Prime number and divisibility
$begingroup$
Let $p$ be prime, prove that for any integer $r$, there at most $2$ solutions to the equation $x^2-r equiv 0pmod p$.
I don't understand the question as if $p=2$, and $r$ odd, then $x^2$ will need to be odd, and we have an infinite number of solutions.
Maybe this is true for $p>2$, but I don't really know how to prove it.
modular-arithmetic
edited Jan 6 at 17:49
amWhy
1
asked Jan 6 at 17:04
Student number xStudent number x
1107
$endgroup$
|
show 1 more comment
$begingroup$
Let $p$ be prime, prove that for any integer $r$, there at most $2$ solutions to the equation $x^2-r equiv 0pmod p$.
I don't understand the question as if $p=2$, and $r$ odd, then $x^2$ will need to be odd, and we have an infinite number of solutions.
Maybe this is true for $p>2$, but I don't really know how to prove it.
modular-arithmetic
edited Jan 6 at 17:49
amWhy
1
asked Jan 6 at 17:04
Student number xStudent number x
1107
$endgroup$
4
$begingroup$
They mean solutions $pmod p$. Thus, if $p=7$ and $r=2$ the only solutions would be $3,4$. For $p=2, r=1$ the only solution is $x=1$.
$endgroup$
– lulu
Jan 6 at 17:06
$begingroup$
makes more sense! thank you - do you have any idea how to prove it though?
$endgroup$
– Student number x
Jan 6 at 17:17
$begingroup$
The mean solutions modulo $p$. There are only $p$ classes. If $p=2$ then are two possible $x$, $0$ and $1$. Any other odd number is $equiv 2 pmod 2$.
$endgroup$
– fleablood
Jan 6 at 17:53
$begingroup$
..... if $p = 2$ and $r=7$ then there is one solution mod $2$. $1^2 - 7equiv 0 pmod 2$ but $0^2 - 7 not equiv 0 pmod 2$. Any other odd number,say $253$ so that $253^2 - 7 equiv 0 pmod 2$ is the same solution as $253 equiv 1 pmod 2$. In the $mod 2$ system $253$ and $1$ are the same class $253 equiv 1$. So $1$ is the only solution.
$endgroup$
– fleablood
Jan 6 at 17:58
$begingroup$
For the record, If $x^2 equiv r pmod p$ then $(kp+x)equiv r pmod p$ so if there are any integer solutions there are an infinite number of INTEGER solutions, but we are not asking for integers solution but EQUIVALENCE solutions. $kp + x equiv x$ so they are the SAME solution. It's only one, even though ther are an infinite number of integers that are all EQUIVALENT to the one solution.
$endgroup$
– fleablood
Jan 6 at 18:16
|
show 1 more comment
$begingroup$
Let $p$ be prime, prove that for any integer $r$, there at most $2$ solutions to the equation $x^2-r equiv 0pmod p$.
I don't understand the question as if $p=2$, and $r$ odd, then $x^2$ will need to be odd, and we have an infinite number of solutions.
Maybe this is true for $p>2$, but I don't really know how to prove it.
modular-arithmetic
edited Jan 6 at 17:49
amWhy
1
asked Jan 6 at 17:04
Student number xStudent number x
1107
$endgroup$
Let $p$ be prime, prove that for any integer $r$, there at most $2$ solutions to the equation $x^2-r equiv 0pmod p$.
I don't understand the question as if $p=2$, and $r$ odd, then $x^2$ will need to be odd, and we have an infinite number of solutions.
Maybe this is true for $p>2$, but I don't really know how to prove it.
modular-arithmetic
modular-arithmetic
edited Jan 6 at 17:49
amWhy
1
asked Jan 6 at 17:04
Student number xStudent number x
1107
edited Jan 6 at 17:49
amWhy
1
asked Jan 6 at 17:04
Student number xStudent number x
1107
edited Jan 6 at 17:49
amWhy
1
edited Jan 6 at 17:49
amWhy
1
edited Jan 6 at 17:49
amWhy
1
1
asked Jan 6 at 17:04
Student number xStudent number x
1107
asked Jan 6 at 17:04
Student number xStudent number x
1107
asked Jan 6 at 17:04
Student number xStudent number x
1107
1107
4
$begingroup$
They mean solutions $pmod p$. Thus, if $p=7$ and $r=2$ the only solutions would be $3,4$. For $p=2, r=1$ the only solution is $x=1$.
