# Tagged Questions

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### Solutions to the diophantine equation: $2a^2 + 2b^2- c^2- d^2 = 0$

As suggested on Mathoverflow (http://mathoverflow.net/questions/168536/solutions-to-the-diophantine-equation-2a2-2b2-c2-d2-0) I am transfering this question to math-stackexchange: I am looking for ...
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### Diophantine like philosophy for computing trigonometric functions with approximation around intervals

I noticed that diophantine expressions are great to approximate constants or simple functions, as far as I know, they are not so great when it comes to approximate and compute transcendental functions ...
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### solving $x^3-2y^3=1$ using cubic number field

I am trying to solve the diophantine equation $x^3-2y^3=1$ using $\mathbb{Q}(\sqrt[3]{2}).$ I've read this link: Solve $x^3 +1 = 2y^3$ The following is what i have tried: Finding all integer ...
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### How to solve diophantine equation $\frac{x^p-y^p}{x-y}=n$

$$\frac{x^p-y^p}{x-y}=n$$ whit $p$ a prime greater than or equal to $3$,for what value to $n$, it's solvable and how to solve,and whether $\frac{x^p-y^p}{x-y}=q_1$ $\frac{x^p-y^p}{x-y}=q_2$ is ...
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### Prove that $x^3 + y^3 = z^3$ has no integer solutions as simply as possible

Can someone prove the special case of Fermat's Last Theorem for $n=3$, i.e., that $$x^3 + y^3 = z^3,$$ has no positive integer solutions, as simply as possible? I have seen some good proofs, but ...
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### Looking for proof of no solution to 4-variable quadratic diophantine equation

Show there are no integers $a,b,c,d$ such that $$\begin{cases}1=9ac+3ad+3bc-16bd \\ 1=3ad+3bc+2bd \end{cases}$$ Motivation: The ideal $I=(3,1+\sqrt{-17})$ in $R=\mathbb{Z}[\sqrt{-17}]$ has the ...
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### Coefficients in expansion of $(\sqrt[3]{2} - 1)^m$

In trying to solve $a^3 - 2b^3 = 1$ over the integers I came across the need to answer the question: when does $(1+ \sqrt[3]{2} + \sqrt[3]{2}^2)^n$ have no $\sqrt[3]{2}^2$ term in it's expansion (in ...
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### $\frac{x^5-y^5}{x-y}=p$,give what p ,the diophantine equation is solvable

for$$\frac{x^3-y^3}{x-y}=x^2+xy+y^2=p$$$p=6k+1$give p prime, On what conditions,the diophantine equation $$\frac{x^5-y^5}{x-y}=p$$ is solvable in integers.does it have a linear expression.for ...
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### Slight generalization of Pell's equation?

Suppose I want to solve $x^2 - ny^2 = k$, where $n$ is square-free and $$n \equiv2,3 \mod 4.$$ If, moreover, $k$ is a prime such that for all $\mathfrak P_i$ lying above it in the field extension ...
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### Find all Integers ($n$) such that $n\neq 6xy\pm x\pm y$

I am interested in proving that there exist an infinite number of positive integers ($n$) which are not of the form $$n=6xy\pm x\pm y$$ for $x,y\in\Bbb Z^+$. [Note: The $\pm$ signs above are ...
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### Geometric intuition behind the Hasse principle

Let $f(X,Y) \in \mathbb{Q}[X,Y]$ be a quadratic polynomial. The Hasse-Minkowski theorem says that $f(X,Y) = 0$ has a solution $(x,y) \in \mathbb{Q}^2$ iff it has a solution in $\mathbb{R}^2$ and ...
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### Finding all solution of a Pell equation

I wanted to solve the equation $x^2-5y^2 = -4$ with $x$ and $y$ integers. Let $\omega=\frac{1+\sqrt5}{2}$ and $A = \mathbb{Z}[\omega]$. One can reduce the Pell equation to finding the elements of $A$ ...
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### Step in a solution of $y^2 = x^3 - 2$

I am reading Algebraic Number Theory notes here by Keith Conrad. In page 9, there is a solution of $y^2=x^3-2$ using unique factorization in $\mathbb{Z}[\sqrt{-2}]$. We start by writing ...
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### Ramanujan-Nagell Theorem Proof Question

I'm currently working through Stewart and Tall's Algebraic Number Theory. In particular, section 4.9 of this book provides a proof of the Ramanujan-Nagell Theorem, which states that the only integer ...
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### Solve: $x^2-py^2=q$

Solve $$x^2-py^2=q$$ for integers $x,y$, here $p,q$ are both given prime numbers. It's obvious that $p,q$ should satisfy $(\frac{p}{q})=(\frac{q}{p})=1,$ here $(\frac{p}{q})$ is the Jacobi symbol. ...
### Way to show $x^n + y^n = z^n$ factorises as $(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n$
For odd $n$ the Fermat equation $x^n + y^n = z^n$ factorises as $$(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n,$$ where $\zeta = e^{2 \pi i/n}$. I tried seeing this was true by multiplying ...