Solve equation in prime numbers












2















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




















  • 2




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
















2















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




















  • 2




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














2












2








2


1






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
















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









greedoid

38.6k114797




38.6k114797










asked Nov 9 '18 at 20:29









LeoxLeox

5,2431423




5,2431423








  • 2




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














  • 2




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








2




2




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




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










2 Answers
2






active

oldest

votes


















6














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































    0














    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

















    • 1




      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, $
      – Will Jagy
      Nov 9 '18 at 23:14











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2991909%2fsolve-equation-in-prime-numbers%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    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




























      6














      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


























        6












        6








        6






        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














        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









        greedoidgreedoid

        38.6k114797




        38.6k114797























            0














            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

















            • 1




              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, $
              – Will Jagy
              Nov 9 '18 at 23:14
















            0














            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

















            • 1




              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, $
              – Will Jagy
              Nov 9 '18 at 23:14














            0












            0








            0






            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












            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

            5,2431423




            5,2431423








            • 1




              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, $
              – Will Jagy
              Nov 9 '18 at 23:14














            • 1




              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, $
              – Will Jagy
              Nov 9 '18 at 23:14








            1




            1




            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, $
            – Will Jagy
            Nov 9 '18 at 23:14




            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, $
            – Will Jagy
            Nov 9 '18 at 23:14


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2991909%2fsolve-equation-in-prime-numbers%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            An IMO inspired problem

            Image
            14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
            Read more

            Management

            Image
            Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
            Read more

            Investment

            Image
            Investment From Wikipedia, the free encyclopedia Jump to navigation Jump to search This article is about investment in finance. For investment in macroeconomics, see Investment (macroeconomics). For other uses, see Investment (disambiguation). "Invest" redirects here. For the term in meteorology, see Invest (meteorology). This article needs additional citations for verification . Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. ( February 2017 ) (Learn how and when to remove this template message) In general, to invest is to distribute money in the expectation of some benefit in the future – for example, investment in durable goods, in real estate by the service industry, in factories for manufacturing, in product development, and in research and development. However, this article focuses specifically on investment in financial
            Read more