$endgroup$
– lulu
Jan 6 at 17:06
$begingroup$
makes more sense! thank you - do you have any idea how to prove it though?
$endgroup$
– Student number x
Jan 6 at 17:17
$begingroup$
The mean solutions modulo $p$. There are only $p$ classes. If $p=2$ then are two possible $x$, $0$ and $1$. Any other odd number is $equiv 2 pmod 2$.
$endgroup$
– fleablood
Jan 6 at 17:53
$begingroup$
..... if $p = 2$ and $r=7$ then there is one solution mod $2$. $1^2 - 7equiv 0 pmod 2$ but $0^2 - 7 not equiv 0 pmod 2$. Any other odd number,say $253$ so that $253^2 - 7 equiv 0 pmod 2$ is the same solution as $253 equiv 1 pmod 2$. In the $mod 2$ system $253$ and $1$ are the same class $253 equiv 1$. So $1$ is the only solution.
$endgroup$
– fleablood
Jan 6 at 17:58
$begingroup$
For the record, If $x^2 equiv r pmod p$ then $(kp+x)equiv r pmod p$ so if there are any integer solutions there are an infinite number of INTEGER solutions, but we are not asking for integers solution but EQUIVALENCE solutions. $kp + x equiv x$ so they are the SAME solution. It's only one, even though ther are an infinite number of integers that are all EQUIVALENT to the one solution.
$endgroup$
– fleablood
Jan 6 at 18:16
|
show 1 more comment
4
$begingroup$
They mean solutions $pmod p$. Thus, if $p=7$ and $r=2$ the only solutions would be $3,4$. For $p=2, r=1$ the only solution is $x=1$.
$endgroup$
– lulu
Jan 6 at 17:06
$begingroup$
makes more sense! thank you - do you have any idea how to prove it though?
$endgroup$
– Student number x
Jan 6 at 17:17
$begingroup$
The mean solutions modulo $p$. There are only $p$ classes. If $p=2$ then are two possible $x$, $0$ and $1$. Any other odd number is $equiv 2 pmod 2$.
$endgroup$
– fleablood
Jan 6 at 17:53
$begingroup$
..... if $p = 2$ and $r=7$ then there is one solution mod $2$. $1^2 - 7equiv 0 pmod 2$ but $0^2 - 7 not equiv 0 pmod 2$. Any other odd number,say $253$ so that $253^2 - 7 equiv 0 pmod 2$ is the same solution as $253 equiv 1 pmod 2$. In the $mod 2$ system $253$ and $1$ are the same class $253 equiv 1$. So $1$ is the only solution.
$endgroup$
– fleablood
Jan 6 at 17:58
$begingroup$
For the record, If $x^2 equiv r pmod p$ then $(kp+x)equiv r pmod p$ so if there are any integer solutions there are an infinite number of INTEGER solutions, but we are not asking for integers solution but EQUIVALENCE solutions. $kp + x equiv x$ so they are the SAME solution. It's only one, even though ther are an infinite number of integers that are all EQUIVALENT to the one solution.
$endgroup$
– fleablood
Jan 6 at 18:16
4
4
$begingroup$
They mean solutions $pmod p$. Thus, if $p=7$ and $r=2$ the only solutions would be $3,4$. For $p=2, r=1$ the only solution is $x=1$.
$endgroup$
– lulu
Jan 6 at 17:06
$begingroup$
They mean solutions $pmod p$. Thus, if $p=7$ and $r=2$ the only solutions would be $3,4$. For $p=2, r=1$ the only solution is $x=1$.
$endgroup$
– lulu
Jan 6 at 17:06
$begingroup$
makes more sense! thank you - do you have any idea how to prove it though?
$endgroup$
– Student number x
Jan 6 at 17:17
$begingroup$
makes more sense! thank you - do you have any idea how to prove it though?
$endgroup$
– Student number x
Jan 6 at 17:17
$begingroup$
The mean solutions modulo $p$. There are only $p$ classes. If $p=2$ then are two possible $x$, $0$ and $1$. Any other odd number is $equiv 2 pmod 2$.
