Find the values of $p$ and $q$












6















If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
Find all the solutions (p, q)




I tried to solve this exercise using that:



$p^2 = -1(text{mod} , q)$ and $q = -1(text{mod} , p)$;
So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.



Then I tried to solve a quadratic equation, but I could not finish the problem






algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations







share|cite|improve this question





























    6















    If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
    Find all the solutions (p, q)




    I tried to solve this exercise using that:



    $p^2 = -1(text{mod} , q)$ and $q = -1(text{mod} , p)$;
    So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.



    Then I tried to solve a quadratic equation, but I could not finish the problem






    algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations







    share|cite|improve this question



























      6












      6








      6








      If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
      Find all the solutions (p, q)




      I tried to solve this exercise using that:



      $p^2 = -1(text{mod} , q)$ and $q = -1(text{mod} , p)$;
      So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.



      Then I tried to solve a quadratic equation, but I could not finish the problem






      algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations







      share|cite|improve this question
















      If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
      Find all the solutions (p, q)




      I tried to solve this exercise using that:



      $p^2 = -1(text{mod} , q)$ and $q = -1(text{mod} , p)$;
      So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.



      Then I tried to solve a quadratic equation, but I could not finish the problem






      algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations




      algebra-precalculus elementary-number-theory polynomials 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 Nov 28 '18 at 1:50









      Servaes

      22.4k33793




      22.4k33793










      asked Oct 6 '18 at 18:43









      Matheus Domingos

      956




      956






















          4 Answers
          4






          active

          oldest

          votes


















          3














          Clearly $pneq q$, and because $p$ and $q$ are prime and
          $$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
          we must have $pmid q+1$ and $qmid p^2+1$. Write
          $$q+1=apqquadtext{ and }qquad p^2+1=bq,$$
          to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmod{p}$, say $b=cp-1$, but then
          $$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
          Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
          $$(ac-1)p=a+c,$$
          and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
          $$p^2+1=bq=(cp-1)q=(p-1)q.$$
          In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.






          share|cite|improve this answer























          • How do I finish this solution? Finding all the solution?
            – Matheus Domingos
            Oct 6 '18 at 19:05










          • @MatheusDomingos I completed the solution for you.
            – Servaes
            Oct 6 '18 at 19:05






          • 1




            Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
            – Matheus Domingos
            Oct 6 '18 at 19:11






          • 1




            I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
            – Servaes
            Oct 6 '18 at 19:12












          • Understood, more elegant this way,thanks again
            – Matheus Domingos
            Oct 6 '18 at 19:14



















          5














          $$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$



          Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to



          $$p^2-k^2p+(k+1)=0,$$



          which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?






          share|cite|improve this answer























          • Oh thanks, now I know how to finish it, thanks
            – Matheus Domingos
            Oct 6 '18 at 18:55










          • Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
            – Matheus Domingos
            Oct 6 '18 at 19:12










          • @MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
            – Carl Schildkraut
            Oct 6 '18 at 19:16






          • 1




            Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
            – Servaes
            Oct 6 '18 at 19:17












          • Sorry, I didn't understood how to finish using it
            – Matheus Domingos
            Oct 6 '18 at 19:24



















          2














          Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?






          share|cite|improve this answer





























            2














            Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.



            From $$p(p^2+1)= q(q+1)implies pmid q;;;{rm or};;;pmid q+1$$



            1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$



            Since $p^2+1geq s$ we have 2 subcases:



            1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq {p^2+1over p+1} <pimplies sleq p-1$$



            So $q+1leq p-1 leq q-1$ and thus no solution.



            1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.



