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Find all [x;y] for which: GCD(x,y) + 5 = LCM(x,y)

Greatest Common Factors and Least Common MultiplesWhy don't all elements of an arithmetic progression divide the lcm of the start and step?LCM. What am I missing?Problem with equation containing $operatorname{lcm}$ and $gcd$$B$-powersmooth number divides $mathrm{lcm}(1,2,3,ldots B)$GCD of $a+b$ and $frac{a^p + b^p}{a+b}$If $gcd(|x|,|y|) = 1$ then $|xy| = mathrm{lcm}(|x|,|y|)$ in an abelian group.How many pairs of numbers exist,which satisfy the following conditions?computing difference between all pairs of numbers which is given in ascending orderProving If and only if gcd(a,b) = gcd(b,c) = 1, then gcd(ab,c) = 1

0

$begingroup$

Suppose we have two numbers (x and y) both of which are from the natural numbers. The task is:

Find all [x;y] pairs for which:

GCD(x,y) + 5 = LCM(x,y)

The result should be:
{[1;6], [6;1], [2;3], [3;2], [5;10], [10;5]}

The problem is I have no idea how to get to the result. I got stuck in a loop of statements. So I'll appreciate any input.

elementary-number-theory

share|cite|improve this question

asked Oct 16 '16 at 17:44

DeritusDeritus

125

$endgroup$

add a comment | 

0

$begingroup$

Suppose we have two numbers (x and y) both of which are from the natural numbers. The task is:

Find all [x;y] pairs for which:

GCD(x,y) + 5 = LCM(x,y)

The result should be:
{[1;6], [6;1], [2;3], [3;2], [5;10], [10;5]}

The problem is I have no idea how to get to the result. I got stuck in a loop of statements. So I'll appreciate any input.

elementary-number-theory

share|cite|improve this question

asked Oct 16 '16 at 17:44

DeritusDeritus

125

$endgroup$

add a comment | 

$begingroup$

Suppose we have two numbers (x and y) both of which are from the natural numbers. The task is:

Find all [x;y] pairs for which:

GCD(x,y) + 5 = LCM(x,y)

The result should be:
{[1;6], [6;1], [2;3], [3;2], [5;10], [10;5]}

The problem is I have no idea how to get to the result. I got stuck in a loop of statements. So I'll appreciate any input.

elementary-number-theory

share|cite|improve this question

asked Oct 16 '16 at 17:44

DeritusDeritus

125

$endgroup$

share|cite|improve this question

asked Oct 16 '16 at 17:44

DeritusDeritus

125

share|cite|improve this question

asked Oct 16 '16 at 17:44

DeritusDeritus

125

asked Oct 16 '16 at 17:44

DeritusDeritus

125

asked Oct 16 '16 at 17:44

DeritusDeritus

125

3 Answers
3

active

oldest

votes

0

$begingroup$

Note that the lcm is a multiple of the gcd, hence (as it is different) at least twice as big. This makes the lcm $le 10$, and leaves only $(1,6), (5,10)$ for $(gcd,operatorname{lcm})$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

Hagen von EitzenHagen von Eitzen

282k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Took me quite a while to see it but I finalyl got it. Thank you!
  $endgroup$
  – Deritus
  Oct 16 '16 at 18:34
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint:
Rewrite $x=md,y=nd$ where $gcd(m,n)=1$ then find that $d|5$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

arberavdullahuarberavdullahu

1,1571513

$endgroup$

add a comment | 

0

$begingroup$

Hint $newcommand{lcm}{operatorname{lcm}}$Since $gcd(x,y)midgcd(x,y)$ and $gcd(x,y) mid lcm(x,y)$ then we gave
$$gcd(x,y)midlcm(x,y)-gcd(x,y)=5$$.

This shows that $gcd(x,y) in {1,5 }$.

