For any prime $p$ not a divisor of $ab$, prove that $ax^2 + by^2 equiv с pmod{p}$ is solvable.
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Given any integers $a,b,c$ and any prime $p$ not a divisor of $ab$, prove that $ax^2 + by^2
equiv с pmod{p} $ is solvable.
$p$ does not divide $ab$ implies $p$ does not divide both $a$ and $b$. But from here how can I show that the aforesaid equation is solvable?
elementary-number-theory
edited Nov 24 at 1:53
Ricardo Largaespada
558212
asked Nov 23 at 23:03
abcdmath
315110
add a comment |
up vote
0
down vote
favorite
Given any integers $a,b,c$ and any prime $p$ not a divisor of $ab$, prove that $ax^2 + by^2
equiv с pmod{p} $ is solvable.
$p$ does not divide $ab$ implies $p$ does not divide both $a$ and $b$. But from here how can I show that the aforesaid equation is solvable?
elementary-number-theory
edited Nov 24 at 1:53
Ricardo Largaespada
558212
asked Nov 23 at 23:03
abcdmath
315110
pigeonhole principle: there are, including $0,$ exactly $frac{p+1}{2}$ distinct values of $t^2 pmod p ; $ ( this is for odd prime $p$ ). There are $frac{p+1}{2}$ distinct values of $a x^2 pmod p ; $ and $frac{p+1}{2}$ distinct values of $c-b y^2 pmod p , ; $ so there is an overlap.
– Will Jagy
Nov 24 at 1:28
2
You can apply the method in one of my answer, and obtain a formula for the number of solutions to your equation: math.stackexchange.com/questions/398200/…
– i707107
Nov 24 at 2:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given any integers $a,b,c$ and any prime $p$ not a divisor of $ab$, prove that $ax^2 + by^2
equiv с pmod{p} $ is solvable.
$p$ does not divide $ab$ implies $p$ does not divide both $a$ and $b$. But from here how can I show that the aforesaid equation is solvable?
elementary-number-theory
edited Nov 24 at 1:53
Ricardo Largaespada
558212
asked Nov 23 at 23:03
abcdmath
315110
Given any integers $a,b,c$ and any prime $p$ not a divisor of $ab$, prove that $ax^2 + by^2
equiv с pmod{p} $ is solvable.
$p$ does not divide $ab$ implies $p$ does not divide both $a$ and $b$. But from here how can I show that the aforesaid equation is solvable?
elementary-number-theory
elementary-number-theory
edited Nov 24 at 1:53
Ricardo Largaespada
558212
asked Nov 23 at 23:03
abcdmath
315110
edited Nov 24 at 1:53
Ricardo Largaespada
558212
asked Nov 23 at 23:03
abcdmath
315110
edited Nov 24 at 1:53
Ricardo Largaespada
558212
edited Nov 24 at 1:53
Ricardo Largaespada
558212
edited Nov 24 at 1:53
Ricardo Largaespada
558212
558212
asked Nov 23 at 23:03
abcdmath
315110
asked Nov 23 at 23:03
abcdmath
315110
asked Nov 23 at 23:03
abcdmath
315110
315110
pigeonhole principle: there are, including $0,$ exactly $frac{p+1}{2}$ distinct values of $t^2 pmod p ; $ ( this is for odd prime $p$ ). There are $frac{p+1}{2}$ distinct values of $a x^2 pmod p ; $ and $frac{p+1}{2}$ distinct values of $c-b y^2 pmod p , ; $ so there is an overlap.
– Will Jagy
Nov 24 at 1:28
2
You can apply the method in one of my answer, and obtain a formula for the number of solutions to your equation: math.stackexchange.com/questions/398200/…
– i707107
Nov 24 at 2:00
add a comment |
pigeonhole principle: there are, including $0,$ exactly $frac{p+1}{2}$ distinct values of $t^2 pmod p ; $ ( this is for odd prime $p$ ). There are $frac{p+1}{2}$ distinct values of $a x^2 pmod p ; $ and $frac{p+1}{2}$ distinct values of $c-b y^2 pmod p , ; $ so there is an overlap.
– Will Jagy
Nov 24 at 1:28
2
You can apply the method in one of my answer, and obtain a formula for the number of solutions to your equation: math.stackexchange.com/questions/398200/…
– i707107
Nov 24 at 2:00
pigeonhole principle: there are, including $0,$ exactly $frac{p+1}{2}$ distinct values of $t^2 pmod p ; $ ( this is for odd prime $p$ ). There are $frac{p+1}{2}$ distinct values of $a x^2 pmod p ; $ and $frac{p+1}{2}$ distinct values of $c-b y^2 pmod p , ; $ so there is an overlap.
– Will Jagy
Nov 24 at 1:28
pigeonhole principle: there are, including $0,$ exactly $frac{p+1}{2}$ distinct values of $t^2 pmod p ; $ ( this is for odd prime $p$ ). There are $frac{p+1}{2}$ distinct values of $a x^2 pmod p ; $ and $frac{p+1}{2}$ distinct values of $c-b y^2 pmod p , ; $ so there is an overlap.
