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Divisibility and number theory in terms of a and b

Simple Division ProofShow that if $x,y,z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square iff $xy +1, yz +1, zx+1$ are all perfect squares.Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$Show that there are infinitely many pairs (a,b) such that both $x^2+ax+b$ and $x^2+2ax+b$ have integer rootsMaximal Consecutive Integer Sequencenumber of coprime divisors of n with their difference divisible by 3For any positive integer $a$, prove that there exists infinitely many composite $n$ such that $a^n-1equiv 1mod n$.

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5

1

$begingroup$

Are there infinitely many pairs of $(a, b)$ of relatively prime integers $a > 1$ and $b > 1$ such that $a^b+b^a$ is divisible by $a+b$?

I've spent almost two hours on this question to no avail. The small cases I have tried, such as $(3, 5)$, $(3, 7)$ and $(5, 7)$ all work, so I'm conjecturing that the answer is yes. However, all methods of proof I have tried have failed. Tried modulo arguments but can't really simplify my results.

Any help would be greatly appreciated.

number-theory divisibility coprime

share|cite|improve this question

asked Sep 14 at 14:26

EthanMathsEthanMaths

5344 bronze badges

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• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you show in detail an example/examples of your attempted proofs?
  $endgroup$
  – TheSimpliFire
  Sep 14 at 14:32
  

  
  
  
  

add a comment
 | 

5

1

$begingroup$

Are there infinitely many pairs of $(a, b)$ of relatively prime integers $a > 1$ and $b > 1$ such that $a^b+b^a$ is divisible by $a+b$?

I've spent almost two hours on this question to no avail. The small cases I have tried, such as $(3, 5)$, $(3, 7)$ and $(5, 7)$ all work, so I'm conjecturing that the answer is yes. However, all methods of proof I have tried have failed. Tried modulo arguments but can't really simplify my results.

Any help would be greatly appreciated.

number-theory divisibility coprime

share|cite|improve this question

asked Sep 14 at 14:26

EthanMathsEthanMaths

5344 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can you show in detail an example/examples of your attempted proofs?
  $endgroup$
  – TheSimpliFire
  Sep 14 at 14:32
  

  
  
  
  

add a comment
 | 

$begingroup$

Are there infinitely many pairs of $(a, b)$ of relatively prime integers $a > 1$ and $b > 1$ such that $a^b+b^a$ is divisible by $a+b$?

I've spent almost two hours on this question to no avail. The small cases I have tried, such as $(3, 5)$, $(3, 7)$ and $(5, 7)$ all work, so I'm conjecturing that the answer is yes. However, all methods of proof I have tried have failed. Tried modulo arguments but can't really simplify my results.

Any help would be greatly appreciated.

number-theory divisibility coprime

share|cite|improve this question

asked Sep 14 at 14:26

EthanMathsEthanMaths

5344 bronze badges

$endgroup$

share|cite|improve this question

asked Sep 14 at 14:26

EthanMathsEthanMaths

5344 bronze badges

share|cite|improve this question

asked Sep 14 at 14:26

EthanMathsEthanMaths

5344 bronze badges

asked Sep 14 at 14:26

EthanMathsEthanMaths

5344 bronze badges

asked Sep 14 at 14:26

EthanMathsEthanMaths

5344 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

Sil posted a now deleted almost correct answer, so this is based on those ideas. (The answer has since been undeleted and corrected)

Let $a=2k-1$, $b=2k+1$ for any integer $k$. $a,b$ are coprime. $a+b=4k$.

Note that
$$ab=4k^2-1 equiv -1 pmod4k,$$
$$a^2 = 4k^2 - 4k +1 equiv 1 pmod 4k,$$
$$b^2 = 4k^2 + 4k +1 equiv 1 pmod 4k.$$
Since $abequiv -1pmod4k$, $b^-1 equiv -apmod4k$.

Then
$$
a^b+b^a
=a^2k+1+b^2k-1
equiv
a^1 + b^-1
equiv
a-a
=0pmod4k.
$$

Thus $a^b+b^a equiv 0pmoda+b$, so $a+bmid a^b+b^a$.

