## Squares of squares, and the group of rational points on the circle

The purpose of this post is to describe a slightly different way of thinking about the existence – or otherwise – of a 3×3 magic square of squares. Of course it may not lead to any real progress, but it does at least suggest some natural weaker questions that may be more amenable. It follows on from part 1 and part 2.

John P Robertson gave, in his paper Magic squares of squares (Mathematics Magazine 69.4 (1996): 289-293), three different formulations of the problem:

To which we now add a fourth:

P4. Prove or disprove that there are two rational points $(x_1, y_1)$ and $(x_2, y_2)$ on the unit circle such that each is double another point, in the group of rational points on the unit circle, and such that the numbers $y_1 + y_2$ and $y_1 - y_2$ are also the y-coordinates of double points on the unit circle.

Let’s unpack that a bit. The group of rational points on the unit circle can be thought of as consisting of the complex numbers z = x+iy whose real and imaginary parts x and y are rational numbers with x²+y²=1. The group operation is complex multiplication.

Using the “line sweep” trick I described in part 2, the rational points on the unit circle can be parametrized by the rational numbers via the bijection:

$\displaystyle t\mapsto \frac{t^2-1}{t^2+1} + \frac{2t}{t^2+1}i$

(It is not coincidental that this is $\bigl(\frac{t+i}{|t+i|}\bigr)^2$, but that’s another story.)

Squaring this expression gives a parametrization of the double points on the unit circle:

$\displaystyle t\mapsto \frac{t^4-6t^2+1}{(t^2+1)^2} + \frac{4(t^3-t)}{(t^2+1)^2}i$

Now notice that the imaginary part of this expression is the common difference of an arithmetic progression of three squares:

$\displaystyle \frac{(t^2-1-2t)^2}{(t^2+1)^2}, 1, \frac{(t^2-1+2t)^2}{(t^2+1)^2}$

and that every arithmetic progression of three squares whose middle term is 1 is of this form. (This follows from the parametrization of arithmetic progressions of three squares that we saw in part 2.)

Recall from part 1 that a magic square can be rearranged into a 3×3 matrix where the rows are all arithmetic progressions with the same common difference, and the columns are all arithmetic progressions with the same common difference. If we divide all entries by the middle entry, the new middle entry is 1 and the magicality is retained. If the entries were all squares to start with, they still are.

But now we have the situation described in P4 above. The central row is a three-term arithmetic progression of squares whose middle term is 1: let’s call its common difference a. The central column is also a three-term arithmetic progression of squares whose middle term is 1: let’s call its common difference b. Then the two diagonals are also three-term arithmetic progression of squares whose middle term is 1, with common differences a+b and ab respectively.

## Weakenings

This naturally suggests several weaker questions, none of which I know the answer to. I suspect the first and last may be easier, since they don’t involve double points.

W1. Are there two rational points $(x_1, y_1)$ and $(x_2, y_2)$ on the unit circle such that the numbers $y_1 + y_2$ and $y_1 - y_2$ are also the y-coordinates of rational points on the unit circle, except where $y_1$ or $y_2$ is zero?

W2. Are there are three double points on the unit circle whose y-coordinates are in arithmetic progression, other than the trivial case where the middle point is on the x-axis?

W3. Are there two double points $(x_1, y_1)$ and $(x_2, y_2)$ on the unit circle such that $y_1 + y_2$ is also the y-coordinate of a double point on the unit circle?

W4. Are there three rational points on the unit circle whose y-coordinates are in arithmetic progression, where the common difference of this arithmetic progression is also the y-coordinate of a rational point on the unit circle, other than the trivial case where the middle point is on the x-axis?

Note that all of these would follow from the truth of P4. The arithmetic progression mentioned in W2 and W4 is $(y_1-y_2, y_1, y_1+y_2)$.

More encouragingly, if any of these questions has a negative answer then we can conclude that there is no 3×3 magic square of squares.

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