle $XP_1P_2$, and hence $\epsilon<c_1/d(P_1,Q)$, so these points lie within a bounded distance $c_2$ of the line $P_1X$. Three of them, not on a line and far apart, then contradict the integrality inequality for the longest side of their triangle, as in the first step.

Dependencies

Within the paper: nothing beyond the two steps above. Outside it: the representation of the square of a prime $p\equiv1\pmod4$ as a sum of two nonzero squares, Ptolemy's theorem, the law of cosines and the triangle inequality.

Bears on

• Problem 213: the problem asks for $n$ points with no three on a line, no four on a circle and all distances integers. The points of the finite half's proof all lie on one circle (p. 598), and those of the second configuration all but one on a line (p. 599), so neither meets the problem's conditions for $n\ge4$; the infinite half shows that no infinite set has the three properties, since such a set is not contained in a line. The theorem settles no instance of the problem.
• Problem 130: in the problem's graph on an infinite set $A$ with no three points on a line and no four on a circle, a complete subgraph on infinitely many vertices would be an infinite set, not all on a line, with all distances integers, which the infinite half rules out. The theorem bounds neither the size of finite complete subgraphs nor the chromatic number.