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 $big[sqrt[x]big] = big[sqrtxbig]$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
Perform and show arithmetic with LuaLaTeX
Modeling an IP Address
RSA: Danger of using p to create q
Why is consensus so controversial in Britain?
Is it possible to run Internet Explorer on OS X El Capitan?
Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?
Does an object always see its latest internal state irrespective of thread?
Rock identification in KY
Did Shadowfax go to Valinor?
How does quantile regression compare to logistic regression with the variable split at the quantile?
Maximum likelihood parameters deviate from posterior distributions
Codimension of non-flat locus
Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?
Is it inappropriate for a student to attend their mentor's dissertation defense?
How is it possible to have an ability score that is less than 3?
Was any UN Security Council vote triple-vetoed?
Java Casting: Java 11 throws LambdaConversionException while 1.8 does not
What does "Puller Prush Person" mean?
meaning of に in 本当に?
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?
Alternative to sending password over mail?
Can you really stack all of this on an Opportunity Attack?
DC-DC converter from low voltage at high current, to high voltage at low current
How does one intimidate enemies without having the capacity for violence?
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 $big[sqrt[x]big] = big[sqrtxbig]$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
$begingroup$
Consider the equation $$f(x) =sqrtx^2 - x + 1$$
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: $$x in mathbbZ$$
elementary-number-theory
asked 5 hours ago
DiehardwalnutDiehardwalnut
257110
$endgroup$
add a comment |
$begingroup$
Consider the equation $$f(x) =sqrtx^2 - x + 1$$
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: $$x in mathbbZ$$
elementary-number-theory
asked 5 hours ago
DiehardwalnutDiehardwalnut
257110
$endgroup$
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
5 hours ago
$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
5 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbbZ$$
$endgroup$
– Diehardwalnut
3 hours ago
add a comment |
$begingroup$
Consider the equation $$f(x) =sqrtx^2 - x + 1$$
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: $$x in mathbbZ$$
elementary-number-theory
asked 5 hours ago
DiehardwalnutDiehardwalnut
257110
$endgroup$
Consider the equation $$f(x) =sqrtx^2 - x + 1$$
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: $$x in mathbbZ$$
elementary-number-theory
elementary-number-theory
asked 5 hours ago
DiehardwalnutDiehardwalnut
257110
asked 5 hours ago
DiehardwalnutDiehardwalnut
257110
edited 3 hours ago
Diehardwalnut
asked 5 hours ago
DiehardwalnutDiehardwalnut
257110
asked 5 hours ago
DiehardwalnutDiehardwalnut
257110
asked 5 hours ago
DiehardwalnutDiehardwalnut
257110
257110
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
5 hours ago
$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
5 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbbZ$$
$endgroup$
– Diehardwalnut
3 hours ago
add a comment |
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
5 hours ago
$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
5 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbbZ$$
$endgroup$
– Diehardwalnut
3 hours ago
4
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
5 hours ago
$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
5 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbbZ$$
$endgroup$
– Diehardwalnut
3 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbbZ$$
$endgroup$
– Diehardwalnut
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $$x^2-x+1=(x-1)^2+colorbluex$$
It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.
Furthermore
beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
x^2-(x-1)^2&=colorblue2x-1tag2
endalign*
Can you end it now?
Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.
answered 5 hours ago
Dr. MathvaDr. Mathva
3,215630
$endgroup$
add a comment |
$begingroup$
Hint: let $y = sqrtx^2-x+1$. 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)?
answered 5 hours ago
ThéophileThéophile
20.4k13047
$endgroup$
add a comment |
$begingroup$
Hint:
For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;
for $x<0$, $x^2<x^2-x+1<(x-1)^2$.
answered 3 hours ago
J. W. TannerJ. W. Tanner
4,4691320
$endgroup$
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
);
);
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
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
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$x^2-x+1=(x-1)^2+colorbluex$$
It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.
Furthermore
beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
x^2-(x-1)^2&=colorblue2x-1tag2
endalign*
Can you end it now?
Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.
answered 5 hours ago
Dr. MathvaDr. Mathva
3,215630
$endgroup$
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 $$x^2-x+1=(x-1)^2+colorbluex$$
It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.
Furthermore
beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
x^2-(x-1)^2&=colorblue2x-1tag2
endalign*
Can you end it now?
Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.
answered 5 hours ago
Dr. MathvaDr. Mathva
3,215630
$endgroup$
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 $$x^2-x+1=(x-1)^2+colorbluex$$
It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.
Furthermore
beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
x^2-(x-1)^2&=colorblue2x-1tag2
endalign*
Can you end it now?
Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.
answered 5 hours ago
Dr. MathvaDr. Mathva
3,215630
$endgroup$
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 $$x^2-x+1=(x-1)^2+colorbluex$$
It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.
Furthermore
beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
x^2-(x-1)^2&=colorblue2x-1tag2
endalign*
Can you end it now?
Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.
answered 5 hours ago
Dr. MathvaDr. Mathva
3,215630
edited 4 hours ago
answered 5 hours ago
Dr. MathvaDr. Mathva
3,215630
answered 5 hours ago
Dr. MathvaDr. Mathva
3,215630
answered 5 hours ago
Dr. MathvaDr. Mathva
3,215630
3,215630
add a comment |
add a comment |
$begingroup$
Hint: let $y = sqrtx^2-x+1$. 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)?
answered 5 hours ago
ThéophileThéophile
20.4k13047
$endgroup$
add a comment |
$begingroup$
Hint: let $y = sqrtx^2-x+1$. 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)?
answered 5 hours ago
ThéophileThéophile
20.4k13047
$endgroup$
add a comment |
$begingroup$
Hint: let $y = sqrtx^2-x+1$. 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)?
answered 5 hours ago
ThéophileThéophile
20.4k13047
$endgroup$
Hint: let $y = sqrtx^2-x+1$. 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)?
answered 5 hours ago
ThéophileThéophile
20.4k13047
answered 5 hours ago
ThéophileThéophile
20.4k13047
answered 5 hours ago
ThéophileThéophile
20.4k13047
answered 5 hours ago
ThéophileThéophile
20.4k13047
20.4k13047
add a comment |
add a comment |
$begingroup$
Hint:
For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;
for $x<0$, $x^2<x^2-x+1<(x-1)^2$.
answered 3 hours ago
J. W. TannerJ. W. Tanner
4,4691320
$endgroup$
add a comment |
$begingroup$
Hint:
For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;
for $x<0$, $x^2<x^2-x+1<(x-1)^2$.
answered 3 hours ago
J. W. TannerJ. W. Tanner
4,4691320
$endgroup$
add a comment |
$begingroup$
Hint:
For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;
for $x<0$, $x^2<x^2-x+1<(x-1)^2$.
answered 3 hours ago
J. W. TannerJ. W. Tanner
4,4691320
$endgroup$
Hint:
For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;
for $x<0$, $x^2<x^2-x+1<(x-1)^2$.
answered 3 hours ago
J. W. TannerJ. W. Tanner
4,4691320
answered 3 hours ago
J. W. TannerJ. W. Tanner
4,4691320
answered 3 hours ago
J. W. TannerJ. W. Tanner
4,4691320
answered 3 hours ago
J. W. TannerJ. W. Tanner
4,4691320
4,4691320
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
5 hours ago
$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
5 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbbZ$$
$endgroup$
– Diehardwalnut
3 hours ago