– Will Jagy
Nov 24 at 1:28
2
2
You can apply the method in one of my answer, and obtain a formula for the number of solutions to your equation: math.stackexchange.com/questions/398200/…
– i707107
Nov 24 at 2:00
You can apply the method in one of my answer, and obtain a formula for the number of solutions to your equation: math.stackexchange.com/questions/398200/…
– i707107
Nov 24 at 2:00
add a comment |
1 Answer
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0
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I might be over simplifying your problem a bit. Let $x,y in mathbb{Z}$. Therefore $x^{2}$ and $y^{2}$ is equal to some integer $q_{1}$ and $q_{2}$ respectively. This is due to the property of multiplicative closure of integers. The left hand side of the equation can now be rewritten as $aq_{1}+bq_{2}$. Once again we can say that $aq_{1}$ is some integer due to integer closure, and $bq_{2}$ is, likewise, an integer by the same argument. By integer closure under addition the two integer products can be rewritten as a single integer called $m$.
Now we can rewrite the equation as follows: $m=c mod(p)$. Which states that there is an integer $m$ that can be rewritten as a distinct congruence class $c$. This can also be rewritten as $m=pq_{3}+c$, for $c<p$ and $q_{3}in mathbb{Z}$
I hope this helps some. Sorry if it was a bit long, I am not sure what direction you wanted to go in.
answered Nov 24 at 3:35
multicusp
211
add a comment |
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I might be over simplifying your problem a bit. Let $x,y in mathbb{Z}$. Therefore $x^{2}$ and $y^{2}$ is equal to some integer $q_{1}$ and $q_{2}$ respectively. This is due to the property of multiplicative closure of integers. The left hand side of the equation can now be rewritten as $aq_{1}+bq_{2}$. Once again we can say that $aq_{1}$ is some integer due to integer closure, and $bq_{2}$ is, likewise, an integer by the same argument. By integer closure under addition the two integer products can be rewritten as a single integer called $m$.
Now we can rewrite the equation as follows: $m=c mod(p)$. Which states that there is an integer $m$ that can be rewritten as a distinct congruence class $c$. This can also be rewritten as $m=pq_{3}+c$, for $c<p$ and $q_{3}in mathbb{Z}$
I hope this helps some. Sorry if it was a bit long, I am not sure what direction you wanted to go in.
answered Nov 24 at 3:35
multicusp
211
add a comment |
up vote
0
down vote
I might be over simplifying your problem a bit. Let $x,y in mathbb{Z}$. Therefore $x^{2}$ and $y^{2}$ is equal to some integer $q_{1}$ and $q_{2}$ respectively. This is due to the property of multiplicative closure of integers. The left hand side of the equation can now be rewritten as $aq_{1}+bq_{2}$. Once again we can say that $aq_{1}$ is some integer due to integer closure, and $bq_{2}$ is, likewise, an integer by the same argument. By integer closure under addition the two integer products can be rewritten as a single integer called $m$.
Now we can rewrite the equation as follows: $m=c mod(p)$. Which states that there is an integer $m$ that can be rewritten as a distinct congruence class $c$. This can also be rewritten as $m=pq_{3}+c$, for $c<p$ and $q_{3}in mathbb{Z}$
I hope this helps some. Sorry if it was a bit long, I am not sure what direction you wanted to go in.
answered Nov 24 at 3:35
multicusp
211
add a comment |
up vote
0
down vote
up vote
0
down vote
I might be over simplifying your problem a bit. Let $x,y in mathbb{Z}$. Therefore $x^{2}$ and $y^{2}$ is equal to some integer $q_{1}$ and $q_{2}$ respectively. This is due to the property of multiplicative closure of integers. The left hand side of the equation can now be rewritten as $aq_{1}+bq_{2}$. Once again we can say that $aq_{1}$ is some integer due to integer closure, and $bq_{2}$ is, likewise, an integer by the same argument. By integer closure under addition the two integer products can be rewritten as a single integer called $m$.
Now we can rewrite the equation as follows: $m=c mod(p)$. Which states that there is an integer $m$ that can be rewritten as a distinct congruence class $c$. This can also be rewritten as $m=pq_{3}+c$, for $c<p$ and $q_{3}in mathbb{Z}$
I hope this helps some. Sorry if it was a bit long, I am not sure what direction you wanted to go in.
answered Nov 24 at 3:35
multicusp
211
I might be over simplifying your problem a bit. Let $x,y in mathbb{Z}$. Therefore $x^{2}$ and $y^{2}$ is equal to some integer $q_{1}$ and $q_{2}$ respectively. This is due to the property of multiplicative closure of integers. The left hand side of the equation can now be rewritten as $aq_{1}+bq_{2}$. Once again we can say that $aq_{1}$ is some integer due to integer closure, and $bq_{2}$ is, likewise, an integer by the same argument. By integer closure under addition the two integer products can be rewritten as a single integer called $m$.
Now we can rewrite the equation as follows: $m=c mod(p)$. Which states that there is an integer $m$ that can be rewritten as a distinct congruence class $c$. This can also be rewritten as $m=pq_{3}+c$, for $c<p$ and $q_{3}in mathbb{Z}$
I hope this helps some. Sorry if it was a bit long, I am not sure what direction you wanted to go in.
answered Nov 24 at 3:35
multicusp
211
answered Nov 24 at 3:35
multicusp
211
answered Nov 24 at 3:35
multicusp
211
answered Nov 24 at 3:35
multicusp
211
211
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pigeonhole principle: there are, including $0,$ exactly $frac{p+1}{2}$ distinct values of $t^2 pmod p ; $ ( this is for odd prime $p$ ). There are $frac{p+1}{2}$ distinct values of $a x^2 pmod p ; $ and $frac{p+1}{2}$ distinct values of $c-b y^2 pmod p , ; $ so there is an overlap.
– Will Jagy
Nov 24 at 1:28
2
You can apply the method in one of my answer, and obtain a formula for the number of solutions to your equation: math.stackexchange.com/questions/398200/…
– i707107
Nov 24 at 2:00