$endgroup$
– fleablood
Jan 6 at 17:53
$begingroup$
The mean solutions modulo $p$. There are only $p$ classes. If $p=2$ then are two possible $x$, $0$ and $1$. Any other odd number is $equiv 2 pmod 2$.
$endgroup$
– fleablood
Jan 6 at 17:53
$begingroup$
..... if $p = 2$ and $r=7$ then there is one solution mod $2$. $1^2 - 7equiv 0 pmod 2$ but $0^2 - 7 not equiv 0 pmod 2$. Any other odd number,say $253$ so that $253^2 - 7 equiv 0 pmod 2$ is the same solution as $253 equiv 1 pmod 2$. In the $mod 2$ system $253$ and $1$ are the same class $253 equiv 1$. So $1$ is the only solution.
$endgroup$
– fleablood
Jan 6 at 17:58
$begingroup$
..... if $p = 2$ and $r=7$ then there is one solution mod $2$. $1^2 - 7equiv 0 pmod 2$ but $0^2 - 7 not equiv 0 pmod 2$. Any other odd number,say $253$ so that $253^2 - 7 equiv 0 pmod 2$ is the same solution as $253 equiv 1 pmod 2$. In the $mod 2$ system $253$ and $1$ are the same class $253 equiv 1$. So $1$ is the only solution.
$endgroup$
– fleablood
Jan 6 at 17:58
$begingroup$
For the record, If $x^2 equiv r pmod p$ then $(kp+x)equiv r pmod p$ so if there are any integer solutions there are an infinite number of INTEGER solutions, but we are not asking for integers solution but EQUIVALENCE solutions. $kp + x equiv x$ so they are the SAME solution. It's only one, even though ther are an infinite number of integers that are all EQUIVALENT to the one solution.
$endgroup$
– fleablood
Jan 6 at 18:16
$begingroup$
For the record, If $x^2 equiv r pmod p$ then $(kp+x)equiv r pmod p$ so if there are any integer solutions there are an infinite number of INTEGER solutions, but we are not asking for integers solution but EQUIVALENCE solutions. $kp + x equiv x$ so they are the SAME solution. It's only one, even though ther are an infinite number of integers that are all EQUIVALENT to the one solution.
$endgroup$
– fleablood
Jan 6 at 18:16
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
If $a$ and $x$ are solutions of $x^2 equiv r pmod p$ then $x^2 - r equiv x^2 - a^2 = (x-a)(x+a) equiv 0 pmod p$.
Since $p$ is prime, $p | (x-a)(x+a)$ means $p | (x-a)$ or $p | (x+a)$, i.e., $x equiv a$ or $-a pmod p$.
Thus there can be at most two solutions.
answered Jan 6 at 19:00
J. W. TannerJ. W. Tanner
2129
$endgroup$
$begingroup$
As noted by others, it is possible there are no solutions. For example $x^2 - 2 equiv 0 pmod 3$ has no solutions.
$endgroup$
– J. W. Tanner
Jan 7 at 18:57
add a comment |
$begingroup$
The integers mod p are a field. In general, a polynomial of degree n over a field has at most n roots in the field. Here n=2.
answered Jan 6 at 18:19
J. W. TannerJ. W. Tanner
2129
$endgroup$
1
$begingroup$
Do you think this question is being at a level where the student can be expected to prove that a polynomial of degree $n$ has at most $n$ solutions? Or to take it as a given?
$endgroup$
– fleablood
Jan 6 at 18:57
$begingroup$
But +1 for bringing in ideas about what the question is actually asking and considering consequences.
$endgroup$
– fleablood
Jan 6 at 18:58
$begingroup$
I provided another answer, which is less abstract
$endgroup$
– J. W. Tanner
Jan 6 at 19:18
add a comment |
$begingroup$
$x^2 equiv r pmod p$ and $0le x < p$
$y^2 equiv r pmod p$ so that $yne x$ and $0 le y < p$
$x^2 - y^2 equiv 0 pmod p$
$(x+y)(x-y) equiv 0 pmod p$.
Now $x ne y $ so $x-y ne 0$ and $|x-y| < p$ so $pnot mid x-y$ hence $p|(x+y)$ but $0 < x+y < 2p$ so $x + y= p$
Now if we had a third option so that
$z^2 equiv r pmod p$ and $0 le z <p$ but $z ne x; z ne y; xne y$ we could use the exact same argument to conclude:
$x+y = x+z = y+z = p$. But that would imply $z = y$ (and $y = x$ and $x = z$).