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Oct 16 '16 at 17:51

N. S.N. S.

104k7114209

$endgroup$

add a comment | 

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0

$begingroup$

Note that the lcm is a multiple of the gcd, hence (as it is different) at least twice as big. This makes the lcm $le 10$, and leaves only $(1,6), (5,10)$ for $(gcd,operatorname{lcm})$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

Hagen von EitzenHagen von Eitzen

282k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Took me quite a while to see it but I finalyl got it. Thank you!
  $endgroup$
  – Deritus
  Oct 16 '16 at 18:34
  

  
  
  
  

add a comment | 

0

$begingroup$

Note that the lcm is a multiple of the gcd, hence (as it is different) at least twice as big. This makes the lcm $le 10$, and leaves only $(1,6), (5,10)$ for $(gcd,operatorname{lcm})$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

Hagen von EitzenHagen von Eitzen

282k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Took me quite a while to see it but I finalyl got it. Thank you!
  $endgroup$
  – Deritus
  Oct 16 '16 at 18:34
  

  
  
  
  

add a comment | 

$begingroup$

Note that the lcm is a multiple of the gcd, hence (as it is different) at least twice as big. This makes the lcm $le 10$, and leaves only $(1,6), (5,10)$ for $(gcd,operatorname{lcm})$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

Hagen von EitzenHagen von Eitzen

282k23272507

$endgroup$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

Hagen von EitzenHagen von Eitzen

282k23272507

answered Oct 16 '16 at 17:48

Hagen von EitzenHagen von Eitzen

282k23272507

answered Oct 16 '16 at 17:48

Hagen von EitzenHagen von Eitzen

282k23272507

0

$begingroup$

Hint:
Rewrite $x=md,y=nd$ where $gcd(m,n)=1$ then find that $d|5$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

arberavdullahuarberavdullahu

1,1571513

$endgroup$

add a comment | 

0

$begingroup$

Hint:
Rewrite $x=md,y=nd$ where $gcd(m,n)=1$ then find that $d|5$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

arberavdullahuarberavdullahu

1,1571513

$endgroup$

add a comment | 

$begingroup$

Hint:
Rewrite $x=md,y=nd$ where $gcd(m,n)=1$ then find that $d|5$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

arberavdullahuarberavdullahu

1,1571513

$endgroup$

share|cite|improve this answer

answered Oct 16 '16 at 17:48

arberavdullahuarberavdullahu

1,1571513

answered Oct 16 '16 at 17:48

arberavdullahuarberavdullahu

1,1571513

answered Oct 16 '16 at 17:48

arberavdullahuarberavdullahu

1,1571513

0

$begingroup$

Hint $newcommand{lcm}{operatorname{lcm}}$Since $gcd(x,y)midgcd(x,y)$ and $gcd(x,y) mid lcm(x,y)$ then we gave
$$gcd(x,y)midlcm(x,y)-gcd(x,y)=5$$.

This shows that $gcd(x,y) in {1,5 }$.

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Oct 16 '16 at 17:51

N. S.N. S.

104k7114209

$endgroup$

add a comment | 

0

$begingroup$

Hint $newcommand{lcm}{operatorname{lcm}}$Since $gcd(x,y)midgcd(x,y)$ and $gcd(x,y) mid lcm(x,y)$ then we gave
$$gcd(x,y)midlcm(x,y)-gcd(x,y)=5$$.

This shows that $gcd(x,y) in {1,5 }$.

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Oct 16 '16 at 17:51

N. S.N. S.

104k7114209

$endgroup$

add a comment | 

$begingroup$

Hint $newcommand{lcm}{operatorname{lcm}}$Since $gcd(x,y)midgcd(x,y)$ and $gcd(x,y) mid lcm(x,y)$ then we gave
$$gcd(x,y)midlcm(x,y)-gcd(x,y)=5$$.

This shows that $gcd(x,y) in {1,5 }$.

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Oct 16 '16 at 17:51

N. S.N. S.

104k7114209

$endgroup$

share|cite|improve this answer

edited yesterday

Martin Sleziak

44.8k10119273

answered Oct 16 '16 at 17:51

N. S.N. S.

104k7114209

edited yesterday

Martin Sleziak

44.8k10119273

edited yesterday

Martin Sleziak

44.8k10119273

answered Oct 16 '16 at 17:51

N. S.N. S.

104k7114209

answered Oct 16 '16 at 17:51

N. S.N. S.

104k7114209