Hence we have found infinitely many such pairs.

share|cite|improve this answer

edited Sep 14 at 15:18

answered Sep 14 at 15:07

jgonjgon

20k33 gold badges2525 silver badges4848 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, this is more elegant then my (now hopefully fixed) argument (I had to use binomial theorem to get from $2k$ modulus to $4k$).
  $endgroup$
  – Sil
  Sep 14 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil The binomial theorem argument is also pretty nice, (+1)
  $endgroup$
  – jgon
  Sep 14 at 15:18
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Sure, choose for example $a=2k-1,b=2k+1$ for any integer $k$. Clearly these pairs are coprime, and we have

beginalign
a^b+b^a&=(2k-1)^2k+1+(2k+1)^2k-1\
&equiv -1+(2k+1)2k+1+(2k-1)2ktag*\
&=8k^2\
&equiv 0 pmod4k.
endalign

where $(*)$ follows from Binomial theorem for example (or see jgon's answer for alternative argument). So $a^b+b^a equiv 0 bmod a+b$ since $a+b=4k$.

share|cite|improve this answer

edited Sep 14 at 15:14

answered Sep 14 at 14:58

SilSil

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add a comment
 | 

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4

$begingroup$

Sil posted a now deleted almost correct answer, so this is based on those ideas. (The answer has since been undeleted and corrected)

Let $a=2k-1$, $b=2k+1$ for any integer $k$. $a,b$ are coprime. $a+b=4k$.

Note that
$$ab=4k^2-1 equiv -1 pmod4k,$$
$$a^2 = 4k^2 - 4k +1 equiv 1 pmod 4k,$$
$$b^2 = 4k^2 + 4k +1 equiv 1 pmod 4k.$$
Since $abequiv -1pmod4k$, $b^-1 equiv -apmod4k$.

Then
$$
a^b+b^a
=a^2k+1+b^2k-1
equiv
a^1 + b^-1
equiv
a-a
=0pmod4k.
$$

Thus $a^b+b^a equiv 0pmoda+b$, so $a+bmid a^b+b^a$.

Hence we have found infinitely many such pairs.

share|cite|improve this answer

edited Sep 14 at 15:18

answered Sep 14 at 15:07

jgonjgon

20k33 gold badges2525 silver badges4848 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, this is more elegant then my (now hopefully fixed) argument (I had to use binomial theorem to get from $2k$ modulus to $4k$).
  $endgroup$
  – Sil
  Sep 14 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil The binomial theorem argument is also pretty nice, (+1)
  $endgroup$
  – jgon
  Sep 14 at 15:18
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Sil posted a now deleted almost correct answer, so this is based on those ideas. (The answer has since been undeleted and corrected)

Let $a=2k-1$, $b=2k+1$ for any integer $k$. $a,b$ are coprime. $a+b=4k$.

Note that
$$ab=4k^2-1 equiv -1 pmod4k,$$
$$a^2 = 4k^2 - 4k +1 equiv 1 pmod 4k,$$
$$b^2 = 4k^2 + 4k +1 equiv 1 pmod 4k.$$
Since $abequiv -1pmod4k$, $b^-1 equiv -apmod4k$.

Then
$$
a^b+b^a
=a^2k+1+b^2k-1
equiv
a^1 + b^-1
equiv
a-a
=0pmod4k.
$$

Thus $a^b+b^a equiv 0pmoda+b$, so $a+bmid a^b+b^a$.

Hence we have found infinitely many such pairs.

share|cite|improve this answer

edited Sep 14 at 15:18

answered Sep 14 at 15:07

jgonjgon

20k33 gold badges2525 silver badges4848 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, this is more elegant then my (now hopefully fixed) argument (I had to use binomial theorem to get from $2k$ modulus to $4k$).
  $endgroup$
  – Sil
  Sep 14 at 15:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Sil The binomial theorem argument is also pretty nice, (+1)
  $endgroup$
  – jgon
  Sep 14 at 15:18
  

  
  
  
  

add a comment
 | 

$begingroup$

Sil posted a now deleted almost correct answer, so this is based on those ideas. (The answer has since been undeleted and corrected)

Let $a=2k-1$, $b=2k+1$ for any integer $k$. $a,b$ are coprime. $a+b=4k$.

