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Solving $2x^3=3$ over $mathbbZ_5$.

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0

$begingroup$

Really new to dealing with $mathbbZ_n$.

I'm trying to calculate $2x^3=3$ over $mathbbZ_5$.

I did a similar question - solving $3x^2+x+1=0$ over $mathbbZ_5$:

$$3x^2+x+1=0 \Rightarrow6x^2+2x+2=0\
Rightarrow x^2+2x+1=-1=4\
Rightarrow(x+1)^2=2^2\
Rightarrow x+1=pm2=2,3$$

But I have no idea how to solve $2x^3=3$. It feels like it should be easier.

elementary-number-theory modular-arithmetic

share|cite|improve this question

edited Mar 17 at 22:42

Foobaz John

22.8k41452

asked Mar 17 at 22:36

vesiivesii

3878

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $2cdot 3=1$ in $Bbb Z_5$, so multiply with $3$. We obtain $x^3=-1$. This is easy to solve. Try out all $5$ elements.
  $endgroup$
  – Dietrich Burde
  Mar 17 at 22:39
  

  
  
  
  

add a comment | 

0

$begingroup$

Really new to dealing with $mathbbZ_n$.

I'm trying to calculate $2x^3=3$ over $mathbbZ_5$.

I did a similar question - solving $3x^2+x+1=0$ over $mathbbZ_5$:

$$3x^2+x+1=0 \Rightarrow6x^2+2x+2=0\
Rightarrow x^2+2x+1=-1=4\
Rightarrow(x+1)^2=2^2\
Rightarrow x+1=pm2=2,3$$

But I have no idea how to solve $2x^3=3$. It feels like it should be easier.

elementary-number-theory modular-arithmetic

share|cite|improve this question

edited Mar 17 at 22:42

Foobaz John

22.8k41452

asked Mar 17 at 22:36

vesiivesii

3878

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $2cdot 3=1$ in $Bbb Z_5$, so multiply with $3$. We obtain $x^3=-1$. This is easy to solve. Try out all $5$ elements.
  $endgroup$
  – Dietrich Burde
  Mar 17 at 22:39
  

  
  
  
  

add a comment | 

$begingroup$

Really new to dealing with $mathbbZ_n$.

I'm trying to calculate $2x^3=3$ over $mathbbZ_5$.

I did a similar question - solving $3x^2+x+1=0$ over $mathbbZ_5$:

$$3x^2+x+1=0 \Rightarrow6x^2+2x+2=0\
Rightarrow x^2+2x+1=-1=4\
Rightarrow(x+1)^2=2^2\
Rightarrow x+1=pm2=2,3$$

But I have no idea how to solve $2x^3=3$. It feels like it should be easier.

elementary-number-theory modular-arithmetic

share|cite|improve this question

edited Mar 17 at 22:42

Foobaz John

22.8k41452

asked Mar 17 at 22:36

vesiivesii

3878

$endgroup$

share|cite|improve this question

edited Mar 17 at 22:42

Foobaz John

22.8k41452

asked Mar 17 at 22:36

vesiivesii

3878

share|cite|improve this question

edited Mar 17 at 22:42

Foobaz John

22.8k41452

asked Mar 17 at 22:36

vesiivesii

3878

edited Mar 17 at 22:42

Foobaz John

22.8k41452

edited Mar 17 at 22:42

Foobaz John

22.8k41452

asked Mar 17 at 22:36

vesiivesii

3878

asked Mar 17 at 22:36

vesiivesii

3878

1 Answer
1

active

oldest

votes

3

$begingroup$

There are only 5 elements to check and $0$ and $1$ are easily ruled out. So just plug in $x = 2, 3, 4$ and see which of those work.

While we're here, let me point out that in $mathbbZ_n$, there is a difference between a polynomial function and a polynomial that is not seen in $mathbbQ$ or $mathbbR$ or $mathbbC$. For example, in $mathbbZ_2$, $x = x^2$ as functions because $0 = 0^2$ and $1 = 1^2$. However, as polynomials they are different because they are made up of different terms. (This is something that was confusing to me when I first learned about polynomials in $mathbbZ_n$.)

