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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). $$In \mathbf R^3, this is also possible if and only if N^2 - M^2 k is not a square the sum of three squares; this comes from simple tricks with integral quaternions. However, we cannot arrange can no longer take one vector as fixed, or bound its length ahead of time. Taking \cos \theta k = \frac{\pm M}{N}. In particular u^2 + v^2 + w^2, and requiring that u,v not both be 0, we cannot arrange find \cos \theta =\frac{\pm 1}{2} as a lattice \theta is the angle in betweenvectors \mathbf R^2. Ivan Niven P_3 = (u,v,0), \; \; T_3 = (C u - v w, Integers of quadratic fields as sums of squaresC v + u w , Trans. Amer. Math. Soc. 48 (1940) pp. 405-417u^2 + v^2 ).$$

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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}.$$But if N^2 - M^2 is not a square we cannot arrange \cos \theta =\frac{\pm M}{N}. In particular we cannot arrange \cos \theta =\frac{\pm 1}{2} as a lattice angle in \mathbf R^2. Ivan Niven, Integers of quadratic fields as sums of squares, Trans. Amer. Math. Soc. 48 (1940) pp. 405-417. 2 edited body The problem in \mathbf R^2 is quite similar to that in \mathbf R^4. Mordell showed in 1932, with a gap filled by Ivan Niven in 1940, that we can solve [ \left( \begin{array}{rr} p & q \\ t & u \end{array} \right) \; \cdot \; \left( \begin{array}{rr} p & t \\ q & u \end{array} \right) = \; \;\; \; \left( \begin{array}{cc} A & B \\ B & C \end{array} \right) ] if and only if AC - B^2 is a square and \gcd(A,B,C) is the sum of two squares. With F^2 < G and$$ \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.

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