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Find the number of different residues mod $p$ of $(x^2+y^2)^2$ where $(x,p)=(y,p)=1$

1

1

$begingroup$

Let $p=4k+3$ be a prime number. Find the number of different residues mod $p$ of $$(x^2+y^2)^2$$ where $(x,p)=(y,p)=1$

This problem is from the (Problems from the book) chapter 18 Quadratic reciprocity. Because this book problems have no answer. so How to use this methods to solve it?

I think the answer is $dfracp-12$.But How to prove it?

number-theory

share|cite|improve this question

edited Mar 27 at 4:07

Robert Shore

3,852324

asked Mar 27 at 4:03

function sugfunction sug

2731439

$endgroup$

add a comment | 

1

1

$begingroup$

Let $p=4k+3$ be a prime number. Find the number of different residues mod $p$ of $$(x^2+y^2)^2$$ where $(x,p)=(y,p)=1$

This problem is from the (Problems from the book) chapter 18 Quadratic reciprocity. Because this book problems have no answer. so How to use this methods to solve it?

I think the answer is $dfracp-12$.But How to prove it?

number-theory

share|cite|improve this question

edited Mar 27 at 4:07

Robert Shore

3,852324

asked Mar 27 at 4:03

function sugfunction sug

2731439

$endgroup$

add a comment | 

$begingroup$

Let $p=4k+3$ be a prime number. Find the number of different residues mod $p$ of $$(x^2+y^2)^2$$ where $(x,p)=(y,p)=1$

This problem is from the (Problems from the book) chapter 18 Quadratic reciprocity. Because this book problems have no answer. so How to use this methods to solve it?

I think the answer is $dfracp-12$.But How to prove it?

number-theory

share|cite|improve this question

edited Mar 27 at 4:07

Robert Shore

3,852324

asked Mar 27 at 4:03

function sugfunction sug

2731439

$endgroup$

share|cite|improve this question

edited Mar 27 at 4:07

Robert Shore

3,852324

asked Mar 27 at 4:03

function sugfunction sug

2731439

share|cite|improve this question

edited Mar 27 at 4:07

Robert Shore

3,852324

asked Mar 27 at 4:03

function sugfunction sug

2731439

edited Mar 27 at 4:07

Robert Shore

3,852324

edited Mar 27 at 4:07

Robert Shore

3,852324

asked Mar 27 at 4:03

function sugfunction sug

2731439

asked Mar 27 at 4:03

function sugfunction sug

2731439

1 Answer
1

active

oldest

votes

2

$begingroup$

Your assumed answer is correct. To see this, since $p = 4k + 3$ and $(x,p) = (y,p) = 1$, then $x^2 + y^2 notequiv 0 pmod p$. As such, the number of possible residues of

$$(x^2+y^2)^2 tag1labeleq1$$

is $le fracp - 12$. To see that it's actually equal, consider the set of residues with $x = y$ for $1 le y le fracp - 12$. In those cases, eqrefeq1 becomes $4y^4$. To confirm they're all unique, let some $1 le z le fracp - 12$, where $z neq y$, be such that

$$4y^4 equiv 4z^4 pmod p ; Rightarrow left(y^2 - z^2right)left(y^2 + z^2right) equiv 0 pmod p tag2labeleq2$$

Since $y^2 + z^2 notequiv 0 pmod p$, this means that $y^2 - z^2 equiv 0 pmod p$. However, since the $fracp-12$ residues of values between $1$ and $fracp-12$, inclusive, are all unique, this can only be the case if $y = z$, which is a contradiction of our earlier assumption. Thus, each of those values must give a unique quadratic residue, showing that the total number of possible such values is $fracp-12$.

