For any prime $p$ not a divisor of $ab$, prove that $ax^2 + by^2 equiv с pmod{p}$ is solvable.












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Will Jagy
    Nov 24 '18 at 1:28






  • 2




    $begingroup$
    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/…
    $endgroup$
    – i707107
    Nov 24 '18 at 2:00
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Will Jagy
    Nov 24 '18 at 1:28






  • 2




    $begingroup$
    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/…
    $endgroup$
    – i707107
    Nov 24 '18 at 2:00














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 24 '18 at 1:53









Ricardo Largaespada

563213




563213










asked Nov 23 '18 at 23:03









abcdmathabcdmath

317110




317110












  • $begingroup$
    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.
    $endgroup$
    – Will Jagy
    Nov 24 '18 at 1:28






  • 2




    $begingroup$
    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/…
    $endgroup$
    – i707107
    Nov 24 '18 at 2:00


















  • $begingroup$
    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.
    $endgroup$
    – Will Jagy
    Nov 24 '18 at 1:28






  • 2




    $begingroup$
    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/…
    $endgroup$
    – i707107
    Nov 24 '18 at 2:00
















$begingroup$
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.
$endgroup$
– Will Jagy
Nov 24 '18 at 1:28




$begingroup$
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.
$endgroup$
– Will Jagy
Nov 24 '18 at 1:28




2




2




$begingroup$
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/…
$endgroup$
– i707107
Nov 24 '18 at 2:00




$begingroup$
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/…
$endgroup$
– i707107
Nov 24 '18 at 2:00










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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.






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    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 3:35









        multicuspmulticusp

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