Fdtxjr

Can a virus destroy the BIOS of a modern computer?

Should I tell management that I intend to leave due to bad software development practices?

Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?

I would say: "You are another teacher", but she is a woman and I am a man

Avoiding the "not like other girls" trope?

How dangerous is XSS?

Examples of smooth manifolds admitting inbetween one and a continuum of complex structures

Detention in 1997

One verb to replace 'be a member of' a club

Dreadful Dastardly Diseases, or Always Atrocious Ailments

Extract rows of a table, that include less than x NULLs

Why would the Red Woman birth a shadow if she worshipped the Lord of the Light?

What is the difference between 仮定 and 想定?

Short story with a alien planet, government officials must wear exploding medallions

What is the most common color to indicate the input-field is disabled?

Can my sorcerer use a spellbook only to collect spells and scribe scrolls, not cast?

Size of subfigure fitting its content (tikzpicture)

How can saying a song's name be a copyright violation?

Avoiding direct proof while writing proof by induction

How would I stat a creature to be immune to everything but the Magic Missile spell? (just for fun)

What mechanic is there to disable a threat instead of killing it?

Is there a hemisphere-neutral way of specifying a season?

Why are the 737's rear doors unusable in a water landing?

What method can I use to design a dungeon difficult enough that the PCs can't make it through without killing them?

There exists a lattice point $(a,b)$ whose distance from every *visible* point is greater than $n$.

Prove that there exists a $m×m$ lattice square in the $x-y$ plane such that none of its coordinates are visibleEvery integer greater than 1 is divisible by a primesucessive primes with distance greater than kProve that for any natural number $n$ there exists a prime number $p$ greater than $n$Every prime divisor of $n! + 1$ is an odd integer greater than $n$A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 -y^2 = 2000^2$A new property of Rowland's sequence or there is a counterexample? Each element $a_i$ is a witness of a greater $a_j$ whose index $j$ is always primeProve that among any five consecutive positive integers there is one integer which is relatively prime to the other four integers.A question about lattice visible point in the planeIs there a prime $p$ whose successor is greater than $2p$?$IMO 2014$ shortlist question. PigeonHole-Principle to solve?

5

2

$begingroup$

Question- A lattice point $(x,y)inmathbbZ^2$ is called visible if $gcd(x,y)=1$. Prove that given a positive integer $n$, there exists a lattice point $(a,b)$ whose distance from every visible point is greater than $n$.

I am totally nowhere near progress on this. I was thinking of trying pigeon hole principle, but cant find any appropriate candidate pigeons. Please give any hints to start.

number-theory elementary-number-theory contest-math

share|cite|improve this question

edited Aug 21 '16 at 18:02

gambler101

asked Aug 17 '16 at 15:53

gambler101gambler101

707214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Anyone please help!
  $endgroup$
  – gambler101
  Aug 21 '16 at 12:33
  

  
  
  
  

add a comment | 

5

2

$begingroup$

Question- A lattice point $(x,y)inmathbbZ^2$ is called visible if $gcd(x,y)=1$. Prove that given a positive integer $n$, there exists a lattice point $(a,b)$ whose distance from every visible point is greater than $n$.

I am totally nowhere near progress on this. I was thinking of trying pigeon hole principle, but cant find any appropriate candidate pigeons. Please give any hints to start.

number-theory elementary-number-theory contest-math

share|cite|improve this question

edited Aug 21 '16 at 18:02

gambler101

asked Aug 17 '16 at 15:53

gambler101gambler101

707214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Anyone please help!
  $endgroup$
  – gambler101
  Aug 21 '16 at 12:33
  

  
  
  
  

add a comment | 

$begingroup$

Question- A lattice point $(x,y)inmathbbZ^2$ is called visible if $gcd(x,y)=1$. Prove that given a positive integer $n$, there exists a lattice point $(a,b)$ whose distance from every visible point is greater than $n$.

I am totally nowhere near progress on this. I was thinking of trying pigeon hole principle, but cant find any appropriate candidate pigeons. Please give any hints to start.

number-theory elementary-number-theory contest-math

share|cite|improve this question

edited Aug 21 '16 at 18:02

gambler101

asked Aug 17 '16 at 15:53

gambler101gambler101

707214

$endgroup$

share|cite|improve this question

edited Aug 21 '16 at 18:02

gambler101

asked Aug 17 '16 at 15:53

gambler101gambler101

707214

share|cite|improve this question

edited Aug 21 '16 at 18:02

gambler101

asked Aug 17 '16 at 15:53

gambler101gambler101

707214

asked Aug 17 '16 at 15:53

gambler101gambler101

707214

asked Aug 17 '16 at 15:53

gambler101gambler101

707214

1 Answer
1

active

oldest

votes

2

$begingroup$

This follows from Theorem 7.3 in the book “Mathematical Journeys,” by Peter D. Schumer. The proof, which uses the Chinese remainder theorem, can be seen on Google Books here or (summarized) by hovering over the spoiler paragraph below.

Schumer constructs an $ntimes n$ square of invisible lattice points as follows. (You can choose $(a,b)$ at the center of such a square of side $2n$.)

