MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 4 added 15 characters in body

The problem lattice angles in $\mathbf R^2$ is quite similar to that R^3$ are the same as those in $\mathbf R^4.$ Mordell showed This seems to be what Michael Beeson is saying in 1932, with a gap filled by Ivan Niven the first page of his 1992 article (link in 1940Gjergji's answer), that we can solve\left( \begin{array}{rr} p & q \\ t & u
\end{array} \right) \; \cdot \;\left( \begin{array}{rr} p & t \\ q & u \end{array} \right) = \; \;\; \; \left( \begin{array}{cc} A & B \\ B & C
\end{array} \right)
if and only if $AC - B^2$ is a square and $\gcd(A,B,C)$ is I have not seen the sum rest of two squaresthe article yet.

With

Using the symbols of Gerhard "Fluid Mechanics Have Side Effects" Paseman, we are attempting to find a lattice angle $F^2 < G$ and\theta$ such that $$ \cos \theta = \frac{\pm F}{\sqrt{G}}, frac{ C}{\sqrt{C^2 + k}},$$with integers $C,k$ and $we must require k >0.$

In $G - F^2$ to be \mathbf R^5,$ this is always possible.

In $\mathbf R^2,$ this is possible if and only if $k$ is a square.

Once again, in which case $\theta$ is the angle betweenvectors $\gcd$ condition does not really matter, as we can take$ $A = 1, \; B = F, \; C = \sqrt{G-F^2}, \; \; P P_2 = (1,0), \; \; T T_2 = (F,\sqrt{G-F^2}).$$

Many rational values are allowed, as any Pythagorean triple (primitive or not)C,\sqrt{k}). $L^2 + M^2 = N^2$ allows us to take $G= N^2$

In $\mathbf R^4,$ this is possible if and only if $F k$ is the sum of three squares. Taking$$ k = M,$ oru^2 + v^2 + w^2,$$ $\theta$ is the angle betweenvectors $\cos \theta $ P_4 = (1,0,0,0), \frac{\pm F}{\sqrt{G}} = ; \frac{\pm M}{N}.$$But ; T_4 = (C,u,v,w). $$

In $\mathbf R^3,$ this is also possible if and only if $N^2 - M^2$ k$ is not a square the sum of three squares; this comes from simple tricks with integral quaternions. However, we cannot arrange can no longer take one vector as fixed, or bound its length ahead of time. Taking$ \cos \theta k = \frac{\pm M}{N}.$ In particular u^2 + v^2 + w^2,$ and requiring that $u,v$ not both be $0,$ we cannot arrange find $\cos \theta =\frac{\pm 1}{2}$ as a lattice \theta$ is the angle in betweenvectors $\mathbf R^2.$

Ivan Niven$ P_3 = (u,v,0), \; \; T_3 = (C u - v w, Integers of quadratic fields as sums of squaresC v + u w , Trans. Amer. Math. Soc. 48 (1940) pp. 405-417u^2 + v^2 ). $$

show/hide this revision's text 3 added 172 characters in body

The problem in $\mathbf R^2$ is quite similar to that in $\mathbf R^4.$ Mordell showed in 1932, with a gap filled by Ivan Niven in 1940, that we can solve [
\left( \begin{array}{rr} p & q \\ t & u
\end{array} \right) \; \cdot \; \left( \begin{array}{rr} p & t \\ q & u \end{array} \right) = \; \;\; \; \left( \begin{array}{cc} A & B \\ B & C
\end{array} \right)
] if and only if $AC - B^2$ is a square and $\gcd(A,B,C)$ is the sum of two squares.

With $F^2 < G$ and $$ \cos \theta = \frac{\pm F}{\sqrt{G}}, $$ we must require $G - F^2$ to be a square.

Once again, the $\gcd$ condition does not really matter, as we can take $$A = 1, \; B = F, \; C = \sqrt{G-F^2}, \; \; P = (1,0), \; \; T = (F,\sqrt{G-F^2}).$$

Again, many

Many rational values are also ruled outallowed, as any Pythagorean triple (primitive or not) $L^2 + M^2 = N^2$ allows us to take $G= N^2$ and $F = L,$ M,$ or $$ \cos \theta = \frac{\pm F}{\sqrt{G}} = \frac{\pm L}{N}.$$M}{N}.$$ But if $N^2 - M^2$ is not a square we cannot arrange $\cos \theta =\frac{\pm M}{N}.$ In particular we cannot arrange $\cos \theta =\frac{\pm 1}{2}$ as a lattice angle in $\mathbf R^2.$

Ivan Niven, Integers of quadratic fields as sums of squares, Trans. Amer. Math. Soc. 48 (1940) pp. 405-417.

show/hide this revision's text 2 edited body

The problem in $\mathbf R^2$ is quite similar to that in $\mathbf R^4.$ Mordell showed in 1932, with a gap filled by Ivan Niven in 1940, that we can solve [
\left( \begin{array}{rr} p & q \\ t & u
\end{array} \right) \; \cdot \; \left( \begin{array}{rr} p & t \\ q & u \end{array} \right) = \; \;\; \; \left( \begin{array}{cc} A & B \\ B & C
\end{array} \right)
] if and only if $AC - B^2$ is a square and $\gcd(A,B,C)$ is the sum of two squares.

With $F^2 < G$ and $$ \cos \theta = \frac{\pm F}{\sqrt{G}}, $$ we must require $G - F^2$ to be a square.

Once again, the $\gcd$ condition does not really matter, as we can take $$A = 1, \; B = F, \; C = \sqrt{G-F^2}, \; \; P = (1,0), \; \; T = (F,\sqrt{G-F^2}).$$

Again, many rational values are also ruled out, as any Pythagorean triple (primitive or not) $L^2 + M^2 = N^2$ allows us to take $G= N^2$ and $F = L,$ or $$ \cos \theta = \frac{\pm F}{\sqrt{G}} = \frac{\pm L}{M}.$$L}{N}.$$

Ivan Niven, Integers of quadratic fields as sums of squares, Trans. Amer. Math. Soc. 48 (1940) pp. 405-417.

show/hide this revision's text 1