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30 votes
Accepted

Primes in solutions to Pell-type equations

Let $\alpha = (2 + \sqrt{5})$ and $\beta = (2 - \sqrt{5})$. Note that $\alpha \beta = -1$. The general solution is given by $$y_n = \frac{\alpha^n - \beta^n}{\alpha - \beta} = \frac{\alpha^n -\beta^n}{...
user865424's user avatar
12 votes
Accepted

Quadratic Diophantine Equation $x^2 + 2y^2 = 2013$

The equation is not solvable modulo $8$, hence there are no integer solutions. You can see this by considering $x^2\equiv 0,1$ or $4$ modulo $8$ for every $x$.
Peter's user avatar
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11 votes
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Does this system of simultaneous Pell-like equations have any non-trivial positive integer solutions?

${\bf Summary:}$ We prove that for any positive integer solution $a$, $b$ and $c$ to the given equations there exist $x,y\in \Bbb{Z}$ such that $$ 4a=x^2+3y^2,\quad b=\frac{|x^2-3y^2|}{2}\quad\text{...
san's user avatar
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11 votes

First 30 solutions of Pell's equation.

Floating point precision is not sufficient for the larger solutions. You should probably look at using a recursive definition for the $k_n$ and $z_n$. They should satisfy $a_{n+2}=18a_{n+1}-a_{n}$ (...
sharding4's user avatar
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11 votes
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Can $n+1$ , $2n+1$ , $3n+1$ all be perfect squares , if $n$ is a positive integer?

If this happens, then $1, n+1, 2n+1, 3n+1$ is a four-term arithmetic progression of perfect squares. But it can be shown (painfully, using elliptic curves, as demonstrated at this link - see Theorem ...
Misha Lavrov's user avatar
11 votes
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Prove $(1+\sqrt{2})^k= \sqrt{n}+\sqrt{n-1}$

This follows from $(1-\sqrt{2})^k\,(1+\sqrt{2})^k=(-1)^k$ for every integer $k$. So, if $a_k$ and $b_k$ are integers such that $(1+\sqrt{2})^k=a_k+b_k\sqrt{2}$, then $a_k-b_k\sqrt{2}=(1-\sqrt{2})^k$, ...
Batominovski's user avatar
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9 votes
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How to find integer solutions to $M^2=5N^2+2N+1$?

One possibility is to rewrite as $$M^2=(2N)^2+(N+1)^2. $$ So you are looking at the Pythagorean triples of the form $(N+1,2N,M) $. Edit: Along the lines of Erick's complaint, I want to make this ...
Martin Argerami's user avatar
9 votes
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Evaluating $\liminf_{n\to\infty}n\{n\sqrt2\}$

For $n > 0$, let $m = \lfloor n\sqrt{2}\rfloor$. Since $\sqrt{2} \not\in \mathbb{Q}$, $$n\sqrt{2} > m \implies 2n^2 > m^2 \implies 2n^2 \ge m^2 + 1 \implies \sqrt{2}n \ge \sqrt{m^2+1}$$ ...
achille hui's user avatar
8 votes
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On $\big(\tfrac{1+\sqrt{5}}{2}\big)^{12}=\small 161+72\sqrt{5}$ and $\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{161+72\sqrt{5}\,x}}$

Starting from $$ \,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-a\big)=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+ax^4}}, $$ $$ (b+\sqrt{b^2-1})^{-1/4}\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{...
Nemo's user avatar
  • 3,903
8 votes

Pairs of $(p,q)$ such that $p^2 + 1 = 2 q^2$?

These are what are called "units of the integer ring $\Bbb Z[\sqrt 2]$ of norm $-1$." It is known that they are exactly the integer coefficients of the numbers $p+q\sqrt 2$ when $$p+q\sqrt 2= (1+\...
Adam Hughes's user avatar
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8 votes
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Integer solutions of $n^2+n = 2x^2+2x$

One way is to set $m=2n+1$ and $y=2x+1$ and get the equivalent Pell equation $m^2=2y^2-1$. The fundamental solution is $m=1, y=1$ and the general solution comes from the odd powers of $1+\sqrt2$. You ...
lhf's user avatar
  • 217k
8 votes
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Diophantine equation $x^3+x+a^2=y^2$

Your first idea is a great idea! When working this out, I had slightly different variables, but it is equivalent to your starting. Suppose positive integers $b, c, d, e$ satisfy $y+a=bc$, $y-a=de$, $...
Sharky Kesa's user avatar
  • 4,187
8 votes
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System of Pell's equations

Let $a$, $b$ and $k$ be positive integers such that \begin{eqnarray} 2k^2+1&=&a^2,\\ 6k^2+1&=&b^2. \end{eqnarray} Then clearly $k$ is coprime to $a$ and $b$ and $$a^2+(2k)^2=b^2,$$ so $...
Servaes's user avatar
  • 63.5k
7 votes

General solution of Pell's equation

It's well know that the solution of the Pell's equation satisfy the reccurence relation: $$x_{k+1} = x_1x_k + y_1ny_k$$ $$y_{k+1} = y_1x_k + x_1y_k$$ This was found by Brahmagupta and it holds ...
Stefan4024's user avatar
  • 35.9k
7 votes

