Solve equation in prime numbers












2












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Solve the equation in prime number $$p^3+q^3+1=p^2 q^2.$$




I have found the solutions $(2,3), (3,2)$ and need to prove that there are no other solutions. I think that there is an inequality $p^3+q^3+1leq p^2 q^2$ for $p,q >2 $ but how to prove it.






elementary-number-theory prime-numbers diophantine-equations







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  • 2




    $begingroup$
    your inequality is not right when $p>q^2$, for then $p^3 > p^2 q^2$.
    $endgroup$
    – dyf
    Nov 9 '18 at 20:36
















2












$begingroup$



Solve the equation in prime number $$p^3+q^3+1=p^2 q^2.$$




I have found the solutions $(2,3), (3,2)$ and need to prove that there are no other solutions. I think that there is an inequality $p^3+q^3+1leq p^2 q^2$ for $p,q >2 $ but how to prove it.






elementary-number-theory prime-numbers diophantine-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    your inequality is not right when $p>q^2$, for then $p^3 > p^2 q^2$.
    $endgroup$
    – dyf
    Nov 9 '18 at 20:36














2












2








2


1



$begingroup$



Solve the equation in prime number $$p^3+q^3+1=p^2 q^2.$$




I have found the solutions $(2,3), (3,2)$ and need to prove that there are no other solutions. I think that there is an inequality $p^3+q^3+1leq p^2 q^2$ for $p,q >2 $ but how to prove it.






elementary-number-theory prime-numbers diophantine-equations







share|cite|improve this question











$endgroup$





Solve the equation in prime number $$p^3+q^3+1=p^2 q^2.$$




I have found the solutions $(2,3), (3,2)$ and need to prove that there are no other solutions. I think that there is an inequality $p^3+q^3+1leq p^2 q^2$ for $p,q >2 $ but how to prove it.






elementary-number-theory prime-numbers diophantine-equations




elementary-number-theory prime-numbers diophantine-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 21:49









Maria Mazur

49.9k1361124




49.9k1361124










asked Nov 9 '18 at 20:29









LeoxLeox

5,3481424




5,3481424








  • 2




    $begingroup$
    your inequality is not right when $p>q^2$, for then $p^3 > p^2 q^2$.
    $endgroup$
    – dyf
    Nov 9 '18 at 20:36














  • 2




    $begingroup$
    your inequality is not right when $p>q^2$, for then $p^3 > p^2 q^2$.
    $endgroup$
    – dyf
    Nov 9 '18 at 20:36








2




2




$begingroup$
your inequality is not right when $p>q^2$, for then $p^3 > p^2 q^2$.
$endgroup$
– dyf
Nov 9 '18 at 20:36




$begingroup$
your inequality is not right when $p>q^2$, for then $p^3 > p^2 q^2$.
$endgroup$
– dyf
Nov 9 '18 at 20:36










2 Answers
2






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oldest

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6












$begingroup$

If $p=q$ we get $2p^3+1=p^4$ so $pmid 1$ which is impossible. So we can assume $pne q$.



Because of simmetry we can also assume that $p>q$.



Now we have $$p^2mid q^3+1 = (q+1)(q^2-q+1)$$



Case 1: $pnot{mid};q+1$, then $$p^2mid q^2-q+1implies p^2leq q^2-q+1<q^2$$
so $p<q$, a contradiction.



Case 2: $pmid q+1$ then $pleq q+1$ so $p=q+1$ (remember that $p>q$, so $pgeq q+1$). So $p$ and $q$ are consecutive primes so one is even, so $q=2$ and $p=3$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    My solution



    Case 1. Let $p mod 3=q mod 3=1$. Then $p^3+q^3+1=3=0 mod 3 neq p^2 q^2=1 mod 3$



    Case 2 $p mod 3=q mod 3=2$. Then $p^3+q^3+1=2+2+1=2 mod 3 neq p^2 q^2=1 mod 3$



    Сase 3. $p mod 3=1, q mod 3=2$. Then $p^3+q^3+1=1+2+1=1 mod 3 neq p^2 q^2=1 cdot 2 =2 mod 3$



    So, must be, say $p=0 mod 3 implies p=3$. We get the equation $q^3+28=9 q^2 $ or $q^3-9 q^2+28=left( q-2 right) left( {q}^{2}-7,q-14 right) $ and $q=2.$ By symmetry another solution is $p=2, q=3.$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      you have an important error, in Case 3, actually both $p^2, q^2 equiv 1 pmod 3, $ therefore $p^2 q^2 equiv 1 pmod 3, $
      $endgroup$
      – Will Jagy
      Nov 9 '18 at 23:14












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    2 Answers
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    2 Answers
    2






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    active

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    6












    $begingroup$

    If $p=q$ we get $2p^3+1=p^4$ so $pmid 1$ which is impossible. So we can assume $pne q$.



