32 votes

Why does this test for Fibonacci work?

Suppose $A = F_n$ is a Fibonacci number. Then we can easily prove by induction that $$ 5 F_n^2 - L_n^2 = \begin{cases} 4 \qquad &\mbox{if $n$ is odd}; \\ -4 \qquad &\mbox{if $n$ is even}. \end{...
user avatar
  • 15.5k
29 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}{...
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 $x$.
user avatar
  • 78.4k
11 votes
Accepted

Solving the nonlinear Diophantine equation $x^2-3x=2y^2$

Multiply through by $4$ and complete the square, $(2x-3)^2 - 8 y^2 = 9,$ so that $(2x-3)^2 + y^2 \equiv 0 \pmod 3.$ It follows that both $(2x-3)$ and $y$ are divisible by $3,$ therefore $x$ as well. ...
user avatar
  • 130k
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 $a_{n+2}=18a_{n+1}-a_{n}$ (...
user avatar
  • 4,757
11 votes
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 ...
user avatar
11 votes
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 $a_k-b_k\sqrt{2}=(1-\sqrt{2})^k$, ...
user avatar
  • 48.4k
10 votes
Accepted

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{...
user avatar
  • 13.7k
9 votes
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 ...
user avatar
8 votes
Accepted

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}{...
user avatar
  • 2,833
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+\...
user avatar
  • 35.1k
8 votes
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 ...
user avatar
  • 208k
8 votes
Accepted

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 $y+a=bc$, $y-a=de$, $...
user avatar
  • 2,673
8 votes
Accepted

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

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

If $(m,n)\in\mathbb Z_+^2$ satisfies $3m^2+m = 4n^2+n$ then $(m-n)$ is a perfect square.

Rewrite the original equation $3m^2+m=4n^2+n$ as $$12m^2+12n^2+m-n-24mn=16n^2+9m^2-24mn.$$ This factors as $$(m-n)(12(m-n)+1)=(4n-3m)^2.$$ Since $\gcd(m-n,12(m-n)+1)=1$, it follows that $m-n$ is ...
user avatar
  • 1,700
7 votes

How to find a fundamental solution to Pell's equation?

(Too long for a comment.) If you're lucky and your discriminant $d$ has a certain form ($d = 5, 13, 21, 29, 53, 61, \dots$), you can use the smaller fundamental solutions of the Pell equation, $$p^2-...
user avatar
7 votes

Integer solutions of $ y^{2} = 5x^{2} + 17 $

Your attempt cannot work, since reducing modulo an integer (here $5$) irreversibly wipes out the distinction between positive an negative numbers; one can never apply arguments based on comparison "...
user avatar
7 votes

Find all primes $p$ and $q$ such that $p^2-2q^2=1.$

From your last expression we have $$\tag1(p+1)(p-1)=2\cdot[\;2(2d^2+2d)+1\;].$$ If $p=2$ then you get from the original expression that $2q^2=3,$ which is impossible. Therefore $p>2$ and hence $p$ ...
user avatar
  • 3,309
7 votes

Find all primes $p$ and $q$ such that $p^2-2q^2=1.$

Since $p^2-2q^2=1$, $p$ must be odd. Furthermore, $q$ must be even; if $q$ were odd, then $p^2-2q^2$ would be $3$ mod $4$. Thus, $q=2$ and then $p=3$.
user avatar
  • 326k
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 ...
user avatar
  • 34.9k
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 $$ x^2 - 37 y^2 \equiv x^2 - y^2 \pmod 4 $$
user avatar
  • 130k
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{...
user avatar
  • 16.2k
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 $x^2 - 6 y^2.$ ...
user avatar
  • 130k
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 $x^2=N$...
user avatar
  • 1,732
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 ...
user avatar
  • 3,112
6 votes

When is $991n^2 +1$ a perfect square?

You can just use Reduce or Solve to find all solutions. ...
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 ...
user avatar
  • 130k
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 ...
user avatar
6 votes

there are infnitely many postive integer $n$ such $ \lfloor \sqrt{7}\cdot n \rfloor=k^2+1(k\in \Bbb{Z})$

I get that there are an infinite number of $n$ such that $\lfloor n\sqrt{d} \rfloor =k^2-1 $, not $k^2+1$. However, for $d$ such that there are solutions to $x^2-dy^2 = -3$, such as $d=7$, then there ...
user avatar

Only top scored, non community-wiki answers of a minimum length are eligible