Found it. Given odd dimension n and quadratic form f = a_1 x_1^2 + a_2 x_2^2 + \cdots + a_n x_n^2, everything in a finite field with odd number of elements q, the count \#\left(f = b\right) \; = \; q^{n-1} + q^{(n-1)/2} \; \; \chi \left( \; (-1)^{(n-1)/2} \; b a_1 a_2 \ldots a_n\right). ...

I can't understand why this solid is getting chopped up into two pieces. We know the \theta goes all the way around, from 0 to 2\pi, then the boundary conditions can be solved as z=r^2, r^2+z^2=r^2+r^4=2 ...

Let x=u^2,\quad y=v^2,\quad z=w^2,\qquad(1) then the inequality becomes (u^2+2v^2+w^2)\left(\dfrac{u^2}{v^2}+\dfrac{2v^2}{w^2}+\dfrac{w^2}{u^2}\right) > 12 for u^4+v^4+w^4=3. Using ...

x^2=-1 is a conjugate complex pair of lines, namely x=\pm i. They intersect at the point at infinity in the y direction, i.e. have the real point (0,1,0) in common. Projectively speaking, X^2+Z^2=0 ...

A hint for i): By contradiction. Assume they are not pairwise disjoint. Then \exists q_1 , q_2 \in Q such that x \in (q_1 + C) \cap (q_2 + C). Then q_1 + c = x = q_2 + c^\prime. So q_1 - q_2 = c^\prime - c ...

There are two curves. They can be parametrized as follows. \gamma_1 :[0,2\pi]\to\mathbb{R}^3 given by \gamma_1(t)=(\cos t,\frac{1}{\sqrt{2}}\sin t,\frac{1}{\sqrt{2}}\sin t), and \gamma_2 :[0,2\pi]\to\mathbb{R}^3 ...