    Because of simmetry we can also assume that $p>q$.



    Now we have $$p^2mid q^3+1 = (q+1)(q^2-q+1)$$



    Case 1: $pnot{mid};q+1$, then $$p^2mid q^2-q+1implies p^2leq q^2-q+1<q^2$$
    so $p<q$, a contradiction.



    Case 2: $pmid q+1$ then $pleq q+1$ so $p=q+1$ (remember that $p>q$, so $pgeq q+1$). So $p$ and $q$ are consecutive primes so one is even, so $q=2$ and $p=3$.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      If $p=q$ we get $2p^3+1=p^4$ so $pmid 1$ which is impossible. So we can assume $pne q$.



      Because of simmetry we can also assume that $p>q$.



      Now we have $$p^2mid q^3+1 = (q+1)(q^2-q+1)$$



      Case 1: $pnot{mid};q+1$, then $$p^2mid q^2-q+1implies p^2leq q^2-q+1<q^2$$
      so $p<q$, a contradiction.



      Case 2: $pmid q+1$ then $pleq q+1$ so $p=q+1$ (remember that $p>q$, so $pgeq q+1$). So $p$ and $q$ are consecutive primes so one is even, so $q=2$ and $p=3$.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        If $p=q$ we get $2p^3+1=p^4$ so $pmid 1$ which is impossible. So we can assume $pne q$.



        Because of simmetry we can also assume that $p>q$.



        Now we have $$p^2mid q^3+1 = (q+1)(q^2-q+1)$$



        Case 1: $pnot{mid};q+1$, then $$p^2mid q^2-q+1implies p^2leq q^2-q+1<q^2$$
        so $p<q$, a contradiction.



        Case 2: $pmid q+1$ then $pleq q+1$ so $p=q+1$ (remember that $p>q$, so $pgeq q+1$). So $p$ and $q$ are consecutive primes so one is even, so $q=2$ and $p=3$.






        share|cite|improve this answer











        $endgroup$



        If $p=q$ we get $2p^3+1=p^4$ so $pmid 1$ which is impossible. So we can assume $pne q$.



        Because of simmetry we can also assume that $p>q$.



        Now we have $$p^2mid q^3+1 = (q+1)(q^2-q+1)$$



        Case 1: $pnot{mid};q+1$, then $$p^2mid q^2-q+1implies p^2leq q^2-q+1<q^2$$
        so $p<q$, a contradiction.



        Case 2: $pmid q+1$ then $pleq q+1$ so $p=q+1$ (remember that $p>q$, so $pgeq q+1$). So $p$ and $q$ are consecutive primes so one is even, so $q=2$ and $p=3$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 9 '18 at 21:26

























        answered Nov 9 '18 at 21:20









        Maria MazurMaria Mazur

        49.9k1361124




        49.9k1361124























            0












            $begingroup$

            My solution



            Case 1. Let $p mod 3=q mod 3=1$. Then $p^3+q^3+1=3=0 mod 3 neq p^2 q^2=1 mod 3$



            Case 2 $p mod 3=q mod 3=2$. Then $p^3+q^3+1=2+2+1=2 mod 3 neq p^2 q^2=1 mod 3$



            Сase 3. $p mod 3=1, q mod 3=2$. Then $p^3+q^3+1=1+2+1=1 mod 3 neq p^2 q^2=1 cdot 2 =2 mod 3$



            So, must be, say $p=0 mod 3 implies p=3$. We get the equation $q^3+28=9 q^2 $ or $q^3-9 q^2+28=left( q-2 right) left( {q}^{2}-7,q-14 right) $ and $q=2.$ By symmetry another solution is $p=2, q=3.$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              you have an important error, in Case 3, actually both $p^2, q^2 equiv 1 pmod 3, $ therefore $p^2 q^2 equiv 1 pmod 3, $
              $endgroup$
              – Will Jagy
              Nov 9 '18 at 23:14
















