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Proof by Cases [discrete mathematics] [closed]

-1

$begingroup$

So I've come across this interesting proof question that I'm trying to solve but I'm rather confused about how to go about starting it. Any help would be greatly appreciated:

Prove that if $n$ is a positive integer, then $n^3+4n+2$ is not
divisible by 4. Hint: divide the proof into two cases.

Thank you!

discrete-mathematics proof-writing proof-explanation

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asked Mar 16 at 18:13

Joe BidenJoe Biden

65

$endgroup$

closed as off-topic by heropup, José Carlos Santos, mrtaurho, John Omielan, YiFan Mar 17 at 0:01

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

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – heropup, José Carlos Santos, mrtaurho, John Omielan, YiFan

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint : First assume that $n$ is odd, then that $n$ is even.
  $endgroup$
  – Peter
  Mar 16 at 18:15
  

  
  
  
  

add a comment | 

-1

$begingroup$

So I've come across this interesting proof question that I'm trying to solve but I'm rather confused about how to go about starting it. Any help would be greatly appreciated:

Prove that if $n$ is a positive integer, then $n^3+4n+2$ is not
divisible by 4. Hint: divide the proof into two cases.

Thank you!

discrete-mathematics proof-writing proof-explanation

share|cite|improve this question

asked Mar 16 at 18:13

Joe BidenJoe Biden

65

$endgroup$

closed as off-topic by heropup, José Carlos Santos, mrtaurho, John Omielan, YiFan Mar 17 at 0:01

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

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – heropup, José Carlos Santos, mrtaurho, John Omielan, YiFan

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint : First assume that $n$ is odd, then that $n$ is even.
  $endgroup$
  – Peter
  Mar 16 at 18:15
  

  
  
  
  

add a comment | 

$begingroup$

So I've come across this interesting proof question that I'm trying to solve but I'm rather confused about how to go about starting it. Any help would be greatly appreciated:

Prove that if $n$ is a positive integer, then $n^3+4n+2$ is not
divisible by 4. Hint: divide the proof into two cases.

Thank you!

discrete-mathematics proof-writing proof-explanation

share|cite|improve this question

asked Mar 16 at 18:13

Joe BidenJoe Biden

65

$endgroup$

share|cite|improve this question

asked Mar 16 at 18:13

Joe BidenJoe Biden

65

share|cite|improve this question

asked Mar 16 at 18:13

Joe BidenJoe Biden

65

asked Mar 16 at 18:13

Joe BidenJoe Biden

65

asked Mar 16 at 18:13

Joe BidenJoe Biden

65

2 Answers
2

active

oldest

votes

0

$begingroup$

Assume

1) $n$ is odd:

$n=2k+1, k=0,1,2,....$

$(2k+1)^3+4(2k+1)+2=$

$1+3(2k)+3(2k)^2 +(2k)^3+$

$8k+6=$

$[3(2k)^2+ (2k)^3]+7(2k+1)$;

The first term (in brackets) is divisible by $4$,

$7(2k+1)$ is not (why?).

Hence $f(n)$ is not divisible by $4$ for odd $n$.

2) Let $n$ be even, $n=2k$, k=1,2,... and complete the proof.

share|cite|improve this answer

edited Mar 17 at 6:21

answered Mar 16 at 18:50

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  egreg.Why delete your answer? Straightforward, nice.While mine is contorted.:)
  $endgroup$
  – Peter Szilas
  Mar 16 at 19:04
  

  
  
  
  

add a comment | 

1

$begingroup$

$$n^3+4n+2equiv n^3+2mod4$$When $n$ is odd, $n^3$ is odd and so is $n^3+2$. When $n$ is even, $2|nimplies 4|n^3implies n^3+2equiv2mod4$.

share|cite|improve this answer

answered Mar 16 at 18:56

Shubham JohriShubham Johri

5,477818

$endgroup$

add a comment | 

0

$begingroup$

Assume

1) $n$ is odd:

$n=2k+1, k=0,1,2,....$

$(2k+1)^3+4(2k+1)+2=$

$1+3(2k)+3(2k)^2 +(2k)^3+$

$8k+6=$

$[3(2k)^2+ (2k)^3]+7(2k+1)$;

The first term (in brackets) is divisible by $4$,

$7(2k+1)$ is not (why?).

