# Tag Info

Accepted

• 13.7k

### 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 $a_{n+2}=18a_{n+1}-a_{n}$ (...
• 4,917
Accepted

### 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 ...
• 144k
Accepted

### 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 $a_k-b_k\sqrt{2}=(1-\sqrt{2})^k$, ...
• 49.7k
Accepted

### 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 ...
• 208k
Accepted

### Evaluating $\liminf_{n\to\infty}n\{n\sqrt2\}$

For $n > 0$, let $m = \lfloor n\sqrt{2}\rfloor$. Since $\sqrt{2} \not\in \mathbb{Q}$, $$n\sqrt{2} > m \implies 2n^2 > m^2 \implies 2n^2 \ge m^2 + 1 \implies \sqrt{2}n \ge \sqrt{m^2+1}$$ ...
• 123k
Accepted

• 36.9k
Accepted

### 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 ...
• 217k
Accepted

• 63.5k

### 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 ...
• 35.9k

### 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 $$x^2 - 37 y^2 \equiv x^2 - y^2 \pmod 4$$
• 140k

• 140k
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 ...
• 41.8k

### 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 ...
• 20.8k
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 ...