How to show that $x^2 - 37y^2 =2$ does not have integer solutions

There are no integer solutions to $$ x^2 - 37 y^2 \equiv 2 \pmod 4 $$ as $$ x^2 - 37 y^2 \equiv x^2 - y^2 \pmod 4 $$
Will Jagy's user avatar
  • 140k
7 votes

Demonstrate that these two Pell's equations have no integer solutions

The second one is easy $$x^2 - 82y^2 = \pm3 \Rightarrow x^2 \equiv \pm3 \pmod{41}$$ but according to Euler's criterion $$\left(\frac{\pm3}{41}\right) \equiv \left(\pm3\right)^{\frac{41-1}{2}} \pmod{...
rtybase's user avatar
  • 17k
7 votes

Solving $X^2-6Y^2=Z^3$ in positive integers

The represented numbers can be described explicitly; this is what becomes possible when there is one class per genus of the discriminant. There are two genera, one form each. We know $x^2 - 6 y^2.$ ...
Will Jagy's user avatar
  • 140k
7 votes
Accepted

$p$-adic structure of Pell-type equations

As $d$ gets larger and $p^d|y_k$, this can be thought of as taking the limit as $y \to 0$ in the p-adics, since $\lim_{d \to \infty}|p^d|_p=0$. So in your equation we're looking at solving just $x^2=N$...
Merosity's user avatar
  • 2,578
7 votes

Solve in $\mathbf{N}$ the equation $9x^2+p=y^2$

First, reformulate the problem as the following: $$ y^2 - 9x^2 = p \Rightarrow(y-3x)(y+3x)= p $$ Now, for any given $p$, find its prime factors. Then, for any 2-partitions of them, solve a simple ...
OmG's user avatar
  • 3,138
6 votes
Accepted

Does the Pell-like equation $X^2-dY^2=k$ have a simple recursion like $X^2-dY^2=1$?

Yes. The recursion is just the Brahmagupta-Fibonacci Identity in disguise, $$(u x + d v y)^2 - d(v x + u y)^2 = (u^2 - d v^2) (x^2 - d y^2) = k$$ The coefficients $u,v$ are determined by the ...
Tito Piezas III's user avatar
6 votes

Does the Pell-like equation $X^2-dY^2=k$ have a simple recursion like $X^2-dY^2=1$?

Make this an answer. It turns out that, using the recursion you describe, the set of all solutions to $x^2 - dy^2 = k$ split into a small number of orbits. The cleanest way to locate the "seed" values ...
Will Jagy's user avatar
  • 140k
6 votes
Accepted

Show that $x^2-dy^2 = -2$ with $d = m^2+2$ has infinitetly many integer solutions

If $d\in\Bbb N$ is not a square, and $$x^2-dy^2=k\tag1$$ has some integer solution for $k\ne 0$ then $(1)$ has infinitely many integer solutions. This is because Pell's equation $$x^2-dy^2=1\tag2$$ ...
Angina Seng's user avatar
6 votes

How to show that $x^2 - 37y^2 =2$ does not have integer solutions

If there were a solution, then you could take the equation mod $37$ and find $$x^2\equiv 2\pmod{37}$$ In particular, $2$ would be a quadratic residue modulo $37$. However, it isn't, since the ...
vadim123's user avatar
  • 82.9k
6 votes

A System of Simultaneous Pell Equations

Added: from Cipu and Mignotte, the first system expected to have two positive solutions, one piece $\color{red}{x^2 - 24 z^2 = 1},$ uses $490^2 - 1 = 240099.$ $$ 5^2 - 24 \cdot 1^2 = 1 \; , \; \; \; ...
Will Jagy's user avatar
  • 140k
6 votes
Accepted

Trouble with Pell equations

You period $n = 1$. The first convergent is $\frac{73}{12}$ which means your fundamental solution can be $$x = 73, y = 12$$ and indeed $$73^2 - 37(12)^2 = 1$$ Assuming your continued fraction is ...
Mark Fischler's user avatar
6 votes

Notice that $\sqrt{51}\approx 7+\frac{\sqrt{2}}{10}$

Well, I was thinking this: $$\frac{d\sqrt{x}}{dx}|_{x=50} = \frac{1}{2\sqrt{x}}|_{x=50} = \frac{1}{10\sqrt{2}}.$$ So you'd expect the difference between $\sqrt{51}$ and $7=\sqrt{49}$ to be something ...
Mike's user avatar
  • 20.8k
6 votes

Missing pattern in solvable negative Pell equation

There is a pattern, but not one you can see just by looking at the values of $a$. Theorem. For a positive nonsquare integer $D$, the equation $x^2 - Dy^2 = -1$ has a solution in integers if and only ...
KCd's user avatar
  • 46.9k
5 votes

Rational solutions of Pell's equation

There is a simple and more general way to look at this problem that applies to finding rational points on any conic section with rational coefficients. i.e. any equation in the form $ax^2 + bxy + cy^2 ...
Philip Gibbs's user avatar
5 votes

Is this $1811$ proof for the negative Pell equation really wrong?

Your sentence Of course, he was most likely wrong since we still today don't know when it is solvable does not correctly apply, as the negative Pell question is known for prime values of $a$ in $...
Will Jagy's user avatar
  • 140k

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