Hence $f(n)$ is not divisible by $4$ for odd $n$.

2) Let $n$ be even, $n=2k$, k=1,2,... and complete the proof.

share|cite|improve this answer

edited Mar 17 at 6:21

answered Mar 16 at 18:50

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  egreg.Why delete your answer? Straightforward, nice.While mine is contorted.:)
  $endgroup$
  – Peter Szilas
  Mar 16 at 19:04
  

  
  
  
  

add a comment | 

0

$begingroup$

Assume

1) $n$ is odd:

$n=2k+1, k=0,1,2,....$

$(2k+1)^3+4(2k+1)+2=$

$1+3(2k)+3(2k)^2 +(2k)^3+$

$8k+6=$

$[3(2k)^2+ (2k)^3]+7(2k+1)$;

The first term (in brackets) is divisible by $4$,

$7(2k+1)$ is not (why?).

Hence $f(n)$ is not divisible by $4$ for odd $n$.

2) Let $n$ be even, $n=2k$, k=1,2,... and complete the proof.

share|cite|improve this answer

edited Mar 17 at 6:21

answered Mar 16 at 18:50

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  egreg.Why delete your answer? Straightforward, nice.While mine is contorted.:)
  $endgroup$
  – Peter Szilas
  Mar 16 at 19:04
  

  
  
  
  

add a comment | 

$begingroup$

Assume

1) $n$ is odd:

$n=2k+1, k=0,1,2,....$

$(2k+1)^3+4(2k+1)+2=$

$1+3(2k)+3(2k)^2 +(2k)^3+$

$8k+6=$

$[3(2k)^2+ (2k)^3]+7(2k+1)$;

The first term (in brackets) is divisible by $4$,

$7(2k+1)$ is not (why?).

Hence $f(n)$ is not divisible by $4$ for odd $n$.

2) Let $n$ be even, $n=2k$, k=1,2,... and complete the proof.

share|cite|improve this answer

edited Mar 17 at 6:21

answered Mar 16 at 18:50

Peter SzilasPeter Szilas

11.6k2822

$endgroup$

share|cite|improve this answer

edited Mar 17 at 6:21

answered Mar 16 at 18:50

Peter SzilasPeter Szilas

11.6k2822

answered Mar 16 at 18:50

Peter SzilasPeter Szilas

11.6k2822

answered Mar 16 at 18:50

Peter SzilasPeter Szilas

11.6k2822

1

$begingroup$

$$n^3+4n+2equiv n^3+2mod4$$When $n$ is odd, $n^3$ is odd and so is $n^3+2$. When $n$ is even, $2|nimplies 4|n^3implies n^3+2equiv2mod4$.

share|cite|improve this answer

answered Mar 16 at 18:56

Shubham JohriShubham Johri

5,477818

$endgroup$

add a comment | 

1

$begingroup$

$$n^3+4n+2equiv n^3+2mod4$$When $n$ is odd, $n^3$ is odd and so is $n^3+2$. When $n$ is even, $2|nimplies 4|n^3implies n^3+2equiv2mod4$.

share|cite|improve this answer

answered Mar 16 at 18:56

Shubham JohriShubham Johri

5,477818

$endgroup$

add a comment | 

$begingroup$

$$n^3+4n+2equiv n^3+2mod4$$When $n$ is odd, $n^3$ is odd and so is $n^3+2$. When $n$ is even, $2|nimplies 4|n^3implies n^3+2equiv2mod4$.

share|cite|improve this answer

answered Mar 16 at 18:56

Shubham JohriShubham Johri

5,477818

$endgroup$

share|cite|improve this answer

answered Mar 16 at 18:56

Shubham JohriShubham Johri

5,477818

answered Mar 16 at 18:56

Shubham JohriShubham Johri

5,477818

answered Mar 16 at 18:56

Shubham JohriShubham Johri

5,477818