            2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$



            so we have $pqleq p^2+q+1$ so $q leq {p^2+1over p-1} leq p+2$ if $pgeq 3$.
            So if $pgeq 3$ and since $pmid q$ that $qin {p,p+1,p+2}$ which is easy to finish by hand.






            share|cite|improve this answer























            • Thanks, got this solution!
              – Matheus Domingos
              Oct 6 '18 at 19:25











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            4 Answers
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            active

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            4 Answers
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            active

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            oldest

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            3














            Clearly $pneq q$, and because $p$ and $q$ are prime and
            $$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
            we must have $pmid q+1$ and $qmid p^2+1$. Write
            $$q+1=apqquadtext{ and }qquad p^2+1=bq,$$
            to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmod{p}$, say $b=cp-1$, but then
            $$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
            Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
            $$(ac-1)p=a+c,$$
            and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
            $$p^2+1=bq=(cp-1)q=(p-1)q.$$
            In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.






            share|cite|improve this answer























            • How do I finish this solution? Finding all the solution?
              – Matheus Domingos
              Oct 6 '18 at 19:05










            • @MatheusDomingos I completed the solution for you.
              – Servaes
              Oct 6 '18 at 19:05






            • 1




              Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
              – Matheus Domingos
              Oct 6 '18 at 19:11






            • 1




              I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
              – Servaes
              Oct 6 '18 at 19:12












            • Understood, more elegant this way,thanks again
              – Matheus Domingos
              Oct 6 '18 at 19:14
















            3














            Clearly $pneq q$, and because $p$ and $q$ are prime and
            $$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
            we must have $pmid q+1$ and $qmid p^2+1$. Write
            $$q+1=apqquadtext{ and }qquad p^2+1=bq,$$
            to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmod{p}$, say $b=cp-1$, but then
            $$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
            Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
            $$(ac-1)p=a+c,$$
            and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
            $$p^2+1=bq=(cp-1)q=(p-1)q.$$
            In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.






            share|cite|improve this answer























            • How do I finish this solution? Finding all the solution?
              – Matheus Domingos
              Oct 6 '18 at 19:05










            • @MatheusDomingos I completed the solution for you.
              – Servaes
              Oct 6 '18 at 19:05






            • 1




              Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
              – Matheus Domingos
              Oct 6 '18 at 19:11






            • 1




              I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
              – Servaes
              Oct 6 '18 at 19:12












            • Understood, more elegant this way,thanks again
              – Matheus Domingos
              Oct 6 '18 at 19:14














            3












            3








            3






            Clearly $pneq q$, and because $p$ and $q$ are prime and
            $$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
            we must have $pmid q+1$ and $qmid p^2+1$. Write
            $$q+1=apqquadtext{ and }qquad p^2+1=bq,$$
            to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmod{p}$, say $b=cp-1$, but then
            $$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
            Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
            $$(ac-1)p=a+c,$$
            and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
            $$p^2+1=bq=(cp-1)q=(p-1)q.$$
            In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.






            share|cite|improve this answer














            Clearly $pneq q$, and because $p$ and $q$ are prime and
            $$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
            we must have $pmid q+1$ and $qmid p^2+1$. Write
            $$q+1=apqquadtext{ and }qquad p^2+1=bq,$$
            to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmod{p}$, say $b=cp-1$, but then
            $$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
            Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
            $$(ac-1)p=a+c,$$
            and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
            $$p^2+1=bq=(cp-1)q=(p-1)q.$$
            In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 6 '18 at 19:13

























            answered Oct 6 '18 at 18:53









            Servaes

            22.4k33793




            22.4k33793












            • How do I finish this solution? Finding all the solution?
              – Matheus Domingos
              Oct 6 '18 at 19:05










            • @MatheusDomingos I completed the solution for you.
              – Servaes
              Oct 6 '18 at 19:05






            • 1




              Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
              – Matheus Domingos
              Oct 6 '18 at 19:11






            • 1




              I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
              – Servaes
              Oct 6 '18 at 19:12












            • Understood, more elegant this way,thanks again
              – Matheus Domingos
              Oct 6 '18 at 19:14


















            • How do I finish this solution? Finding all the solution?
              – Matheus Domingos
              Oct 6 '18 at 19:05










            • @MatheusDomingos I completed the solution for you.
              – Servaes
              Oct 6 '18 at 19:05






            • 1




              Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
              – Matheus Domingos
              Oct 6 '18 at 19:11






            • 1




              I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
              – Servaes
              Oct 6 '18 at 19:12












            • Understood, more elegant this way,thanks again
              – Matheus Domingos
              Oct 6 '18 at 19:14
















