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 Jul11 accepted $\alpha x^2+\beta y^2=\gamma$ solvable over $\mathbb Q$ iff $ax^2+by^2=z^2$ solvable over $\mathbb Z$ with coprime $x,y,z$? Jul11 asked $\alpha x^2+\beta y^2=\gamma$ solvable over $\mathbb Q$ iff $ax^2+by^2=z^2$ solvable over $\mathbb Z$ with coprime $x,y,z$? Jun27 comment algorithm for solving diagonal quadratic equations over real or complex numbers Yes I thought that it's not in the spirit of the paper, but I want to be very precise at this point. You say that solving linear equations in exact arithmetic has $O(n^3)$ (by Gaussian elimination of course). And speaking of which, I realize that I forgot a square :-/ (even though I wrote "quadratic equation" in the title) - now corrected. So I'm in the case of a quadratic equation where one cannot apply Gauss' algorithm. The next question is: what do you mean with "exact arithmetic"? Working symbolically wherever necessary? Thanks for all your time btw! Jun27 revised algorithm for solving diagonal quadratic equations over real or complex numbers added 2 characters in body Jun27 comment algorithm for solving diagonal quadratic equations over real or complex numbers I added a link to the paper. It is a rather algebraic context. What you say is, concerning my first follow-up question, equivalent to extending the alphabet to $\mathbb F$ and adding special states for adding and multiplying two cells to the set of states of the turing machine, right? Jun27 revised algorithm for solving diagonal quadratic equations over real or complex numbers added 86 characters in body Jun27 asked algorithm for solving diagonal quadratic equations over real or complex numbers Jun25 awarded Scholar Jun25 accepted Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ Jun25 comment Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ So either $\dim(P)<2$ or ($(b.b)=0$ and $(a.b)=0$). The later implies that $b\in\mathop{\rm rad}$. Now if $b=0$, $\dim(P)<2$ and if $b\neq 0$, $P$ is degenerate. Thank you very much! Jun25 comment Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ looking at this element (let's call it $c=(b.b)a-(a.b)b$) i get $c.a=0$ and $c.b=0$ yielding that $c$ is in the radical of $P$. This implies that $c$ is zero or $P$ is degenerate. In the later case we are done. If $c$ is zero, we have that $(b.b)a-(a.b)b=0$ which is a linear combination of $0$ in $P$. So either $\dim(P)<2$ or $(b.b)=0$ and $(a.b)=0$. Jun25 awarded Editor Jun25 comment Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ of course "is zero"... sorry (edited). Jun25 revised Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ added 2 characters in body Jun25 asked Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ Jun10 awarded Supporter May28 awarded Teacher May28 answered Is there a rational univariat polynomial of degree 3 with 3 irrational roots? May27 awarded Student May27 comment Is there a rational univariat polynomial of degree 3 with 3 irrational roots? @RagibZaman yes, that's absolutely right and it works out perfectly. Thank you all! I'll post the answer to this question in around 6 hours, when the software here allows me to.