**Method for finding 3x3 nearly-magic squares
of squares with close sums**

I looked up a suitable formula for three
Pythagorean Triangles having equal areas. I found one in Dickson's **History
of the Theory of Numbers**, Chapter IV, Rational Right Triangles, near the
end of the section, "Right Triangles of Equal Area". It is **Hillyer's
formula**.

The generators of the three triangles are

p = K^2 + KL + L^2, q = K^2 - L^2

r = K^2 + KL + L^2,
s = 2KL + L^2

v = K^2 + KL + L^2, u = 2KL + K^2

Here is how to use Hillyer's formula to form a 3x3 nearly-magic square of squares.

A^2 B^2 C^2

D^2 E^2
F^2

G^2 H^2 I^2

A,E,I will
be the magic diagonal.

C,E,G
will be the non-magic diagonal.

A = u^2 + v^2

B = |p^2 -
q^2 - 2pq|

C = r^2 - s^2 + 2rs

D = p^2 - q^2 + 2pq

E = r^2 +
s^2

F = |u^2 - v^2 - 2uv|

G = |r^2 - s^2 - 2rs|

H = u^2 - v^2
+ 2uv

I = p^2 + q^2

For example, suppose L = 7, K = 10.

p = r = v = 100 + 70 + 49 =
219

q = K^2 - L^2 = 100 - 49 = 51

s = 2KL + L^2 = 140 + 49 =
189

u = 2KL + K^2 = 140 + 100 = 240

A = 240^2 +
219^2 = 105561

B = |219^2 - 51^2 - 2x219x51| =
23022

C = 219^2 - 189^2 + 2x219x189 = 95022

D = 219^2 - 51^2 +
2x219x51 = 67698

E = 219^2 + 189^2 = 83682

F =
|240^2 - 219^2 - 2x240x219| = 95481

G = |219^2 - 189^2 - 2x219x189| =
70542

H = 240^2 - 219^2 + 2x240x219 = 114759

I = 219^2 +
51^2 = 50562

The magic sum is about 2.07023 x 10^10 and the non-magic diagonal sum = 3E^2 is about 2.10080 x 10^10 for a ratio of about 1.015.

In order for the ratio to be exactly 1, we need to have

A^2 + I^2 = 2E^2.

Plugging in the expressions for K and L yields the polynomial

L^8 + 12L^7K + 34L^6K^2 + 44L^5K^3 + 20L^4K^4 - 8L^3K^5 - 12L^2K^6 - 8LK^7 - 2K^8 = 0

If we divide by K^8 and set x = L/K, we get

f(x) = x^8 + 12x^7 + 34x^6 + 44x^5 + 20x^4 - 8x^3 - 12x^2 - 8x - 2 = 0

The first derivative of this is

f'(x) = 8x^7 + 84x^6 + 204x^5 + 220x^4 + 80x^3 - 24x^2 - 24x - 8

Using Newton's iteration for root finding, if x[n] is a good guess at a root, then a better guess will be

x[n+1] = x[n] - f(x) / f'(x).

If you graph f(x) between 0 and 1, there is one root at about 0.7. Using a few Newton iterations, you can improve that to 0.68793. If you convert x back to L/K, which can be done simply by setting L = 68793, K = 100000, you get a nearly-magic square of integer squares with a diagonal sum ratio of 1.000004.

More Newton interations will get a better approximation and the diagonal sum ratio will get closer to 1 to any desired accuracy.

Note that Hillyer's formula can never produce a fully-magic square of squares.

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