Note that
$$ab=4k^2-1 equiv -1 pmod4k,$$
$$a^2 = 4k^2 - 4k +1 equiv 1 pmod 4k,$$
$$b^2 = 4k^2 + 4k +1 equiv 1 pmod 4k.$$
Since $abequiv -1pmod4k$, $b^-1 equiv -apmod4k$.

Then
$$
a^b+b^a
=a^2k+1+b^2k-1
equiv
a^1 + b^-1
equiv
a-a
=0pmod4k.
$$

Thus $a^b+b^a equiv 0pmoda+b$, so $a+bmid a^b+b^a$.

Hence we have found infinitely many such pairs.

share|cite|improve this answer

edited Sep 14 at 15:18

answered Sep 14 at 15:07

jgonjgon

20k33 gold badges2525 silver badges4848 bronze badges

$endgroup$

share|cite|improve this answer

edited Sep 14 at 15:18

answered Sep 14 at 15:07

jgonjgon

20k33 gold badges2525 silver badges4848 bronze badges

answered Sep 14 at 15:07

jgonjgon

20k33 gold badges2525 silver badges4848 bronze badges

answered Sep 14 at 15:07

jgonjgon

20k33 gold badges2525 silver badges4848 bronze badges

4

$begingroup$

Sure, choose for example $a=2k-1,b=2k+1$ for any integer $k$. Clearly these pairs are coprime, and we have

beginalign
a^b+b^a&=(2k-1)^2k+1+(2k+1)^2k-1\
&equiv -1+(2k+1)2k+1+(2k-1)2ktag*\
&=8k^2\
&equiv 0 pmod4k.
endalign

where $(*)$ follows from Binomial theorem for example (or see jgon's answer for alternative argument). So $a^b+b^a equiv 0 bmod a+b$ since $a+b=4k$.

share|cite|improve this answer

edited Sep 14 at 15:14

answered Sep 14 at 14:58

SilSil

7,05622 gold badges1919 silver badges4646 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Sure, choose for example $a=2k-1,b=2k+1$ for any integer $k$. Clearly these pairs are coprime, and we have

beginalign
a^b+b^a&=(2k-1)^2k+1+(2k+1)^2k-1\
&equiv -1+(2k+1)2k+1+(2k-1)2ktag*\
&=8k^2\
&equiv 0 pmod4k.
endalign

where $(*)$ follows from Binomial theorem for example (or see jgon's answer for alternative argument). So $a^b+b^a equiv 0 bmod a+b$ since $a+b=4k$.

share|cite|improve this answer

edited Sep 14 at 15:14

answered Sep 14 at 14:58

SilSil

7,05622 gold badges1919 silver badges4646 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Sure, choose for example $a=2k-1,b=2k+1$ for any integer $k$. Clearly these pairs are coprime, and we have

beginalign
a^b+b^a&=(2k-1)^2k+1+(2k+1)^2k-1\
&equiv -1+(2k+1)2k+1+(2k-1)2ktag*\
&=8k^2\
&equiv 0 pmod4k.
endalign

where $(*)$ follows from Binomial theorem for example (or see jgon's answer for alternative argument). So $a^b+b^a equiv 0 bmod a+b$ since $a+b=4k$.

share|cite|improve this answer

edited Sep 14 at 15:14

answered Sep 14 at 14:58

SilSil

7,05622 gold badges1919 silver badges4646 bronze badges

$endgroup$

share|cite|improve this answer

edited Sep 14 at 15:14

answered Sep 14 at 14:58

SilSil

7,05622 gold badges1919 silver badges4646 bronze badges

answered Sep 14 at 14:58

SilSil

7,05622 gold badges1919 silver badges4646 bronze badges

answered Sep 14 at 14:58

SilSil

7,05622 gold badges1919 silver badges4646 bronze badges