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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

1

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 3 hours ago

Diehardwalnut

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 $x=(1+sqrt{13})/2$, 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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

1

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 3 hours ago

Diehardwalnut

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 $x=(1+sqrt{13})/2$, 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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Justed edited, yes I meant to say $$x in mathbb{Z}$$
  $endgroup$
  – Diehardwalnut
  3 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 3 hours ago

Diehardwalnut

asked 5 hours ago

DiehardwalnutDiehardwalnut

257110

$endgroup$

share|cite|improve this question

edited 3 hours ago

Diehardwalnut

asked 5 hours ago

DiehardwalnutDiehardwalnut

257110

share|cite|improve this question

edited 3 hours ago

Diehardwalnut

asked 5 hours ago

DiehardwalnutDiehardwalnut

257110

asked 5 hours ago

DiehardwalnutDiehardwalnut

257110

asked 5 hours ago

DiehardwalnutDiehardwalnut

257110

3 Answers
3

active

oldest

votes

3

$begingroup$

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 $$mid;{2x-3}mid>mid x;mid text{ unless } xin[1, 3]$$ $$mid;{2x-1}mid>mid x;mid text{ unless } xin[frac{1}{3}, 1]$$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

answered 5 hours ago

Dr. MathvaDr. Mathva

3,215630

$endgroup$

add a comment | 

2

$begingroup$

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 5 hours ago

ThéophileThéophile

20.4k13047

$endgroup$

add a comment | 

0

$begingroup$

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 3 hours ago

J. W. TannerJ. W. Tanner

4,4691320

$endgroup$

add a comment | 

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3

$begingroup$

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 $$mid;{2x-3}mid>mid x;mid text{ unless } xin[1, 3]$$ $$mid;{2x-1}mid>mid x;mid text{ unless } xin[frac{1}{3}, 1]$$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

answered 5 hours ago

Dr. MathvaDr. Mathva

3,215630

$endgroup$

add a comment | 

3

$begingroup$

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 $$mid;{2x-3}mid>mid x;mid text{ unless } xin[1, 3]$$ $$mid;{2x-1}mid>mid x;mid text{ unless } xin[frac{1}{3}, 1]$$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

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 $$x^2-x+1=(x-1)^2+color{blue}x$$

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 $$mid;{2x-3}mid>mid x;mid text{ unless } xin[1, 3]$$ $$mid;{2x-1}mid>mid x;mid text{ unless } xin[frac{1}{3}, 1]$$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

answered 5 hours ago

Dr. MathvaDr. Mathva

3,215630

$endgroup$

share|cite|improve this answer

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

2

$begingroup$

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 5 hours ago

ThéophileThéophile

20.4k13047

$endgroup$

add a comment | 

2

$begingroup$

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 5 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 5 hours ago

ThéophileThéophile

20.4k13047

$endgroup$

share|cite|improve this answer

answered 5 hours ago

ThéophileThéophile

20.4k13047

answered 5 hours ago

ThéophileThéophile

20.4k13047

answered 5 hours ago

ThéophileThéophile

20.4k13047

0

$begingroup$

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 3 hours ago

J. W. TannerJ. W. Tanner

4,4691320

$endgroup$

add a comment | 

0

$begingroup$

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

share|cite|improve this answer

answered 3 hours ago

J. W. TannerJ. W. Tanner

4,4691320

$endgroup$

share|cite|improve this answer

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