share|cite|improve this answer

answered Mar 17 at 22:42

Trevor GunnTrevor Gunn

15k32047

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the reply. What do you mean by plug?
  $endgroup$
  – vesii
  Mar 17 at 22:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vesii I mean compute $2 cdot 2^3, 2 cdot 3^3, 2 cdot 4^3$ and see which are equal to $3$.
  $endgroup$
  – Trevor Gunn
  Mar 17 at 22:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so the answer is $x=2,4$?
  $endgroup$
  – vesii
  Mar 17 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vesii just $4$, $2^4 =16 equiv 1 pmod 4$.
  $endgroup$
  – Trevor Gunn
  Mar 17 at 22:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ops, yes you are right, thank you!
  $endgroup$
  – vesii
  Mar 17 at 23:05
  

  
  
  
  

add a comment | 

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3

$begingroup$

There are only 5 elements to check and $0$ and $1$ are easily ruled out. So just plug in $x = 2, 3, 4$ and see which of those work.

While we're here, let me point out that in $mathbbZ_n$, there is a difference between a polynomial function and a polynomial that is not seen in $mathbbQ$ or $mathbbR$ or $mathbbC$. For example, in $mathbbZ_2$, $x = x^2$ as functions because $0 = 0^2$ and $1 = 1^2$. However, as polynomials they are different because they are made up of different terms. (This is something that was confusing to me when I first learned about polynomials in $mathbbZ_n$.)

share|cite|improve this answer

answered Mar 17 at 22:42

Trevor GunnTrevor Gunn

15k32047

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the reply. What do you mean by plug?
  $endgroup$
  – vesii
  Mar 17 at 22:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vesii I mean compute $2 cdot 2^3, 2 cdot 3^3, 2 cdot 4^3$ and see which are equal to $3$.
  $endgroup$
  – Trevor Gunn
  Mar 17 at 22:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so the answer is $x=2,4$?
  $endgroup$
  – vesii
  Mar 17 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vesii just $4$, $2^4 =16 equiv 1 pmod 4$.
  $endgroup$
  – Trevor Gunn
  Mar 17 at 22:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ops, yes you are right, thank you!
  $endgroup$
  – vesii
  Mar 17 at 23:05
  

  
  
  
  

add a comment | 

3

$begingroup$

There are only 5 elements to check and $0$ and $1$ are easily ruled out. So just plug in $x = 2, 3, 4$ and see which of those work.

While we're here, let me point out that in $mathbbZ_n$, there is a difference between a polynomial function and a polynomial that is not seen in $mathbbQ$ or $mathbbR$ or $mathbbC$. For example, in $mathbbZ_2$, $x = x^2$ as functions because $0 = 0^2$ and $1 = 1^2$. However, as polynomials they are different because they are made up of different terms. (This is something that was confusing to me when I first learned about polynomials in $mathbbZ_n$.)

share|cite|improve this answer

answered Mar 17 at 22:42

Trevor GunnTrevor Gunn

15k32047

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the reply. What do you mean by plug?
  $endgroup$
  – vesii
  Mar 17 at 22:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vesii I mean compute $2 cdot 2^3, 2 cdot 3^3, 2 cdot 4^3$ and see which are equal to $3$.
  $endgroup$
  – Trevor Gunn
  Mar 17 at 22:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so the answer is $x=2,4$?
  $endgroup$
  – vesii
  Mar 17 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vesii just $4$, $2^4 =16 equiv 1 pmod 4$.
  $endgroup$
  – Trevor Gunn
  Mar 17 at 22:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ops, yes you are right, thank you!
  $endgroup$
  – vesii
  Mar 17 at 23:05
  

  
  
  
  

add a comment | 

$begingroup$

There are only 5 elements to check and $0$ and $1$ are easily ruled out. So just plug in $x = 2, 3, 4$ and see which of those work.

While we're here, let me point out that in $mathbbZ_n$, there is a difference between a polynomial function and a polynomial that is not seen in $mathbbQ$ or $mathbbR$ or $mathbbC$. For example, in $mathbbZ_2$, $x = x^2$ as functions because $0 = 0^2$ and $1 = 1^2$. However, as polynomials they are different because they are made up of different terms. (This is something that was confusing to me when I first learned about polynomials in $mathbbZ_n$.)

share|cite|improve this answer

answered Mar 17 at 22:42

Trevor GunnTrevor Gunn

15k32047

$endgroup$

share|cite|improve this answer

answered Mar 17 at 22:42

Trevor GunnTrevor Gunn

15k32047

answered Mar 17 at 22:42

Trevor GunnTrevor Gunn

15k32047

answered Mar 17 at 22:42

Trevor GunnTrevor Gunn

15k32047