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I think that A=R implies (P) for Euclidean domains. Basically, once you have $GL_N(Z)$ and the diagonal matrices in $GL_N(R)$, you get all the elementary matrices -- the generators of the Chevalley group $SL_N(R)$. Diagonal matrices in $GL_N(R)$ give you all possible determinants in $R^\times$, so that does it.
...An approximation to the solution of this equation is not difficult to find, but Khayyam also generates a direct geometric solution: he uses the numbers in the equation to determine intersecting curves of two conic sections (a circle and a hyperbola), and demonstrates that the solution $x$ is equal to the length of a particular line segment in the diagram." (plato.stanford.edu/entries/umar-khayyam/#SolCubEqu)
The Stanford Encyclopedia of Philosophy discusses this document further: "Khayyam seems to have been attracted to cubic equations originally through his consideration of the following geometric problem: in a quadrant of a circle, drop a perpendicular from some point on the circumference to one of the radii so that the ratio of the perpendicular to the radius is equal to the ratio of the two parts of the radius on which the perpendicular falls. In a short, untitled treatise, Khayyam leads us from one case of this problem to the equation $x^3 + 200x = 20x^2 + 2000$...
Brin's answer below seems to be the best reference, judging from MathSciNet reviews. The translation of Al-Khayyam into French by Rashed and Vahabzadeh seems to be a scholarly critical edition...
By $X$ I meant the set of algebraic homs defined over an algebraic closure. I figured that even though each $K_x$ would not be defined over the base field, the intersection $K$ would would be defined over the base field, and all would work out alright. But judging from the extensive answer below, I have a poor intuition for non-perfect fields.
Let $X$ be the set of algebraic homs from $H$ to the multiplicative group $G_m$. For each $x \in X$, let $K_x$ be its kernel. Let $K$ be the intersection of all such kernels. Then $H/K$ fits the bill.
I appreciate the shout out, and I agree that looking modulo the orthogonal group of the form is a good idea. The topograph gives a good way of seeing a fundamental domain for this orthogonal group acting on primitive pairs $(x,y)$ but I'd need to run some experiments to make a guess on heuristics of $P'(B)$.
I have been very happy with TikZ/PGF. See texample.net/tikz/examples It's a very very good tool to produce high-quality pictures. I think that it's been worth my time, not only for my writing/research, but also for my teaching and presentation (e.g. producing good handouts, beamer slides, etc.).
First, there are only countably many such sums $\sum f(n)$ with $f$ rational with integer coefficients. So this is far from equality-testing on the real numbers. Here's an analogue which might shed light. Consider instead the decidability of $f(x) = 0$, when $f$ is a G-function (in the sense of Siegel) and $x$ is a rational number. Such decidability would be very close to the decidability of equality of Kontsevich-Zagier periods. This is at least as strong as Schanuel's conjecture, I'd guess.
