Is it possible for a square root function,f(x), to map to a finite number of integers for all x in domain of...
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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
$begingroup$
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
asked 10 hours ago
DiehardwalnutDiehardwalnut
262110
$endgroup$
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
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
9 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
asked 10 hours ago
DiehardwalnutDiehardwalnut
262110
$endgroup$
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
elementary-number-theory
asked 10 hours ago
DiehardwalnutDiehardwalnut
262110
asked 10 hours ago
DiehardwalnutDiehardwalnut
262110
edited 9 hours ago
Diehardwalnut
asked 10 hours ago
DiehardwalnutDiehardwalnut
262110
asked 10 hours ago
DiehardwalnutDiehardwalnut
262110
asked 10 hours ago
DiehardwalnutDiehardwalnut
262110
262110
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
9 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
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
9 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
10 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
10 hours ago
1
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
$begingroup$
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
9 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
9 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 $$x^2-x+1=(x-1)^2+color{blue}x$$
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 $xin{1, 2, 3}$.
edited 5 hours ago
Carmeister
2,8592924
answered 10 hours ago
Dr. MathvaDr. Mathva
3,225630
$endgroup$
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)?
answered 10 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 9 hours ago
J. W. TannerJ. W. Tanner
4,4791320
$endgroup$
add a comment |
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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 $$x^2-x+1=(x-1)^2+color{blue}x$$
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 $xin{1, 2, 3}$.
edited 5 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 $xin{1, 2, 3}$.
edited 5 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 $xin{1, 2, 3}$.
edited 5 hours ago
Carmeister
2,8592924
answered 10 hours ago
Dr. MathvaDr. Mathva
3,225630
$endgroup$
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 $$x^2-x+1=(x-1)^2+color{blue}x$$
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 $xin{1, 2, 3}$.
edited 5 hours ago
Carmeister
2,8592924
answered 10 hours ago
Dr. MathvaDr. Mathva
3,225630
edited 5 hours ago
Carmeister
2,8592924
edited 5 hours ago
Carmeister
2,8592924
edited 5 hours ago
Carmeister
2,8592924
2,8592924
answered 10 hours ago
Dr. MathvaDr. Mathva
3,225630
answered 10 hours ago
Dr. MathvaDr. Mathva
3,225630
answered 10 hours ago
Dr. MathvaDr. Mathva
3,225630
3,225630
add a comment |
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)?
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)?
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)?
answered 10 hours ago
ThéophileThéophile
20.4k13047
$endgroup$
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)?
answered 10 hours ago
ThéophileThéophile
20.4k13047
answered 10 hours ago
ThéophileThéophile
20.4k13047
answered 10 hours ago
ThéophileThéophile
20.4k13047
answered 10 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 9 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 $x<0$, $x^2<x^2-x+1<(x-1)^2$.
answered 9 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 $x<0$, $x^2<x^2-x+1<(x-1)^2$.
answered 9 hours ago
J. W. TannerJ. W. Tanner
4,4791320
$endgroup$
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 9 hours ago
J. W. TannerJ. W. Tanner
4,4791320
answered 9 hours ago
J. W. TannerJ. W. Tanner
4,4791320
answered 9 hours ago
J. W. TannerJ. W. Tanner
4,4791320
answered 9 hours ago
J. W. TannerJ. W. Tanner
4,4791320
4,4791320
add a comment |
add a comment |
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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
9 hours ago