Proving prime divisibility relation between $a^2-a+3$ and $b^2-b+25$.












2












$begingroup$


Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.



I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:



Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.



I had the idea of multiplying everything by $9$ because $3times 9approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.



According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).






number-theory prime-numbers modular-arithmetic contest-math divisibility







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
    $endgroup$
    – Doug M
    Jan 5 at 1:43










  • $begingroup$
    @DougM thanks, fixed
    $endgroup$
    – user574848
    Jan 5 at 1:58










  • $begingroup$
    What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
    $endgroup$
    – Will Jagy
    Jan 5 at 2:43


















2












$begingroup$


Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.



I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:



Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.



I had the idea of multiplying everything by $9$ because $3times 9approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.



According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).






number-theory prime-numbers modular-arithmetic contest-math divisibility







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
    $endgroup$
    – Doug M
    Jan 5 at 1:43










  • $begingroup$
    @DougM thanks, fixed
    $endgroup$
    – user574848
    Jan 5 at 1:58










  • $begingroup$
    What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
    $endgroup$
    – Will Jagy
    Jan 5 at 2:43
















2












2








2


1



$begingroup$


Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.



I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:



Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.



I had the idea of multiplying everything by $9$ because $3times 9approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.



According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).






number-theory prime-numbers modular-arithmetic contest-math divisibility







share|cite|improve this question











$endgroup$




Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.



I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:



Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.



I had the idea of multiplying everything by $9$ because $3times 9approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.



According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).






number-theory prime-numbers modular-arithmetic contest-math divisibility




number-theory prime-numbers modular-arithmetic contest-math divisibility






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 2:37







user574848

















asked Jan 5 at 1:29









user574848user574848

689118




689118








  • 2




    $begingroup$
    You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
    $endgroup$
    – Doug M
    Jan 5 at 1:43










  • $begingroup$
    @DougM thanks, fixed
    $endgroup$
    – user574848
    Jan 5 at 1:58










  • $begingroup$
    What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
    $endgroup$
    – Will Jagy
    Jan 5 at 2:43
















  • 2




    $begingroup$
    You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
    $endgroup$
    – Doug M
    Jan 5 at 1:43










  • $begingroup$
    @DougM thanks, fixed
    $endgroup$
    – user574848
    Jan 5 at 1:58










  • $begingroup$
    What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
    $endgroup$
    – Will Jagy
    Jan 5 at 2:43










2




2




$begingroup$
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
$endgroup$
– Doug M
Jan 5 at 1:43




$begingroup$
You have a flipped sign... $9a^2 - 6a + 1 = 3a - 26$
$endgroup$
– Doug M
Jan 5 at 1:43












$begingroup$
@DougM thanks, fixed
$endgroup$
– user574848
Jan 5 at 1:58




$begingroup$
@DougM thanks, fixed
$endgroup$
– user574848
Jan 5 at 1:58












$begingroup$
What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
$endgroup$
– Will Jagy
Jan 5 at 2:43






$begingroup$
What this is saying is that any (odd) prime for which $-11$ is a quadratic residue can be expressed as $x^2 + xy + 3y^2.$ If $p equiv 1 pmod 3,$ we also have $p = u^2 + uv + 25 v^2.$ If $p equiv 2 pmod 3,$ we also have $p = 5s^2 + st + 5 t^2.$ Of course, $3$ does not work right, there is an imprimitive form $3x^2 + 3xy+9y^2$ of discriminant $-99$
$endgroup$
– Will Jagy
Jan 5 at 2:43












3 Answers
3






active

oldest

votes


















3












$begingroup$

$3a - 1equiv bpmod p$



for any prime $p ne 3$



$3$ has a multiplicative inverse.



$a equiv 3^{-1}b + 3^{-1}pmod p$



And if $p=3$ let $a = 1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What's a multiplicative inverse?
    $endgroup$
    – user574848
    Jan 5 at 2:50










  • $begingroup$
    The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
    $endgroup$
    – Doug M
    Jan 5 at 3:34



















2












$begingroup$

$a^2-a+3equiv0pmod p$



$iff(2a-1)^2equiv-11$ for odd $p$



$4(b^2-b+25)=(2b-1)^2+99$



So, we need $(2b-1)^2equiv-99pmod p$



$implies(2b-1)^2equiv3^2(2a-1)^2pmod p$



$iff2b-1equivpm3(2a-1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
    $endgroup$
    – lab bhattacharjee
    Jan 5 at 10:00



















0












$begingroup$

Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
$$
exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
$$
which is equivalent to proving,
$$
left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
$$
where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.






share|cite|improve this answer









$endgroup$














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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    $3a - 1equiv bpmod p$



    for any prime $p ne 3$



    $3$ has a multiplicative inverse.



    $a equiv 3^{-1}b + 3^{-1}pmod p$



    And if $p=3$ let $a = 1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What's a multiplicative inverse?
      $endgroup$
      – user574848
      Jan 5 at 2:50










    • $begingroup$
      The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
      $endgroup$
      – Doug M
      Jan 5 at 3:34
















    3












    $begingroup$

    $3a - 1equiv bpmod p$



    for any prime $p ne 3$



    $3$ has a multiplicative inverse.



    $a equiv 3^{-1}b + 3^{-1}pmod p$



    And if $p=3$ let $a = 1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What's a multiplicative inverse?
      $endgroup$
      – user574848
      Jan 5 at 2:50










    • $begingroup$
      The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
      $endgroup$
      – Doug M
      Jan 5 at 3:34














    3












    3








    3





    $begingroup$

    $3a - 1equiv bpmod p$



    for any prime $p ne 3$



    $3$ has a multiplicative inverse.