The proof goes like this:

Arrange the first $n^2$ primes in an $ntimes n$ array and denote by $m_i$ the product across the $i$-th row and by $M_i$ the product down the $i$-th column. Then consider two systems of congruences: The system $xequiv -i mod m_i$ and the system $yequiv -imod M_i$. Modulo $P=prod m_i=prod M_i$, these systems have unique solutions $xequiv_P a$ and $yequiv_P b$ (by the Chinese Reminder Theorem). Then the square with lower left corner $(a+1,b+1)$ and upper right corner $(a+n,b+n)$ contains no visible lattice points, because for any integers $i,j$ between $1$ and $n$, both $a+i$ and $b+j$ will be divisible by the prime $p$ in row $i$ and column $j$ of the array of primes.

share|cite|improve this answer

edited Aug 21 '16 at 19:12

answered Aug 21 '16 at 19:00

Steve KassSteve Kass

11.4k11530

$endgroup$

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%2f1895206%2fthere-exists-a-lattice-point-a-b-whose-distance-from-every-visible-point-i%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

This follows from Theorem 7.3 in the book “Mathematical Journeys,” by Peter D. Schumer. The proof, which uses the Chinese remainder theorem, can be seen on Google Books here or (summarized) by hovering over the spoiler paragraph below.

Schumer constructs an $ntimes n$ square of invisible lattice points as follows. (You can choose $(a,b)$ at the center of such a square of side $2n$.)

The proof goes like this:

Arrange the first $n^2$ primes in an $ntimes n$ array and denote by $m_i$ the product across the $i$-th row and by $M_i$ the product down the $i$-th column. Then consider two systems of congruences: The system $xequiv -i mod m_i$ and the system $yequiv -imod M_i$. Modulo $P=prod m_i=prod M_i$, these systems have unique solutions $xequiv_P a$ and $yequiv_P b$ (by the Chinese Reminder Theorem). Then the square with lower left corner $(a+1,b+1)$ and upper right corner $(a+n,b+n)$ contains no visible lattice points, because for any integers $i,j$ between $1$ and $n$, both $a+i$ and $b+j$ will be divisible by the prime $p$ in row $i$ and column $j$ of the array of primes.

share|cite|improve this answer

edited Aug 21 '16 at 19:12

answered Aug 21 '16 at 19:00

Steve KassSteve Kass

11.4k11530

$endgroup$

add a comment | 

2

$begingroup$

This follows from Theorem 7.3 in the book “Mathematical Journeys,” by Peter D. Schumer. The proof, which uses the Chinese remainder theorem, can be seen on Google Books here or (summarized) by hovering over the spoiler paragraph below.

Schumer constructs an $ntimes n$ square of invisible lattice points as follows. (You can choose $(a,b)$ at the center of such a square of side $2n$.)

The proof goes like this:

Arrange the first $n^2$ primes in an $ntimes n$ array and denote by $m_i$ the product across the $i$-th row and by $M_i$ the product down the $i$-th column. Then consider two systems of congruences: The system $xequiv -i mod m_i$ and the system $yequiv -imod M_i$. Modulo $P=prod m_i=prod M_i$, these systems have unique solutions $xequiv_P a$ and $yequiv_P b$ (by the Chinese Reminder Theorem). Then the square with lower left corner $(a+1,b+1)$ and upper right corner $(a+n,b+n)$ contains no visible lattice points, because for any integers $i,j$ between $1$ and $n$, both $a+i$ and $b+j$ will be divisible by the prime $p$ in row $i$ and column $j$ of the array of primes.

share|cite|improve this answer

edited Aug 21 '16 at 19:12

answered Aug 21 '16 at 19:00

Steve KassSteve Kass

11.4k11530

$endgroup$

add a comment | 

$begingroup$

This follows from Theorem 7.3 in the book “Mathematical Journeys,” by Peter D. Schumer. The proof, which uses the Chinese remainder theorem, can be seen on Google Books here or (summarized) by hovering over the spoiler paragraph below.

Schumer constructs an $ntimes n$ square of invisible lattice points as follows. (You can choose $(a,b)$ at the center of such a square of side $2n$.)

The proof goes like this:

Arrange the first $n^2$ primes in an $ntimes n$ array and denote by $m_i$ the product across the $i$-th row and by $M_i$ the product down the $i$-th column. Then consider two systems of congruences: The system $xequiv -i mod m_i$ and the system $yequiv -imod M_i$. Modulo $P=prod m_i=prod M_i$, these systems have unique solutions $xequiv_P a$ and $yequiv_P b$ (by the Chinese Reminder Theorem). Then the square with lower left corner $(a+1,b+1)$ and upper right corner $(a+n,b+n)$ contains no visible lattice points, because for any integers $i,j$ between $1$ and $n$, both $a+i$ and $b+j$ will be divisible by the prime $p$ in row $i$ and column $j$ of the array of primes.

share|cite|improve this answer

edited Aug 21 '16 at 19:12

answered Aug 21 '16 at 19:00

Steve KassSteve Kass

11.4k11530

$endgroup$

share|cite|improve this answer

edited Aug 21 '16 at 19:12

answered Aug 21 '16 at 19:00

Steve KassSteve Kass

11.4k11530

answered Aug 21 '16 at 19:00

Steve KassSteve Kass

11.4k11530

answered Aug 21 '16 at 19:00

Steve KassSteve Kass

11.4k11530