It is easy to construct a quadrilateral in which all sides, and all diagonals, have integer lengths. Just take a rectangle with integer sides a & b, such that a^2 + b^2 = c^2 (where c is also an integer).

It is not as obvious how to construct a quadrilateral with non-equal integer side and diagonal lengths. However, such a quadrilateral can be constructed, using similar right-angle triangles.

A quadrilateral ABCD (with centre O) is divided by it's diagonals (AC and BD) into four triangles (ABO, BCO, CDO and DAO). Let the diagonals intersect each other perpendicularly. Simply select any two pythagorean triplets, for instance (3,4,5) and (5,12,13). Now construct triangles DAO and BCO to be similar - each a 3-4-5 triangle, with DAO multiplied by 5 and BCO by 12. ABO and CDO are now similar right angle triangles too - each is a 5-12-13 triangle, with ABO multiplied by 3 and CDO by 4.

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We now have a quadrilateral with non-equal integer sides (39, 60, 52, 25) and diagonals (56 and 63). Infinite such polynomials exist.

My question: is it possible to construct polynomials of higher order, with the same property - that is, all sides and all diagonals (a connection between a vertex and any other non-adjacent vertex) are integer lengths? If so, how? If not, why not?

For instance, consider this attempt with a pentagon. The pentagon was created by taking a rectangle, AB, and slicing a right-angle triangle off one corner (the non-hypotenuse sides of this triangle, C and D, are shown in red). The pentagon's five diagonals are shown in blue.

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By choosing A and B to be part of a pythagorean triplet, and C and D to be part of another, we ensure that the pentagon's 5 sides have integer side lengths, and one integer diagonal (a^2 + b^2). We can ensure that two more diagonals have integer lengths, using the same trick as described with the quadrilateral above: Let A = 3*12, B = 4*12, C=4*5, D=3*5. Now A and D are also part of a pythagorean triplet (hence their deiagonal is an integer), as are C and B.

All that remains is to ensure that two diagonals are integers: that sqrt(A^2 + (B-D)^2) and sqrt(B^2+(A-C)^2) are integers. But I haven't yet been able to construct such a pentagon. Any ideas, or a proof of impossibility?


It is possible to get as many points in the plane as you wish, each at an integer distance from all the others, provided they all lie on a line or all lie on a circle. If you forbid three-on-a-line and four-on-a-circle, then it is known how to get seven points in the plane, each at an integer distance from the others, but it is not known whether one can get eight. See Points at integer distance


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