    $a equiv 3^{-1}b + 3^{-1}pmod p$



    And if $p=3$ let $a = 1$






    share|cite|improve this answer









    $endgroup$



    $3a - 1equiv bpmod p$



    for any prime $p ne 3$



    $3$ has a multiplicative inverse.



    $a equiv 3^{-1}b + 3^{-1}pmod p$



    And if $p=3$ let $a = 1$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 5 at 2:10









    Doug MDoug M

    45.3k31954




    45.3k31954












    • $begingroup$
      What's a multiplicative inverse?
      $endgroup$
      – user574848
      Jan 5 at 2:50










    • $begingroup$
      The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
      $endgroup$
      – Doug M
      Jan 5 at 3:34


















    • $begingroup$
      What's a multiplicative inverse?
      $endgroup$
      – user574848
      Jan 5 at 2:50










    • $begingroup$
      The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
      $endgroup$
      – Doug M
      Jan 5 at 3:34
















    $begingroup$
    What's a multiplicative inverse?
    $endgroup$
    – user574848
    Jan 5 at 2:50




    $begingroup$
    What's a multiplicative inverse?
    $endgroup$
    – user574848
    Jan 5 at 2:50












    $begingroup$
    The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
    $endgroup$
    – Doug M
    Jan 5 at 3:34




    $begingroup$
    The integers modulo $p$ forms a group under multiplication. ($mathbb Z_p^times$) Which means that every member of the group (excluding 0) has a counterpart such that their product equals 1.
    $endgroup$
    – Doug M
    Jan 5 at 3:34











    2












    $begingroup$

    $a^2-a+3equiv0pmod p$



    $iff(2a-1)^2equiv-11$ for odd $p$



    $4(b^2-b+25)=(2b-1)^2+99$



    So, we need $(2b-1)^2equiv-99pmod p$



    $implies(2b-1)^2equiv3^2(2a-1)^2pmod p$



    $iff2b-1equivpm3(2a-1)$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
      $endgroup$
      – lab bhattacharjee
      Jan 5 at 10:00
















    2












    $begingroup$

    $a^2-a+3equiv0pmod p$



    $iff(2a-1)^2equiv-11$ for odd $p$



    $4(b^2-b+25)=(2b-1)^2+99$



    So, we need $(2b-1)^2equiv-99pmod p$



    $implies(2b-1)^2equiv3^2(2a-1)^2pmod p$



    $iff2b-1equivpm3(2a-1)$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
      $endgroup$
      – lab bhattacharjee
      Jan 5 at 10:00














    2












    2








    2





    $begingroup$

    $a^2-a+3equiv0pmod p$



    $iff(2a-1)^2equiv-11$ for odd $p$



    $4(b^2-b+25)=(2b-1)^2+99$



    So, we need $(2b-1)^2equiv-99pmod p$



    $implies(2b-1)^2equiv3^2(2a-1)^2pmod p$



    $iff2b-1equivpm3(2a-1)$






    share|cite|improve this answer









    $endgroup$



    $a^2-a+3equiv0pmod p$



    $iff(2a-1)^2equiv-11$ for odd $p$



    $4(b^2-b+25)=(2b-1)^2+99$



    So, we need $(2b-1)^2equiv-99pmod p$



    $implies(2b-1)^2equiv3^2(2a-1)^2pmod p$



    $iff2b-1equivpm3(2a-1)$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 5 at 2:28









    lab bhattacharjeelab bhattacharjee

    228k15158279




    228k15158279












    • $begingroup$
      @user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
      $endgroup$
      – lab bhattacharjee
      Jan 5 at 10:00


















    • $begingroup$
      @user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
      $endgroup$
      – lab bhattacharjee
      Jan 5 at 10:00
















    $begingroup$
    @user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
    $endgroup$
    – lab bhattacharjee
    Jan 5 at 10:00




    $begingroup$
    @user574848, $$4(a^2-a+3)=(2a-1)^2+11$$
    $endgroup$
    – lab bhattacharjee
    Jan 5 at 10:00











    0












    $begingroup$

    Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
    $$
    exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
    $$
    which is equivalent to proving,
    $$
    left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
    $$
    where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
      $$
      exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
      $$
      which is equivalent to proving,
      $$
      left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
      $$
      where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
        $$
        exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
        $$
        which is equivalent to proving,
        $$
        left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
        $$
        where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.






        share|cite|improve this answer









        $endgroup$



        Here is another proof, via quadratic residues. Note that, $a^2-a+3$ is odd, hence $pneq 2$. Now, $pmid a^2-a+3iff pmid (2a-1)^2+11$. Similarly, $pmid b^2-b+25 iff pmid (2b-1)^2+99$. Namely, the goal is to prove,
        $$
        exists a: pmid (2a-1)^2+11 iff exists b: pmid (2b-1)^2+99
        $$
        which is equivalent to proving,
        $$
        left(frac{-11}{p}right) =1 iff left(frac{-99}{p}right) =1,
        $$
        where, $(a/p)$ is the standard Legendre's symbol. But this last statement is obvious, since $(-99/p)=(9/p)(-11/p)=(-11/p)$, as $(9/p)=1$, due to the fact that $9$ is a quadratic residue modulo $p$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 15:14









        AaronAaron

        1,992415




        1,992415






























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