| > If a rectangle's sides are odd integers, then there is no point in the
> interior of the rectangle such that the distances from this point to the
> vertices of the rectangle are all integers.
I don't see the connection with 1844, but there is a note which treated
the special case where the rectangle is a unit square. The full solution of
the unit square problem is one of the top entries on my Most Wanted list.
However, to tackle the question here. Let's say that the rectangle
is A by B, that the coordinates of the interior point are (a,b), and that
the distance from the vertices is P,Q,R & S. Upper case letters denote
integers, and A & B are by hypothesis odd. Then:
a^2 + b^2 = P^2
a^2 + (B-b)^2 = Q^2
(A-a)^2 + b^2 = R^2
(A-a)^2 + (B-b)^2 = S^2
which when we get rid of a & b yields:
P^2 + S^2 = Q^2 + R^2
B^2*(A^2-R^2+P^2)^2 + A^2*(B^2-Q^2+P^2) = 4*P^2*A^2&B^2
The last equation is the key. The RHS is an even square, so the two
summands, since they are squares, must be even. A&B are odd, so EITHER
P is even and Q & R are odd OR P is odd and Q & R are even. But consider
the previous equation mod 8... The slimy tissue of lies falls apart, and
so our assumption, that A&B are both odd, is revealed as a canard.
Note this is true whatever a & b are, so they can be negative. Thus we are
not just restricted to *internal* points.
Love,
Andrew.
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