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A Diophantine rational function [closed]

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3

$begingroup$

For how many integers $N$ is the rational function $fracN^2-2N-15N^2-N-12$ also an integer?

mathematics

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asked Jun 6 at 9:53

Rand al'ThorRand al'Thor

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$endgroup$

closed as off-topic by greenturtle3141, Brandon_J, gabbo1092, Gamow, Glorfindel Jun 6 at 19:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – greenturtle3141, Brandon_J, gabbo1092, Gamow, Glorfindel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this is off-topic. The "obvious" way to approach this would be to write $N^2−2N−15=M(N^2−N−12)$ and then try to reduce possibilities for $M$ and $N$, perhaps by using modular arithmetic or even brute-forcing. The "aha moment" required by the linked policy is to factorise the polynomials, and the "unexpected result" is that only two values of $N$ work and it's easy to find which ones.
  $endgroup$
  – Rand al'Thor
  Jun 7 at 13:02
  

  
  
  
  

add a comment | 

3

$begingroup$

For how many integers $N$ is the rational function $fracN^2-2N-15N^2-N-12$ also an integer?

mathematics

share|improve this question

asked Jun 6 at 9:53

Rand al'ThorRand al'Thor

73.7k1515 gold badges244244 silver badges490490 bronze badges

$endgroup$

closed as off-topic by greenturtle3141, Brandon_J, gabbo1092, Gamow, Glorfindel Jun 6 at 19:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – greenturtle3141, Brandon_J, gabbo1092, Gamow, Glorfindel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this is off-topic. The "obvious" way to approach this would be to write $N^2−2N−15=M(N^2−N−12)$ and then try to reduce possibilities for $M$ and $N$, perhaps by using modular arithmetic or even brute-forcing. The "aha moment" required by the linked policy is to factorise the polynomials, and the "unexpected result" is that only two values of $N$ work and it's easy to find which ones.
  $endgroup$
  – Rand al'Thor
  Jun 7 at 13:02
  

  
  
  
  

add a comment | 

$begingroup$

For how many integers $N$ is the rational function $fracN^2-2N-15N^2-N-12$ also an integer?

mathematics

share|improve this question

asked Jun 6 at 9:53

Rand al'ThorRand al'Thor

73.7k1515 gold badges244244 silver badges490490 bronze badges

$endgroup$

share|improve this question

asked Jun 6 at 9:53

Rand al'ThorRand al'Thor

73.7k1515 gold badges244244 silver badges490490 bronze badges

share|improve this question

asked Jun 6 at 9:53

Rand al'ThorRand al'Thor

73.7k1515 gold badges244244 silver badges490490 bronze badges

asked Jun 6 at 9:53

Rand al'ThorRand al'Thor

73.7k1515 gold badges244244 silver badges490490 bronze badges

asked Jun 6 at 9:53

Rand al'ThorRand al'Thor

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1 Answer
1

active

oldest

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11

$begingroup$

Since the given expression can be simplified to

$frac(N-5)(N+3)(N-4)(N+3)=fracN-5N-4=1-frac1N-4$,

we simply need to make sure

$N-4$ divides $1$, i.e., $N-4=pm 1$, which leads to $N=5$ or $N=3$. We can check that both of them work.

So

There are two such $N$.

share|improve this answer

answered Jun 6 at 10:03

AnkoganitAnkoganit

11.7k33 gold badges5757 silver badges109109 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect! I was hoping to lead people on a wild goose chase with $N^2-2N-15=M(N^2-N-12)$ or suchlike, but you're too good.
  $endgroup$
  – Rand al'Thor
  Jun 6 at 10:09
  

  
  
  
  

add a comment | 

11

$begingroup$

Since the given expression can be simplified to

$frac(N-5)(N+3)(N-4)(N+3)=fracN-5N-4=1-frac1N-4$,

we simply need to make sure

$N-4$ divides $1$, i.e., $N-4=pm 1$, which leads to $N=5$ or $N=3$. We can check that both of them work.

So

There are two such $N$.

share|improve this answer

answered Jun 6 at 10:03

AnkoganitAnkoganit

11.7k33 gold badges5757 silver badges109109 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect! I was hoping to lead people on a wild goose chase with $N^2-2N-15=M(N^2-N-12)$ or suchlike, but you're too good.
  $endgroup$
  – Rand al'Thor
  Jun 6 at 10:09
  

  
  
  
  

add a comment | 

11

$begingroup$

Since the given expression can be simplified to

$frac(N-5)(N+3)(N-4)(N+3)=fracN-5N-4=1-frac1N-4$,

we simply need to make sure

$N-4$ divides $1$, i.e., $N-4=pm 1$, which leads to $N=5$ or $N=3$. We can check that both of them work.

So

There are two such $N$.

share|improve this answer

answered Jun 6 at 10:03

AnkoganitAnkoganit

11.7k33 gold badges5757 silver badges109109 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect! I was hoping to lead people on a wild goose chase with $N^2-2N-15=M(N^2-N-12)$ or suchlike, but you're too good.
  $endgroup$
  – Rand al'Thor
  Jun 6 at 10:09
  

  
  
  
  

add a comment | 

$begingroup$

Since the given expression can be simplified to

$frac(N-5)(N+3)(N-4)(N+3)=fracN-5N-4=1-frac1N-4$,

we simply need to make sure

$N-4$ divides $1$, i.e., $N-4=pm 1$, which leads to $N=5$ or $N=3$. We can check that both of them work.

So

There are two such $N$.

share|improve this answer

answered Jun 6 at 10:03

AnkoganitAnkoganit

11.7k33 gold badges5757 silver badges109109 bronze badges

$endgroup$

share|improve this answer

answered Jun 6 at 10:03

AnkoganitAnkoganit

11.7k33 gold badges5757 silver badges109109 bronze badges

answered Jun 6 at 10:03

AnkoganitAnkoganit

11.7k33 gold badges5757 silver badges109109 bronze badges

answered Jun 6 at 10:03

AnkoganitAnkoganit

11.7k33 gold badges5757 silver badges109109 bronze badges