share|cite|improve this answer

edited Mar 27 at 21:52

answered Mar 27 at 20:40

John OmielanJohn Omielan

5,1142218

$endgroup$

add a comment | 

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2

$begingroup$

Your assumed answer is correct. To see this, since $p = 4k + 3$ and $(x,p) = (y,p) = 1$, then $x^2 + y^2 notequiv 0 pmod p$. As such, the number of possible residues of

$$(x^2+y^2)^2 tag1labeleq1$$

is $le fracp - 12$. To see that it's actually equal, consider the set of residues with $x = y$ for $1 le y le fracp - 12$. In those cases, eqrefeq1 becomes $4y^4$. To confirm they're all unique, let some $1 le z le fracp - 12$, where $z neq y$, be such that

$$4y^4 equiv 4z^4 pmod p ; Rightarrow left(y^2 - z^2right)left(y^2 + z^2right) equiv 0 pmod p tag2labeleq2$$

Since $y^2 + z^2 notequiv 0 pmod p$, this means that $y^2 - z^2 equiv 0 pmod p$. However, since the $fracp-12$ residues of values between $1$ and $fracp-12$, inclusive, are all unique, this can only be the case if $y = z$, which is a contradiction of our earlier assumption. Thus, each of those values must give a unique quadratic residue, showing that the total number of possible such values is $fracp-12$.

share|cite|improve this answer

edited Mar 27 at 21:52

answered Mar 27 at 20:40

John OmielanJohn Omielan

5,1142218

$endgroup$

add a comment | 

2

$begingroup$

Your assumed answer is correct. To see this, since $p = 4k + 3$ and $(x,p) = (y,p) = 1$, then $x^2 + y^2 notequiv 0 pmod p$. As such, the number of possible residues of

$$(x^2+y^2)^2 tag1labeleq1$$

is $le fracp - 12$. To see that it's actually equal, consider the set of residues with $x = y$ for $1 le y le fracp - 12$. In those cases, eqrefeq1 becomes $4y^4$. To confirm they're all unique, let some $1 le z le fracp - 12$, where $z neq y$, be such that

$$4y^4 equiv 4z^4 pmod p ; Rightarrow left(y^2 - z^2right)left(y^2 + z^2right) equiv 0 pmod p tag2labeleq2$$

Since $y^2 + z^2 notequiv 0 pmod p$, this means that $y^2 - z^2 equiv 0 pmod p$. However, since the $fracp-12$ residues of values between $1$ and $fracp-12$, inclusive, are all unique, this can only be the case if $y = z$, which is a contradiction of our earlier assumption. Thus, each of those values must give a unique quadratic residue, showing that the total number of possible such values is $fracp-12$.

share|cite|improve this answer

edited Mar 27 at 21:52

answered Mar 27 at 20:40

John OmielanJohn Omielan

5,1142218

$endgroup$

add a comment | 

$begingroup$

Your assumed answer is correct. To see this, since $p = 4k + 3$ and $(x,p) = (y,p) = 1$, then $x^2 + y^2 notequiv 0 pmod p$. As such, the number of possible residues of

$$(x^2+y^2)^2 tag1labeleq1$$

is $le fracp - 12$. To see that it's actually equal, consider the set of residues with $x = y$ for $1 le y le fracp - 12$. In those cases, eqrefeq1 becomes $4y^4$. To confirm they're all unique, let some $1 le z le fracp - 12$, where $z neq y$, be such that

$$4y^4 equiv 4z^4 pmod p ; Rightarrow left(y^2 - z^2right)left(y^2 + z^2right) equiv 0 pmod p tag2labeleq2$$

Since $y^2 + z^2 notequiv 0 pmod p$, this means that $y^2 - z^2 equiv 0 pmod p$. However, since the $fracp-12$ residues of values between $1$ and $fracp-12$, inclusive, are all unique, this can only be the case if $y = z$, which is a contradiction of our earlier assumption. Thus, each of those values must give a unique quadratic residue, showing that the total number of possible such values is $fracp-12$.

share|cite|improve this answer

edited Mar 27 at 21:52

answered Mar 27 at 20:40

John OmielanJohn Omielan

5,1142218

$endgroup$

share|cite|improve this answer

edited Mar 27 at 21:52

answered Mar 27 at 20:40

John OmielanJohn Omielan

5,1142218

answered Mar 27 at 20:40

John OmielanJohn Omielan

5,1142218

answered Mar 27 at 20:40

John OmielanJohn Omielan

5,1142218