            How do I finish this solution? Finding all the solution?
            – Matheus Domingos
            Oct 6 '18 at 19:05




            How do I finish this solution? Finding all the solution?
            – Matheus Domingos
            Oct 6 '18 at 19:05












            @MatheusDomingos I completed the solution for you.
            – Servaes
            Oct 6 '18 at 19:05




            @MatheusDomingos I completed the solution for you.
            – Servaes
            Oct 6 '18 at 19:05




            1




            1




            Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
            – Matheus Domingos
            Oct 6 '18 at 19:11




            Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
            – Matheus Domingos
            Oct 6 '18 at 19:11




            1




            1




            I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
            – Servaes
            Oct 6 '18 at 19:12






            I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
            – Servaes
            Oct 6 '18 at 19:12














            Understood, more elegant this way,thanks again
            – Matheus Domingos
            Oct 6 '18 at 19:14




            Understood, more elegant this way,thanks again
            – Matheus Domingos
            Oct 6 '18 at 19:14











            5














            $$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$



            Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to



            $$p^2-k^2p+(k+1)=0,$$



            which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?






            share|cite|improve this answer























            • Oh thanks, now I know how to finish it, thanks
              – Matheus Domingos
              Oct 6 '18 at 18:55










            • Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
              – Matheus Domingos
              Oct 6 '18 at 19:12










            • @MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
              – Carl Schildkraut
              Oct 6 '18 at 19:16






            • 1




              Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
              – Servaes
              Oct 6 '18 at 19:17












            • Sorry, I didn't understood how to finish using it
              – Matheus Domingos
              Oct 6 '18 at 19:24
















            5














            $$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$



            Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to



            $$p^2-k^2p+(k+1)=0,$$



            which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?






            share|cite|improve this answer























            • Oh thanks, now I know how to finish it, thanks
              – Matheus Domingos
              Oct 6 '18 at 18:55










            • Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
              – Matheus Domingos
              Oct 6 '18 at 19:12










            • @MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
              – Carl Schildkraut
              Oct 6 '18 at 19:16






            • 1




              Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
              – Servaes
              Oct 6 '18 at 19:17












            • Sorry, I didn't understood how to finish using it
              – Matheus Domingos
              Oct 6 '18 at 19:24














            5












            5








            5






            $$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$



            Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to



            $$p^2-k^2p+(k+1)=0,$$



            which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?






            share|cite|improve this answer














            $$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$



            Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to



            $$p^2-k^2p+(k+1)=0,$$



            which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 6 '18 at 19:01

























            answered Oct 6 '18 at 18:54









            Carl Schildkraut

            11.2k11441




            11.2k11441












            • Oh thanks, now I know how to finish it, thanks
              – Matheus Domingos
              Oct 6 '18 at 18:55










            • Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
              – Matheus Domingos
              Oct 6 '18 at 19:12










            • @MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
              – Carl Schildkraut
              Oct 6 '18 at 19:16






            • 1




              Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
              – Servaes
              Oct 6 '18 at 19:17












            • Sorry, I didn't understood how to finish using it
              – Matheus Domingos
              Oct 6 '18 at 19:24


















            • Oh thanks, now I know how to finish it, thanks
              – Matheus Domingos
              Oct 6 '18 at 18:55










            • Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
              – Matheus Domingos
              Oct 6 '18 at 19:12










            • @MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
              – Carl Schildkraut
              Oct 6 '18 at 19:16






            • 1




              Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
              – Servaes
              Oct 6 '18 at 19:17












            • Sorry, I didn't understood how to finish using it
              – Matheus Domingos
              Oct 6 '18 at 19:24
















            Oh thanks, now I know how to finish it, thanks
            – Matheus Domingos
            Oct 6 '18 at 18:55




            Oh thanks, now I know how to finish it, thanks
            – Matheus Domingos
            Oct 6 '18 at 18:55












            Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
            – Matheus Domingos
            Oct 6 '18 at 19:12




            Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
            – Matheus Domingos
            Oct 6 '18 at 19:12












            @MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
            – Carl Schildkraut
            Oct 6 '18 at 19:16




            @MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
            – Carl Schildkraut
            Oct 6 '18 at 19:16