So $2$ distinct solutions is a possibility (but not a certainty!) but three distinct solutions is not.
(Bear in mind it's possible that there are no solutions.)
(Also not if $x$ is one solution then so is $p-x$ and if $p$ is odd then $x$ and $p-x$ are two distinct solutions. But that does not mean there is a solution at all.)
answered Jan 6 at 18:13
fleabloodfleablood
68.9k22685
$endgroup$
$begingroup$
In the second line, did you mean $y ne x$ and $0 le$ y $lt p$?
$endgroup$
– J. W. Tanner
Jan 7 at 18:47
$begingroup$
Yes, I did.......
$endgroup$
– fleablood
Jan 7 at 21:40
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $a$ and $x$ are solutions of $x^2 equiv r pmod p$ then $x^2 - r equiv x^2 - a^2 = (x-a)(x+a) equiv 0 pmod p$.
Since $p$ is prime, $p | (x-a)(x+a)$ means $p | (x-a)$ or $p | (x+a)$, i.e., $x equiv a$ or $-a pmod p$.
Thus there can be at most two solutions.
answered Jan 6 at 19:00
J. W. TannerJ. W. Tanner
2129
$endgroup$
$begingroup$
As noted by others, it is possible there are no solutions. For example $x^2 - 2 equiv 0 pmod 3$ has no solutions.
$endgroup$
– J. W. Tanner
Jan 7 at 18:57
add a comment |
$begingroup$
If $a$ and $x$ are solutions of $x^2 equiv r pmod p$ then $x^2 - r equiv x^2 - a^2 = (x-a)(x+a) equiv 0 pmod p$.
Since $p$ is prime, $p | (x-a)(x+a)$ means $p | (x-a)$ or $p | (x+a)$, i.e., $x equiv a$ or $-a pmod p$.
Thus there can be at most two solutions.
answered Jan 6 at 19:00
J. W. TannerJ. W. Tanner
2129
$endgroup$
$begingroup$
As noted by others, it is possible there are no solutions. For example $x^2 - 2 equiv 0 pmod 3$ has no solutions.
$endgroup$
– J. W. Tanner
Jan 7 at 18:57
add a comment |
$begingroup$
If $a$ and $x$ are solutions of $x^2 equiv r pmod p$ then $x^2 - r equiv x^2 - a^2 = (x-a)(x+a) equiv 0 pmod p$.
Since $p$ is prime, $p | (x-a)(x+a)$ means $p | (x-a)$ or $p | (x+a)$, i.e., $x equiv a$ or $-a pmod p$.
Thus there can be at most two solutions.
answered Jan 6 at 19:00
J. W. TannerJ. W. Tanner
2129
$endgroup$
If $a$ and $x$ are solutions of $x^2 equiv r pmod p$ then $x^2 - r equiv x^2 - a^2 = (x-a)(x+a) equiv 0 pmod p$.
Since $p$ is prime, $p | (x-a)(x+a)$ means $p | (x-a)$ or $p | (x+a)$, i.e., $x equiv a$ or $-a pmod p$.
Thus there can be at most two solutions.
answered Jan 6 at 19:00
J. W. TannerJ. W. Tanner
2129
edited Jan 7 at 0:34
answered Jan 6 at 19:00
J. W. TannerJ. W. Tanner
2129
answered Jan 6 at 19:00
J. W. TannerJ. W. Tanner
2129
answered Jan 6 at 19:00
J. W. TannerJ. W. Tanner
2129
2129
$begingroup$
As noted by others, it is possible there are no solutions. For example $x^2 - 2 equiv 0 pmod 3$ has no solutions.
$endgroup$
– J. W. Tanner
Jan 7 at 18:57
add a comment |
$begingroup$
As noted by others, it is possible there are no solutions. For example $x^2 - 2 equiv 0 pmod 3$ has no solutions.
$endgroup$
– J. W. Tanner
Jan 7 at 18:57
$begingroup$
As noted by others, it is possible there are no solutions. For example $x^2 - 2 equiv 0 pmod 3$ has no solutions.
