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 4 hours ago
DiehardwalnutDiehardwalnut
257110
$endgroup$
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 4 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
4 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
4 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
2 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 4 hours ago
DiehardwalnutDiehardwalnut
257110
$endgroup$
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 4 hours ago
DiehardwalnutDiehardwalnut
257110
asked 4 hours ago
DiehardwalnutDiehardwalnut
257110
edited 2 hours ago
Diehardwalnut
asked 4 hours ago
DiehardwalnutDiehardwalnut
257110
asked 4 hours ago
DiehardwalnutDiehardwalnut
257110
asked 4 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
4 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
4 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
2 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
4 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
4 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
2 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
4 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
4 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
4 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
4 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
2 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
2 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 $$mid;{2x-3}mid>mid x;mid text{ unless } xin[1, 3]$$ $$mid;{2x-1}mid>mid x;mid text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.
answered 4 hours ago
Dr. MathvaDr. Mathva
3,215630
$endgroup$
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 4 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 2 hours ago
J. W. TannerJ. W. Tanner
4,4691320
$endgroup$
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 $$mid;{2x-3}mid>mid x;mid text{ unless } xin[1, 3]$$ $$mid;{2x-1}mid>mid x;mid text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.
answered 4 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 $xin{1, 2, 3}$.
answered 4 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 $xin{1, 2, 3}$.
answered 4 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 $$x^2-x+1=(x-1)^2+color{blue}x$$
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 $xin{1, 2, 3}$.
answered 4 hours ago
Dr. MathvaDr. Mathva
3,215630
edited 3 hours ago
answered 4 hours ago
Dr. MathvaDr. Mathva
3,215630
answered 4 hours ago
Dr. MathvaDr. Mathva
3,215630
answered 4 hours ago
Dr. MathvaDr. Mathva
3,215630
3,215630
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 4 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 4 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 4 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 4 hours ago
ThéophileThéophile
20.4k13047
answered 4 hours ago
ThéophileThéophile
20.4k13047
answered 4 hours ago
ThéophileThéophile
20.4k13047
answered 4 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 2 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 2 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 2 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 2 hours ago
J. W. TannerJ. W. Tanner
4,4691320
answered 2 hours ago
J. W. TannerJ. W. Tanner
4,4691320
answered 2 hours ago
J. W. TannerJ. W. Tanner
4,4691320
answered 2 hours ago
J. W. TannerJ. W. Tanner
4,4691320
4,4691320
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
4 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
4 hours ago
$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
2 hours ago