Problem 6

Due in DSC 203 by 3 PM, Friday, October 05, 2001

When a,b,c are the three rational numbers $\frac{1}{2},\frac{1}{3}, \frac{1}{5}$, the value of

$$\begin{array}{rl} &\displaystyle \frac{1}{(a-b)^2}+\frac{1}{... ...36} =\frac{61^2}{6^2}=\left(\frac{61}{6}\right)^2, \end{array}$$

is the square of a rational number.

Prove this is no accident, that, in fact, whenever a,b,c are any three different rational numbers, the quantity

$$\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}$$

is always the square of a rational number.
Solutions, questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Thu Sep 27 19:29:37 CDT 2001