# Integer equation

• Apr 18th 2009, 03:43 PM
maria18
Integer equation
Hello, I need your help, thanks

- Solve in $\mathbb{R}$ the equation : $\lfloor x^2 - 2x\rfloor + 2\lfloor x\rfloor = {\lfloor x\rfloor}^2$ .
• Apr 18th 2009, 04:24 PM
hlolli
$x\in\mathbb{R}$
Try any number, this equation will obiously always end with x=x
• Apr 18th 2009, 05:07 PM
NonCommAlg
Quote:

Originally Posted by maria18
Hello, I need your help, thanks

- Solve in $\mathbb{R}$ the equation : $\lfloor x^2 - 2x\rfloor + 2\lfloor x\rfloor = {\lfloor x\rfloor}^2$ .

clearly every integer is a solution. so i'll assume that $x \notin \mathbb{Z}.$ let $[x]=n$ and $x=n+\theta,$ where $0 < \theta < 1.$ then $x$ is a solution if and only if: $0 \leq \theta^2 + 2(n-1) \theta < 1.$

so $\theta^2 + 2(n-1) \theta = \theta(\theta + 2n - 2) \geq 0.$ hence $0 \leq \theta + 2n - 2 < 2n - 1.$ thus we must have $n \geq 1.$ the converse is obviously true, i.e. if $n \geq 1,$ then $\theta^2 + 2(n-1) \theta \geq 0.$

now, from quadratic formula it's clear that $\theta^2 + 2(n-1)\theta - 1 < 0,$ if and only if $0 < \theta < 1 - n + \sqrt{n^2 - 2n + 2}.$ thus $n < x < 1+ \sqrt{n^2 - 2n + 2}.$ so the general solution of

your equation is: $\mathbb{Z} \cup \bigcup_{n=1}^{\infty} (n, 1 + \sqrt{n^2 - 2n + 2}).$

by the way questions of this type have nothing to do with calculus and they should be posted in high school pre-algebra subforum.
• Apr 18th 2009, 05:31 PM
hlolli
Quote:

Originally Posted by NonCommAlg
clearly every integer is a solution. so i'll assume that $x \notin \mathbb{Z}.$ let $[x]=n$ and $x=n+\theta,$ where $0 < \theta < 1.$ then $x$ is a solution if and only if: $0 \leq \theta^2 + 2(n-1) \theta < 1.$

so $\theta^2 + 2(n-1) \theta = \theta(\theta + 2n - 2) \geq 0.$ hence $0 \leq \theta + 2n - 2 < 2n - 1.$ thus we must have $n \geq 1.$ the converse is obviously true, i.e. if $n \geq 1,$ then $\theta^2 + 2(n-1) \theta \geq 0.$

now, from quadratic formula it's clear that $\theta^2 + 2(n-1)\theta - 1 < 0,$ if and only if $0 < \theta < 1 - n + \sqrt{n^2 - 2n + 2}.$ thus $n < x < 1+ \sqrt{n^2 - 2n + 2}.$ so the general solution of

your equation is: $\mathbb{Z} \cup \bigcup_{n=1}^{\infty} (n, 1 + \sqrt{n^2 - 2n + 2}).$

by the way questions of this type have nothing to do with calculus and they should be posted in high school pre-algebra subforum.

• Apr 19th 2009, 03:11 PM
maria18
Quote:

Originally Posted by NonCommAlg
clearly every integer is a solution. so i'll assume that $x \notin \mathbb{Z}.$ let $[x]=n$ and $x=n+\theta,$ where $0 < \theta < 1.$ then $x$ is a solution if and only if: $0 \leq \theta^2 + 2(n-1) \theta < 1.$

so $\theta^2 + 2(n-1) \theta = \theta(\theta + 2n - 2) \geq 0.$ hence $0 \leq \theta + 2n - 2 < 2n - 1.$ thus we must have $n \geq 1.$ the converse is obviously true, i.e. if $n \geq 1,$ then $\theta^2 + 2(n-1) \theta \geq 0.$

now, from quadratic formula it's clear that $\theta^2 + 2(n-1)\theta - 1 < 0,$ if and only if $0 < \theta < 1 - n + \sqrt{n^2 - 2n + 2}.$ thus $n < x < 1+ \sqrt{n^2 - 2n + 2}.$ so the general solution of

your equation is: $\mathbb{Z} \cup \bigcup_{n=1}^{\infty} (n, 1 + \sqrt{n^2 - 2n + 2}).$

by the way questions of this type have nothing to do with calculus and they should be posted in high school pre-algebra subforum.

1) Hello NonCommAlg and thanks, so i dont understand why $x$ is a solution if and only if: $0 \leq \theta^2 + 2(n-1) \theta < 1.$ ? Thanks

2) I'm seeing that $\lfloor x^2-2x\rfloor + 2\lfloor x\rfloor = {\lfloor x\rfloor}^2 \iff \lfloor (x - 1)^2 \rfloor = (\lfloor x - 1 \rfloor)^2$ so Mr NonCommAlg we can use the same proof that $\lfloor x^2 \rfloor = (\lfloor x \rfloor)^2$ ? look here ==> http://www.mathhelpforum.com/math-he...equations.html