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\address
Professor Michael Beeson
185 View Court
Aptos, CA  95003
\body

Dear Michael:

	Your paper on the triangle problem inspired
me to look for some simplifications.  Here they are.

	1. Instead of thinking about what triangles are possible
on integer lattices, (call them lattice triangles), think about
what lattice angles are possible.  Your main results
become

		a. In two dimensions, they are angles whose tangents
are rational.

		b. In 3 and 4 dimensions, they are angles whose tangents
are rational multiples of $\sqrt{k}$
where $k$ is the sum of three squares.

	c. In five or more dimensions, angles whose tangents are
rational multiples of $\sqrt{k}$ where $k$ can be any positive
integer.

	2. $k$ is determined by the plane of the angle, i.e. all
lattice angles in the same lattice plane have the same $k$.  This
follows from your result that the angles of a lattice triangle
have the same $k$.  Any two angles can be extended to two triangles
that share an angle.  From the present point of view that emphasizes
angles not triangles, this should have an independent proof.

	3. The fact that 3 and 4 dimensions give the same condition
on angles is a consequence of some algebraic identities.  Locate
the vertex of a lattice angle $\theta$ at the origin.  Let it be
determined by vectors $x$ and $y$.  In that case we have

$$\eqalign{\tan ↑2\theta &= \sec ↑2\theta - 1\cr
&= {1\over {{(x\cdot y)↑2}\over {|x|↑2|y|↑2}}} -1\cr
&= {{(\Sigma x↓i↑2)(\Sigma y↓i↑2) - (\Sigma x↓iy↓i)↑2}
\over {(x \cdot y)↑2}}\cr
}.$$

a. In two dimensions
%
$$\eqalign{{{\tan ↑2 \theta}\over {(x\cdot y)↑2}} 
&= (x↓1↑2+x↓2↑2)(y↓1↑2+y↓2↑2) - (x↓1y↓1 +x↓2y↓2)↑2\cr
%
&=(x↓1y↓2 - x↓2y↓1)↑2\cr
},$$
%
which explains why the tangent is rational.

b. In three dimensions
%
$$\eqalign{{{\tan ↑2 \theta}\over {(x\cdot y)↑2}} 
&= (x↓1↑2+x↓2↑2+x↓3↑2)(y↓1↑2+y↓2↑2+y↓3↑2) - (x↓1y↓1 + x↓2y↓2 + x↓3y↓3)↑2\cr
%
&=(x↓1y↓2 - x↓2y↓1)↑2 + (x↓1y↓3 - x↓3y↓1)↑2 + (x↓2y↓3 - x↓3y↓2)↑2\cr
},$$
%
so that $\tan \theta = {\rm rational}\times \sqrt{k}$, where $k$ is the
sum of three squares.  This is just the identity
%
$$|x|↑2|y|↑2 = |x|↑2|y|↑2(\cos ↑2 \theta + \sin ↑2 \theta) = (x\cdot y)↑2
+ |x \times y|↑2.$$

c. In four dimensions
%
$$\eqalign{{{\tan ↑2 \theta}\over {(x\cdot y)↑2}} 
=& (x↓1↑2+x↓2↑2+x↓3↑2+x↓4↑2)(y↓1↑2+y↓2↑2+y↓3↑2+y↓4↑2) -
 (x↓1y↓1 + x↓2y↓2 + x↓3y↓3 + x↓4y↓4)↑2\cr
%
=&(x↓1y↓2 - x↓2y↓1 + x↓3y↓4 - x↓4y↓3)↑2\cr
&+(x↓1y↓3 - x↓3y↓1 - x↓2y↓4 + x↓4y↓2)↑2\cr
&+(x↓1y↓4 - x↓4y↓1 + x↓2y↓3 - x↓3y↓2)↑2\cr
},$$
%
again giving the result that
$\tan \theta = {\rm rational}\times \sqrt{k}$, where $k$ is the
sum of three squares.

	I would not have expected the 4 variable identity but
for your theorems.  They establish that for any integers
$x↓1,x↓2,x↓3,x↓4,y↓1,y↓2,y↓3,y↓4$, $(\Sigma x↑2)(\Sigma y↑2) -(\Sigma xy)↑2$
is a sum of three squares.  Therefore, I was led to look for a
corresponding algebraic identity, which I found by fiddling with
the 3 variable identity (to which the 4 variable identity reduces
when $x↓4 = 0$).  Perhaps the 4 variable identity is well known in
some quaternion guise.

	You are welcome to use these results in improving your paper.

\closing
Sincerely,
John McCarthy
\endletter
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