## January 2003

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It is impossible.

We can suppose that one of the points is at (0,0).

Let the other three points be (x1,y1), (x2,y2), (x3,y3).

(The coordinates need not be integers.)

Arrange these coordinates into a 3x2 matrix M,

whose ith row is the coordinates of the ith point:

| x1 y1 |

| x2 y2 | = M.

| x3 y3 |

Multiply 2M by the transpose of M, obtaining a 3x3 matrix P

whose (i,j) entry is given by 2*(xi*xj+yi*yj).

Since the rank of M is at most 2, the rank of P is also at most 2.

Each diagonal entry of P is twice the square of an odd integer,

so is equivalent to 2 modulo 16.

Each off-diagonal entry of P is

2*(xi*xj+yi*yj)

= (xi^{2}+yi^{2}) + (xj^{2}+yj^{2}) - ((xi-xj)^{2}+(yi-yj)^{2})

= +1 +1 -1 (mod 8)

= 1 mod 8.

So P is equivalent, modulo 8, to the matrix

| 2 1 1 |

| 1 2 1 | = P'.

| 1 1 2 |

We compute that the determinant of P' is 4, so that the determinant

of P is equivalent to 4 modulo 8.

In particular P is nonsingular, and has rank 3.

This contradicts the statement above that P has rank at most 2.

Several readers pointed out to us that this puzzle is Problem B5 of the 1993 Putnam exam.

References:

R. L. Graham, B. L. Rothschild, and E. G. Straus : Are there n+2 points

in E^{n} with pairwise odd integral distances? American Math Monthly

81(1974)

L. Piepmeyer : The maximum number of odd integral distances between

points in the plane. Discrete Computational Geometry. 1996.

Dr. Warren Smith at NEC labs has also extended the results to odd

squares instance.

*If you have any problems you think we might enjoy, please send them in. All replies should be sent to:* ponder@il.ibm.com