$endgroup$
– J. W. Tanner
Jan 7 at 18:57
$begingroup$
As noted by others, it is possible there are no solutions. For example $x^2 - 2 equiv 0 pmod 3$ has no solutions.
$endgroup$
– J. W. Tanner
Jan 7 at 18:57
add a comment |
$begingroup$
The integers mod p are a field. In general, a polynomial of degree n over a field has at most n roots in the field. Here n=2.
answered Jan 6 at 18:19
J. W. TannerJ. W. Tanner
2129
$endgroup$
1
$begingroup$
Do you think this question is being at a level where the student can be expected to prove that a polynomial of degree $n$ has at most $n$ solutions? Or to take it as a given?
$endgroup$
– fleablood
Jan 6 at 18:57
$begingroup$
But +1 for bringing in ideas about what the question is actually asking and considering consequences.
$endgroup$
– fleablood
Jan 6 at 18:58
$begingroup$
I provided another answer, which is less abstract
$endgroup$
– J. W. Tanner
Jan 6 at 19:18
add a comment |
$begingroup$
The integers mod p are a field. In general, a polynomial of degree n over a field has at most n roots in the field. Here n=2.
answered Jan 6 at 18:19
J. W. TannerJ. W. Tanner
2129
$endgroup$
1
$begingroup$
Do you think this question is being at a level where the student can be expected to prove that a polynomial of degree $n$ has at most $n$ solutions? Or to take it as a given?
$endgroup$
– fleablood
Jan 6 at 18:57
$begingroup$
But +1 for bringing in ideas about what the question is actually asking and considering consequences.
$endgroup$
– fleablood
Jan 6 at 18:58
$begingroup$
I provided another answer, which is less abstract
$endgroup$
– J. W. Tanner
Jan 6 at 19:18
add a comment |
$begingroup$
The integers mod p are a field. In general, a polynomial of degree n over a field has at most n roots in the field. Here n=2.
answered Jan 6 at 18:19
J. W. TannerJ. W. Tanner
2129
$endgroup$
The integers mod p are a field. In general, a polynomial of degree n over a field has at most n roots in the field. Here n=2.
answered Jan 6 at 18:19
J. W. TannerJ. W. Tanner
2129
answered Jan 6 at 18:19
J. W. TannerJ. W. Tanner
2129
answered Jan 6 at 18:19
J. W. TannerJ. W. Tanner
2129
answered Jan 6 at 18:19
J. W. TannerJ. W. Tanner
2129
2129
1
$begingroup$
Do you think this question is being at a level where the student can be expected to prove that a polynomial of degree $n$ has at most $n$ solutions? Or to take it as a given?
$endgroup$
– fleablood
Jan 6 at 18:57
$begingroup$
But +1 for bringing in ideas about what the question is actually asking and considering consequences.
$endgroup$
– fleablood
Jan 6 at 18:58
$begingroup$
I provided another answer, which is less abstract
$endgroup$
– J. W. Tanner
Jan 6 at 19:18
add a comment |
1
$begingroup$
Do you think this question is being at a level where the student can be expected to prove that a polynomial of degree $n$ has at most $n$ solutions? Or to take it as a given?
$endgroup$
– fleablood
Jan 6 at 18:57
$begingroup$
But +1 for bringing in ideas about what the question is actually asking and considering consequences.
$endgroup$
– fleablood
Jan 6 at 18:58
$begingroup$
I provided another answer, which is less abstract
$endgroup$
– J. W. Tanner
Jan 6 at 19:18
1
1
$begingroup$
Do you think this question is being at a level where the student can be expected to prove that a polynomial of degree $n$ has at most $n$ solutions? Or to take it as a given?
$endgroup$
– fleablood
Jan 6 at 18:57
$begingroup$
Do you think this question is being at a level where the student can be expected to prove that a polynomial of degree $n$ has at most $n$ solutions? Or to take it as a given?
$endgroup$
– fleablood
Jan 6 at 18:57
$begingroup$
But +1 for bringing in ideas about what the question is actually asking and considering consequences.
$endgroup$
– fleablood
Jan 6 at 18:58
$begingroup$
But +1 for bringing in ideas about what the question is actually asking and considering consequences.
