Rational distance problem

Suppose we have a unit square ABCD. Is it possible to place a point P in the plane of ABCD, such that PA, PB, PC and PD are all rational? It’s not too difficult to show that such a point must necessarily have rational coordinates with respect to the natural choice of axes.

In the latest BMO2 paper, this question was posed with the additional restriction that P must lie on the circle inscribed in the square. This greatly simplifies the problem (not least because the original problem is apparently unsolved!), and it’s quite easy to show that no such configuration exists.

Similarly, if P is constrained to lie on one of the sides of the square, it becomes equivalent to showing that there are no non-trivial rational points on the elliptic curve y^2 = x^3 - 7x - 6. This is a considerably more difficult problem (it cannot be bashed by modular arithmetic) than the BMO2 counterpart, yet elliptic curves are understood sufficiently well that questions such as this one can usually be solved.

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0 Responses to Rational distance problem

  1. Caleb Swabel says:

    How do i prove there is no such combination?

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