            1




            1




            Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
            – Servaes
            Oct 6 '18 at 19:17






            Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
            – Servaes
            Oct 6 '18 at 19:17














            Sorry, I didn't understood how to finish using it
            – Matheus Domingos
            Oct 6 '18 at 19:24




            Sorry, I didn't understood how to finish using it
            – Matheus Domingos
            Oct 6 '18 at 19:24











            2














            Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?






            share|cite|improve this answer


























              2














              Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?






              share|cite|improve this answer
























                2












                2








                2






                Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?






                share|cite|improve this answer












                Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 6 '18 at 18:52









                Dietrich Burde

                77.7k64386




                77.7k64386























                    2














                    Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.



                    From $$p(p^2+1)= q(q+1)implies pmid q;;;{rm or};;;pmid q+1$$



                    1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$



                    Since $p^2+1geq s$ we have 2 subcases:



                    1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq {p^2+1over p+1} <pimplies sleq p-1$$



                    So $q+1leq p-1 leq q-1$ and thus no solution.



                    1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.



                    2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$



                    so we have $pqleq p^2+q+1$ so $q leq {p^2+1over p-1} leq p+2$ if $pgeq 3$.
                    So if $pgeq 3$ and since $pmid q$ that $qin {p,p+1,p+2}$ which is easy to finish by hand.






                    share|cite|improve this answer























                    • Thanks, got this solution!
                      – Matheus Domingos
                      Oct 6 '18 at 19:25
















                    2














                    Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.



                    From $$p(p^2+1)= q(q+1)implies pmid q;;;{rm or};;;pmid q+1$$



                    1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$



                    Since $p^2+1geq s$ we have 2 subcases:



                    1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq {p^2+1over p+1} <pimplies sleq p-1$$



                    So $q+1leq p-1 leq q-1$ and thus no solution.



                    1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.



                    2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$



                    so we have $pqleq p^2+q+1$ so $q leq {p^2+1over p-1} leq p+2$ if $pgeq 3$.
                    So if $pgeq 3$ and since $pmid q$ that $qin {p,p+1,p+2}$ which is easy to finish by hand.






                    share|cite|improve this answer























                    • Thanks, got this solution!
                      – Matheus Domingos
                      Oct 6 '18 at 19:25














                    2












                    2








                    2






                    Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.



                    From $$p(p^2+1)= q(q+1)implies pmid q;;;{rm or};;;pmid q+1$$



                    1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$



                    Since $p^2+1geq s$ we have 2 subcases:



                    1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq {p^2+1over p+1} <pimplies sleq p-1$$



                    So $q+1leq p-1 leq q-1$ and thus no solution.



                    1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.



                    2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$



                    so we have $pqleq p^2+q+1$ so $q leq {p^2+1over p-1} leq p+2$ if $pgeq 3$.
                    So if $pgeq 3$ and since $pmid q$ that $qin {p,p+1,p+2}$ which is easy to finish by hand.






                    share|cite|improve this answer














                    Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.



                    From $$p(p^2+1)= q(q+1)implies pmid q;;;{rm or};;;pmid q+1$$



                    1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$



                    Since $p^2+1geq s$ we have 2 subcases:



                    1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq {p^2+1over p+1} <pimplies sleq p-1$$



                    So $q+1leq p-1 leq q-1$ and thus no solution.



                    1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.



                    2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$



                    so we have $pqleq p^2+q+1$ so $q leq {p^2+1over p-1} leq p+2$ if $pgeq 3$.
                    So if $pgeq 3$ and since $pmid q$ that $qin {p,p+1,p+2}$ which is easy to finish by hand.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Oct 6 '18 at 19:11

























                    answered Oct 6 '18 at 18:54









                    greedoid

                    38k114794




                    38k114794












                    • Thanks, got this solution!
                      – Matheus Domingos
                      Oct 6 '18 at 19:25


















                    • Thanks, got this solution!
                      – Matheus Domingos
                      Oct 6 '18 at 19:25
















                    Thanks, got this solution!
                    – Matheus Domingos
                    Oct 6 '18 at 19:25




                    Thanks, got this solution!
                    – Matheus Domingos
                    Oct 6 '18 at 19:25


















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