$endgroup$
– fleablood
Jan 6 at 18:58
$begingroup$
I provided another answer, which is less abstract
$endgroup$
– J. W. Tanner
Jan 6 at 19:18
$begingroup$
I provided another answer, which is less abstract
$endgroup$
– J. W. Tanner
Jan 6 at 19:18
add a comment |
$begingroup$
$x^2 equiv r pmod p$ and $0le x < p$
$y^2 equiv r pmod p$ so that $yne x$ and $0 le y < p$
$x^2 - y^2 equiv 0 pmod p$
$(x+y)(x-y) equiv 0 pmod p$.
Now $x ne y $ so $x-y ne 0$ and $|x-y| < p$ so $pnot mid x-y$ hence $p|(x+y)$ but $0 < x+y < 2p$ so $x + y= p$
Now if we had a third option so that
$z^2 equiv r pmod p$ and $0 le z <p$ but $z ne x; z ne y; xne y$ we could use the exact same argument to conclude:
$x+y = x+z = y+z = p$. But that would imply $z = y$ (and $y = x$ and $x = z$).
So $2$ distinct solutions is a possibility (but not a certainty!) but three distinct solutions is not.
(Bear in mind it's possible that there are no solutions.)
(Also not if $x$ is one solution then so is $p-x$ and if $p$ is odd then $x$ and $p-x$ are two distinct solutions. But that does not mean there is a solution at all.)
answered Jan 6 at 18:13
fleabloodfleablood
68.9k22685
$endgroup$
$begingroup$
In the second line, did you mean $y ne x$ and $0 le$ y $lt p$?
$endgroup$
– J. W. Tanner
Jan 7 at 18:47
$begingroup$
Yes, I did.......
$endgroup$
– fleablood
Jan 7 at 21:40
add a comment |
$begingroup$
$x^2 equiv r pmod p$ and $0le x < p$
$y^2 equiv r pmod p$ so that $yne x$ and $0 le y < p$
$x^2 - y^2 equiv 0 pmod p$
$(x+y)(x-y) equiv 0 pmod p$.
Now $x ne y $ so $x-y ne 0$ and $|x-y| < p$ so $pnot mid x-y$ hence $p|(x+y)$ but $0 < x+y < 2p$ so $x + y= p$
Now if we had a third option so that
$z^2 equiv r pmod p$ and $0 le z <p$ but $z ne x; z ne y; xne y$ we could use the exact same argument to conclude:
$x+y = x+z = y+z = p$. But that would imply $z = y$ (and $y = x$ and $x = z$).
So $2$ distinct solutions is a possibility (but not a certainty!) but three distinct solutions is not.
(Bear in mind it's possible that there are no solutions.)
(Also not if $x$ is one solution then so is $p-x$ and if $p$ is odd then $x$ and $p-x$ are two distinct solutions. But that does not mean there is a solution at all.)
answered Jan 6 at 18:13
fleabloodfleablood
68.9k22685
$endgroup$
$begingroup$
In the second line, did you mean $y ne x$ and $0 le$ y $lt p$?
$endgroup$
– J. W. Tanner
Jan 7 at 18:47
$begingroup$
Yes, I did.......
$endgroup$
– fleablood
Jan 7 at 21:40
add a comment |
$begingroup$
$x^2 equiv r pmod p$ and $0le x < p$
$y^2 equiv r pmod p$ so that $yne x$ and $0 le y < p$
$x^2 - y^2 equiv 0 pmod p$
$(x+y)(x-y) equiv 0 pmod p$.
Now $x ne y $ so $x-y ne 0$ and $|x-y| < p$ so $pnot mid x-y$ hence $p|(x+y)$ but $0 < x+y < 2p$ so $x + y= p$
Now if we had a third option so that
$z^2 equiv r pmod p$ and $0 le z <p$ but $z ne x; z ne y; xne y$ we could use the exact same argument to conclude:
$x+y = x+z = y+z = p$. But that would imply $z = y$ (and $y = x$ and $x = z$).
So $2$ distinct solutions is a possibility (but not a certainty!) but three distinct solutions is not.
(Bear in mind it's possible that there are no solutions.)
