In number theory, the law of quadratic reciprocity is a theorem about modular arithmetic that gives conditions for the solvability of quadratic equations modulo prime numbers. (Ref: http://en.m.wikipedia.org/wiki/Quadratic_reciprocity)

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Simplifying a reciprocal of a product of two quadratics using modular arithmetic

This is the fraction I'm trying to simplify mod q: $\frac{1}{(k^{2a} + k^a + 1)(k^{2b} + k^b +1)}$ and we have that $k^5 = 1 \bmod q$, $k ≠ 1 \bmod q$, and that $a$ and $b$ are integers where $a ≠ b$ ...
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About the solutions of $\dfrac{x^p - y^p}{x - y} = a^2+pb^2$

Using theorem $IV$ from this article, is possible to prove that when $p$ is a prime $p ≡ 3\bmod4$, $x ≢ y\bmod{p}$ and $\gcd(x,y) = 1$, then the equation $\dfrac{x^p - y^p}{x - y} = a^2+pb^2$ ever ...
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How to prove the roots of $ax^2 + bx + a$ are reciprocals of each other? [closed]

How to prove the roots of $ax^2 + bx + a$ are reciprocals of each other? I tried using quadratic formula to find the two roots but got stuck at the part on how to prove they are reciprocal. Please ...
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How to show a quadratic polynomial with complex roots has solutions mod $p$?

When the discriminant is $-k$ or ${-1\over k}$: As long as $-1$ and $k$ are both squares $\bmod p$ (trivial) or are both not squares, there are solutions mod $p$. How can this be explained/proven? e.g....
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Conceptual explanation for extra/missing $p$ solutions to $x^2+y^2=a \pmod p$ at $a=0$

Throughout, $p$ will denote a prime integer, and $k$ an arbitrary integer. I have worked through V. Lebesgue's proof of quadratic reciprocity outlined by Keith Conrad in this MO thread, and I feel ...
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Show ${\sum_{i=1}^{\frac{p-1}{2}} \lfloor \frac{2iq}{p} \rfloor + {\sum_{i=1}^{\frac{q-1}{2}}} {\lfloor \frac{2ip}{q} \rfloor}}=(p-1)/2\cdot(q-1)/2$

Let $E = {\sum_{i=1}^{\frac{p-1}{2}}} {\lfloor \frac{2iq}{p} \rfloor + {\sum_{i=1}^{\frac{q-1}{2}}} {\lfloor \frac{2ip}{q} \rfloor}}$ and $E'= \frac{p-1}{2} \frac{q-1}{2}$ Im trying to show that they ...
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How to find squares mod $m$, i.e $x$ in $x^2 \equiv a \mod m$, without factoring $m$?

I have very large integers $m$, where ( $\log_2(m)> 630$), and I need to find square roots modulo m. I am aware of several theorems that allow me to find the roots $\mod m$ when m is a power of a ...
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$2^n$-th rational reciprocity laws

Let $p,q$ be odd coprime primes. We are familiar with the quadratic reciprocity law: $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{(p-1)(q-1)}{4}}$. This is generalized (see for ...
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Finite set of quadratic forms representing all prime numbers

I ask if one can determine a finite set of quadratic forms $$mx^2+ny^2$$ with $\,m\,$ and $\,n\,$ positive integers s.t. $(m,n)=1$, capable to represent all prime numbers. We could choose, for ...
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Question 9.27 david M burton elementary number theory

This question is from David burton section: Quadratic reciprocity law, page 184. Question: If p is an odd prime, show that $\sum_{a=1}^{p-2} (a(a+1) /p) =-1$ . I can only say that a, a+1 are always ...
If an integer is odd, it is square modulo all powers of $2$ if and only if it is a square modulo $8$ if and only if it is equivalent to $1$ modulo $8$. How do we determine if an integer $a$ is square ...