# All Questions

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### Legendre symbol problem

Let $p,q$ different odd prime numbers and $a$ a positive odd number, Prove that : $$p \equiv - q\ \ \ \ (mod \ \ 4a )\Longrightarrow \Bigg(\frac{a}{p}\Bigg) =\Bigg(\frac{a}{q}\Bigg)$$ Where ...
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### Bound on sum of complex summands involving binomial coefficients

I am trying to find the asymptotic behavior of the sum: $$\sum^n_{i=0} \begin{pmatrix} 2n \\ i \end{pmatrix} x^i y^{2n-i}$$ as $n\rightarrow\infty$. Here $x$, $y$ are complex numbers and I have ...
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### Canonical divisor for conic bundles

Is there a general formula for the canonical divisor $K_X$ of a smooth conic bundle $X$? Motivation: for smooth hypersurfaces of degree $d$ in $\mathbb{P}^n$, $K_X = \mathcal{O}_X(d-n-1)$. But ...
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### Formula for unequal share distribution [on hold]

What formula would I use to distribute $M$ shares among $N$ shareholders, such that shareholder $X_i$ has 3/2 as many shares as shareholder $X_{i+1}$? P.S. I apologize if the tag isn't relevant. I ...
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### Exactly 2 Girls - Conditional Probability [on hold]

This is very confusing to me. I am really new with this stuff. A couple wants to have 3 or 4 children, including exactly 2 girls. Is it more likely that they will get their wish with 3 children or ...
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### Integral of (2-x)/(x-1) Really stumped [on hold]

So I tried doing this: I have integral (2-x)/(x-1) I used a substitution ; u = x-1 x= u+1 du = dx So then (2-u-1)/u du then : 1/u - 1 Then I integrate and get ln u - u But when I plug ...
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### Canonical identification between 3-manifolds cohomology and group cohomology [on hold]

I am trying to understand why this 3-manifold cohomology is equal to this group-cohomology. $$H_\ast (\mathbb{H}^3/PGL_2(\mathbb{Z})) \simeq H_\ast (PGL_2(\mathbb{Z}))$$ In both cases, use the base ...
### Counting Boolean Normal Matrices of size $2n \times 2n$
Fix $n$ a natural number. Consider the set of all $2n \times 2n$ matrices with entries from {0,1}. This is clearly a finite set. I would like to count the number of such normal matrices for fixed ...
For finite groups $G$, we have the group ring $k[G]$, and we can think of $G$-representations as $k[G]$-modules. It is known that for $G$ virtually polycyclic, $k[G]$ is a Noetherian ring, which means ...