(Also not if $x$ is one solution then so is $p-x$ and if $p$ is odd then $x$ and $p-x$ are two distinct solutions. But that does not mean there is a solution at all.)
answered Jan 6 at 18:13
fleabloodfleablood
68.9k22685
$endgroup$
$x^2 equiv r pmod p$ and $0le x < p$
$y^2 equiv r pmod p$ so that $yne x$ and $0 le y < p$
$x^2 - y^2 equiv 0 pmod p$
$(x+y)(x-y) equiv 0 pmod p$.
Now $x ne y $ so $x-y ne 0$ and $|x-y| < p$ so $pnot mid x-y$ hence $p|(x+y)$ but $0 < x+y < 2p$ so $x + y= p$
Now if we had a third option so that
$z^2 equiv r pmod p$ and $0 le z <p$ but $z ne x; z ne y; xne y$ we could use the exact same argument to conclude:
$x+y = x+z = y+z = p$. But that would imply $z = y$ (and $y = x$ and $x = z$).
So $2$ distinct solutions is a possibility (but not a certainty!) but three distinct solutions is not.
(Bear in mind it's possible that there are no solutions.)
(Also not if $x$ is one solution then so is $p-x$ and if $p$ is odd then $x$ and $p-x$ are two distinct solutions. But that does not mean there is a solution at all.)
answered Jan 6 at 18:13
fleabloodfleablood
68.9k22685
edited Jan 7 at 21:41
answered Jan 6 at 18:13
fleabloodfleablood
68.9k22685
answered Jan 6 at 18:13
fleabloodfleablood
68.9k22685
answered Jan 6 at 18:13
fleabloodfleablood
68.9k22685
68.9k22685
$begingroup$
In the second line, did you mean $y ne x$ and $0 le$ y $lt p$?
$endgroup$
– J. W. Tanner
Jan 7 at 18:47
$begingroup$
Yes, I did.......
$endgroup$
– fleablood
Jan 7 at 21:40
add a comment |
$begingroup$
In the second line, did you mean $y ne x$ and $0 le$ y $lt p$?
$endgroup$
– J. W. Tanner
Jan 7 at 18:47
$begingroup$
Yes, I did.......
$endgroup$
– fleablood
Jan 7 at 21:40
$begingroup$
In the second line, did you mean $y ne x$ and $0 le$ y $lt p$?
$endgroup$
– J. W. Tanner
Jan 7 at 18:47
$begingroup$
In the second line, did you mean $y ne x$ and $0 le$ y $lt p$?
$endgroup$
– J. W. Tanner
Jan 7 at 18:47
$begingroup$
Yes, I did.......
$endgroup$
– fleablood
Jan 7 at 21:40
$begingroup$
Yes, I did.......
$endgroup$
– fleablood
Jan 7 at 21:40
add a comment |
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4
$begingroup$
They mean solutions $pmod p$. Thus, if $p=7$ and $r=2$ the only solutions would be $3,4$. For $p=2, r=1$ the only solution is $x=1$.
$endgroup$
– lulu
Jan 6 at 17:06
$begingroup$
makes more sense! thank you - do you have any idea how to prove it though?
$endgroup$
– Student number x
Jan 6 at 17:17
$begingroup$
The mean solutions modulo $p$. There are only $p$ classes. If $p=2$ then are two possible $x$, $0$ and $1$. Any other odd number is $equiv 2 pmod 2$.
$endgroup$
– fleablood
Jan 6 at 17:53
$begingroup$
..... if $p = 2$ and $r=7$ then there is one solution mod $2$. $1^2 - 7equiv 0 pmod 2$ but $0^2 - 7 not equiv 0 pmod 2$. Any other odd number,say $253$ so that $253^2 - 7 equiv 0 pmod 2$ is the same solution as $253 equiv 1 pmod 2$. In the $mod 2$ system $253$ and $1$ are the same class $253 equiv 1$. So $1$ is the only solution.
$endgroup$
– fleablood
Jan 6 at 17:58
$begingroup$
For the record, If $x^2 equiv r pmod p$ then $(kp+x)equiv r pmod p$ so if there are any integer solutions there are an infinite number of INTEGER solutions, but we are not asking for integers solution but EQUIVALENCE solutions. $kp + x equiv x$ so they are the SAME solution. It's only one, even though ther are an infinite number of integers that are all EQUIVALENT to the one solution.
$endgroup$
– fleablood
Jan 6 at 18:16