            0












            $begingroup$

            My solution



            Case 1. Let $p mod 3=q mod 3=1$. Then $p^3+q^3+1=3=0 mod 3 neq p^2 q^2=1 mod 3$



            Case 2 $p mod 3=q mod 3=2$. Then $p^3+q^3+1=2+2+1=2 mod 3 neq p^2 q^2=1 mod 3$



            Сase 3. $p mod 3=1, q mod 3=2$. Then $p^3+q^3+1=1+2+1=1 mod 3 neq p^2 q^2=1 cdot 2 =2 mod 3$



            So, must be, say $p=0 mod 3 implies p=3$. We get the equation $q^3+28=9 q^2 $ or $q^3-9 q^2+28=left( q-2 right) left( {q}^{2}-7,q-14 right) $ and $q=2.$ By symmetry another solution is $p=2, q=3.$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              you have an important error, in Case 3, actually both $p^2, q^2 equiv 1 pmod 3, $ therefore $p^2 q^2 equiv 1 pmod 3, $
              $endgroup$
              – Will Jagy
              Nov 9 '18 at 23:14














            0












            0








            0





            $begingroup$

            My solution



            Case 1. Let $p mod 3=q mod 3=1$. Then $p^3+q^3+1=3=0 mod 3 neq p^2 q^2=1 mod 3$



            Case 2 $p mod 3=q mod 3=2$. Then $p^3+q^3+1=2+2+1=2 mod 3 neq p^2 q^2=1 mod 3$



            Сase 3. $p mod 3=1, q mod 3=2$. Then $p^3+q^3+1=1+2+1=1 mod 3 neq p^2 q^2=1 cdot 2 =2 mod 3$



            So, must be, say $p=0 mod 3 implies p=3$. We get the equation $q^3+28=9 q^2 $ or $q^3-9 q^2+28=left( q-2 right) left( {q}^{2}-7,q-14 right) $ and $q=2.$ By symmetry another solution is $p=2, q=3.$






            share|cite|improve this answer









            $endgroup$



            My solution



            Case 1. Let $p mod 3=q mod 3=1$. Then $p^3+q^3+1=3=0 mod 3 neq p^2 q^2=1 mod 3$



            Case 2 $p mod 3=q mod 3=2$. Then $p^3+q^3+1=2+2+1=2 mod 3 neq p^2 q^2=1 mod 3$



            Сase 3. $p mod 3=1, q mod 3=2$. Then $p^3+q^3+1=1+2+1=1 mod 3 neq p^2 q^2=1 cdot 2 =2 mod 3$



            So, must be, say $p=0 mod 3 implies p=3$. We get the equation $q^3+28=9 q^2 $ or $q^3-9 q^2+28=left( q-2 right) left( {q}^{2}-7,q-14 right) $ and $q=2.$ By symmetry another solution is $p=2, q=3.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 9 '18 at 21:59









            LeoxLeox

            5,3481424




            5,3481424








            • 1




              $begingroup$
              you have an important error, in Case 3, actually both $p^2, q^2 equiv 1 pmod 3, $ therefore $p^2 q^2 equiv 1 pmod 3, $
              $endgroup$
              – Will Jagy
              Nov 9 '18 at 23:14














            • 1




              $begingroup$
              you have an important error, in Case 3, actually both $p^2, q^2 equiv 1 pmod 3, $ therefore $p^2 q^2 equiv 1 pmod 3, $
              $endgroup$
              – Will Jagy
              Nov 9 '18 at 23:14








            1




            1




            $begingroup$
            you have an important error, in Case 3, actually both $p^2, q^2 equiv 1 pmod 3, $ therefore $p^2 q^2 equiv 1 pmod 3, $
            $endgroup$
            – Will Jagy
            Nov 9 '18 at 23:14




            $begingroup$
            you have an important error, in Case 3, actually both $p^2, q^2 equiv 1 pmod 3, $ therefore $p^2 q^2 equiv 1 pmod 3, $
            $endgroup$
            – Will Jagy
            Nov 9 '18 at 23:14


















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