Kdtyjb

Can you really stack all of this on an Opportunity Attack?

How to draw a waving flag in TikZ

When a company launches a new product do they "come out" with a new product or do they "come up" with a new product?

Meaning of に in 本当に

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

RSA: Danger of using p to create q

Codimension of non-flat locus

Replacing matching entries in one column of a file by another column from a different file

Is it possible to record a short contained sound no longer than 60 milliseconds?

Approximately how much travel time was saved by the opening of the Suez Canal in 1869?

Why is Minecraft giving an OpenGL error?

What is a clear way to write a bar that has an extra beat?

Watching something be written to a file live with tail

Add text to same line using sed

How to determine what difficulty is right for the game?

How to format long polynomial?

Languages that we cannot (dis)prove to be Context-Free

Why can't I see bouncing of a switch on an oscilloscope?

Perform and show arithmetic with LuaLaTeX

How to efficiently unroll a matrix by value with numpy?

dbcc cleantable batch size explanation

How do I gain back my faith in my PhD degree?

Modeling an IP Address

How can bays and straits be determined in a procedurally generated map?

Is it possible for a square root function,f(x), to map to a finite number of integers for all x in domain of f?

For integers $a$ and $b gt 0$, and $n^2$ a sum of two square integers, does this strategy find the largest integer $x | x^2 lt n^2(a^2 + b^2)$?Do roots of a polynomial with coefficients from a Collatz sequence all fall in a disk of radius 1.5?A fun little problemWhat is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer?on roots of an equationAn integer sequence with integer $k$ normsProof Verification: If $x$ is a nonnegative real number, then $big[sqrt{[x]}big] = big[sqrt{x}big]$Digit after decimal point of radicalsProving that there does not exist an infinite descending sequence of naturals using minimal counterexamplenumber of different ways to represent a positive integer as a binomial coefficient

2

1

$begingroup$

Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$

Using python I checked x for $$ -100000000 leq x leq 100000000$$

and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?

Edit: $$x in mathbb{Z}$$

elementary-number-theory

share|cite|improve this question

edited 8 hours ago

Diehardwalnut

asked 10 hours ago

DiehardwalnutDiehardwalnut

262110

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
  $endgroup$
  – Théophile
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
  $endgroup$
  – Arturo Magidin
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Justed edited, yes I meant to say $$x in mathbb{Z}$$
  $endgroup$
  – Diehardwalnut
  8 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$

Using python I checked x for $$ -100000000 leq x leq 100000000$$

and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?

Edit: $$x in mathbb{Z}$$

elementary-number-theory

share|cite|improve this question

edited 8 hours ago

Diehardwalnut

asked 10 hours ago

DiehardwalnutDiehardwalnut

262110

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
  $endgroup$
  – Théophile
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
  $endgroup$
  – Arturo Magidin
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Justed edited, yes I meant to say $$x in mathbb{Z}$$
  $endgroup$
  – Diehardwalnut
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$

Using python I checked x for $$ -100000000 leq x leq 100000000$$

and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?

Edit: $$x in mathbb{Z}$$

elementary-number-theory

share|cite|improve this question

edited 8 hours ago

Diehardwalnut

asked 10 hours ago

DiehardwalnutDiehardwalnut

262110

$endgroup$

share|cite|improve this question

edited 8 hours ago

Diehardwalnut

asked 10 hours ago

DiehardwalnutDiehardwalnut

262110

share|cite|improve this question

edited 8 hours ago

Diehardwalnut

asked 10 hours ago

DiehardwalnutDiehardwalnut

262110

asked 10 hours ago

DiehardwalnutDiehardwalnut

262110

asked 10 hours ago

DiehardwalnutDiehardwalnut

262110

3 Answers
3

active

oldest

votes

4

$begingroup$

First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.

Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$

It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.

Furthermore
begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
end{align*}

Can you end it now?

Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.

share|cite|improve this answer

edited 4 hours ago

Carmeister

2,8592924

answered 10 hours ago

Dr. MathvaDr. Mathva

3,225630

$endgroup$

add a comment | 

3

$begingroup$

Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
$$y^2 = x^2-x+1,$$
so $y^2-1=x^2-x$. That is,
$$(y+1)(y-1) = x(x-1).$$

So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.4k13047

$endgroup$

add a comment | 

1

$begingroup$

Hint:

For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;

for $x<0$, $x^2<x^2-x+1<(x-1)^2$.

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

4,4791320

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3176345%2fis-it-possible-for-a-square-root-function-fx-to-map-to-a-finite-number-of-int%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.

Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$

It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.

Furthermore
begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
end{align*}

Can you end it now?

Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.

share|cite|improve this answer

edited 4 hours ago

Carmeister

2,8592924

answered 10 hours ago

Dr. MathvaDr. Mathva

3,225630

$endgroup$

add a comment | 

4

$begingroup$

First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.

Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$

It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.

Furthermore
begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
end{align*}

Can you end it now?

Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.

share|cite|improve this answer

edited 4 hours ago

Carmeister

2,8592924

answered 10 hours ago

Dr. MathvaDr. Mathva

3,225630

$endgroup$

add a comment | 

$begingroup$

First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.

Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$

It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.

Furthermore
begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
end{align*}

Can you end it now?

Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.

share|cite|improve this answer

edited 4 hours ago

Carmeister

2,8592924

answered 10 hours ago

Dr. MathvaDr. Mathva

3,225630

$endgroup$

share|cite|improve this answer

edited 4 hours ago

Carmeister

2,8592924

answered 10 hours ago

Dr. MathvaDr. Mathva

3,225630

edited 4 hours ago

Carmeister

2,8592924

edited 4 hours ago

Carmeister

2,8592924

answered 10 hours ago

Dr. MathvaDr. Mathva

3,225630

answered 10 hours ago

Dr. MathvaDr. Mathva

3,225630

3

$begingroup$

Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
$$y^2 = x^2-x+1,$$
so $y^2-1=x^2-x$. That is,
$$(y+1)(y-1) = x(x-1).$$

So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.4k13047

$endgroup$

add a comment | 

3

$begingroup$

Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
$$y^2 = x^2-x+1,$$
so $y^2-1=x^2-x$. That is,
$$(y+1)(y-1) = x(x-1).$$

So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.4k13047

$endgroup$

add a comment | 

$begingroup$

Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
$$y^2 = x^2-x+1,$$
so $y^2-1=x^2-x$. That is,
$$(y+1)(y-1) = x(x-1).$$

So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.4k13047

$endgroup$

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.4k13047

answered 10 hours ago

ThéophileThéophile

20.4k13047

answered 10 hours ago

ThéophileThéophile

20.4k13047

1

$begingroup$

Hint:

For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;

for $x<0$, $x^2<x^2-x+1<(x-1)^2$.

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

4,4791320

$endgroup$

add a comment | 

1

$begingroup$

Hint:

For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;

for $x<0$, $x^2<x^2-x+1<(x-1)^2$.

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

4,4791320

$endgroup$

add a comment | 

$begingroup$

Hint:

For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;

for $x<0$, $x^2<x^2-x+1<(x-1)^2$.

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

4,4791320

$endgroup$

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

4,4791320

answered 8 hours ago

J. W. TannerJ. W. Tanner

4,4791320

answered 8 hours ago

J. W. TannerJ. W. Tanner

4,4791320