Of course since the relation is symmetric but not reflexive it must fail to be transitive, but I am of the view that any triple (x,y,z) put forward as a counterexample to transitivity must have distinct values for x, y and z.

(There are a couple of misprints:

In the paragraph beginning “Our example can’t be too trivial”, “all three of x and y to be distinct” should probably read “all three of x and y and z to be distinct”.

In the update, there is a minus sign missing before 3^2 and an unwanted minus sign before 4^2.)

Is there a way to solve 1 without using the fact that .

Consider a square of side n(n+1)/2. If n is even, this is n/2 (n+1)s So you can lay n/2 (n+1)-sided squares across the top and the same along the right hand side and then fill in one more for the top right corner. You have now added n+1 (n+1)-sided squares.

If n is odd you can lay (n+1)/2 across the top (the last one sticks half-way out) and (n-1)/2 along the right-hand side (stopping short by half). So you have n (n+1) sided squares and and L-shape made out of 2 half-squares.

Here’s a picture of it

https://docs.google.com/spreadsheet/ccc?key=0AhbtqdywlrGadGFpaFQ4cGlveko3UUx0V3VINzhDYUE#gid=0

I’m curious how much credit this proof would get.

On the other hand, the first N odd numbers add up to N². Putting the two together gives the result, and with care it can be done without ever mentioning ½n(n+1).

For the details, see Heath. He also quotes al-Karkhī, who gives a presumably Greek-originated geometrical proof.

I believe there is an erroneous “^2″ at the end of the sentence beginning “The right-hand side is a difference of two squares…”

Many thanks — corrected now.