Checking the proof of: find all primes $p$ such that $p^2mid 5^{p^2} +1$












2












$begingroup$



Find all primes $p$ such that $$p^2 mid 5^{p^2} +1$$




Okay, so I got this:



$x = 5^{p^2} +1 = (5^p)^p +1$. In order to use Eulers theorem, I checked that $(5^p, p^2) = 1$, which is true when $p ne 5$.
So, because $varphi(p^2) = p$, it follows that $(5^p)^p + 1 equiv 1 + 1 equiv 2 (modp)$. So, when $pne5$, $x$ is not divisible by $p^2$.
Easy to check that it's not divisible for $p=5$.



My textbook says that answer is $p=3$, which is true when I plug it into $x$.
So what's wrong with my try ?
If the whole approach is wrong, please post solution to this problem, as my textbook only gives final result.






elementary-number-theory divisibility







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Find all primes $p$ such that $$p^2 mid 5^{p^2} +1$$




    Okay, so I got this:



    $x = 5^{p^2} +1 = (5^p)^p +1$. In order to use Eulers theorem, I checked that $(5^p, p^2) = 1$, which is true when $p ne 5$.
    So, because $varphi(p^2) = p$, it follows that $(5^p)^p + 1 equiv 1 + 1 equiv 2 (modp)$. So, when $pne5$, $x$ is not divisible by $p^2$.
    Easy to check that it's not divisible for $p=5$.



    My textbook says that answer is $p=3$, which is true when I plug it into $x$.
    So what's wrong with my try ?
    If the whole approach is wrong, please post solution to this problem, as my textbook only gives final result.






    elementary-number-theory divisibility







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      Find all primes $p$ such that $$p^2 mid 5^{p^2} +1$$




      Okay, so I got this:



      $x = 5^{p^2} +1 = (5^p)^p +1$. In order to use Eulers theorem, I checked that $(5^p, p^2) = 1$, which is true when $p ne 5$.
      So, because $varphi(p^2) = p$, it follows that $(5^p)^p + 1 equiv 1 + 1 equiv 2 (modp)$. So, when $pne5$, $x$ is not divisible by $p^2$.
      Easy to check that it's not divisible for $p=5$.



      My textbook says that answer is $p=3$, which is true when I plug it into $x$.
      So what's wrong with my try ?
      If the whole approach is wrong, please post solution to this problem, as my textbook only gives final result.






      elementary-number-theory divisibility







      share|cite|improve this question











      $endgroup$





      Find all primes $p$ such that $$p^2 mid 5^{p^2} +1$$




      Okay, so I got this:



      $x = 5^{p^2} +1 = (5^p)^p +1$. In order to use Eulers theorem, I checked that $(5^p, p^2) = 1$, which is true when $p ne 5$.
      So, because $varphi(p^2) = p$, it follows that $(5^p)^p + 1 equiv 1 + 1 equiv 2 (modp)$. So, when $pne5$, $x$ is not divisible by $p^2$.
      Easy to check that it's not divisible for $p=5$.



      My textbook says that answer is $p=3$, which is true when I plug it into $x$.
      So what's wrong with my try ?
      If the whole approach is wrong, please post solution to this problem, as my textbook only gives final result.






      elementary-number-theory divisibility




      elementary-number-theory divisibility






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 17:55









      greedoid

      40.1k114799




      40.1k114799










      asked Dec 19 '18 at 17:31







      user626177





























          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Clearly, $pne5$. Then Fermat gives
          $$
          -1 equiv 5^{p^2} equiv (5^{p}){^p} equiv 5^{p} equiv 5 bmod p
          $$
          and so $p$ divides $6$. Now $p=2$ does not work but $p=3$ does work.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's right, thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 17:46










          • $begingroup$
            $p=2$ does not work because $a^2not equiv 2 pmod 4$
            $endgroup$
            – Maged Saeed
            Dec 20 '18 at 4:39





















          0












          $begingroup$

          It is not correct since $varphi(p^2) = p(color{red}{p-1})$, but you can repair it easly.



          All congruences are modulo $p^2$. So we have by Euler (since $pne 5$)



          $$ 5^{p^2-p} equiv -1 implies 5^pequiv 5^{p^2}equiv -1$$



          So $pmid 5^p+1$, but by Fermat little theorem we have $pmid 5^p-5$ so $$pmid (5^p+1)-(5^p-5) = 6 implies p=2;;{rm or};;p=3$$



          Only $p=3$ works...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Holy **** that was stupid. I'll try fixing it now
            $endgroup$
            – user626177
            Dec 19 '18 at 17:36



















          0












          $begingroup$

          If $5^{p^2} + 1$ is divisible by $p^2$, it must also be divisible by $p$



          Freshman's dream...



          Over a finite fields:



          $(a + b)^{p^n}equiv a^{p^n} + b^{p^n} pmod{p}$



          $5^{p^2} + 1^{p^2} equiv (5+1)^{p^2}equiv 0 pmod p$



          $p$ divides $6$



          leaving us with only 2 candidates to check.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Never heard of it, will check it out. Seems like an overkill for problems I need to do. Thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 19:13






          • 1




            $begingroup$
            Usually, the Freshman's dream is written as $(a+b)^2 = a^2+ b^2$ which is of course the mistake that a high-school freshman would make in his algebra class. But, oddly enough is true over rings with the right characteristic.
            $endgroup$
            – Doug M
            Dec 19 '18 at 20:55













          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%2f3046648%2fchecking-the-proof-of-find-all-primes-p-such-that-p2-mid-5p2-1%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown
























          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Clearly, $pne5$. Then Fermat gives
          $$
          -1 equiv 5^{p^2} equiv (5^{p}){^p} equiv 5^{p} equiv 5 bmod p
          $$
          and so $p$ divides $6$. Now $p=2$ does not work but $p=3$ does work.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's right, thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 17:46










          • $begingroup$
            $p=2$ does not work because $a^2not equiv 2 pmod 4$
            $endgroup$
            – Maged Saeed
            Dec 20 '18 at 4:39


















          2












          $begingroup$

          Clearly, $pne5$. Then Fermat gives
          $$
          -1 equiv 5^{p^2} equiv (5^{p}){^p} equiv 5^{p} equiv 5 bmod p
          $$
          and so $p$ divides $6$. Now $p=2$ does not work but $p=3$ does work.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's right, thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 17:46










          • $begingroup$
            $p=2$ does not work because $a^2not equiv 2 pmod 4$
            $endgroup$
            – Maged Saeed
            Dec 20 '18 at 4:39
















          2












          2








          2





          $begingroup$

          Clearly, $pne5$. Then Fermat gives
          $$
          -1 equiv 5^{p^2} equiv (5^{p}){^p} equiv 5^{p} equiv 5 bmod p
          $$
          and so $p$ divides $6$. Now $p=2$ does not work but $p=3$ does work.






          share|cite|improve this answer











          $endgroup$



          Clearly, $pne5$. Then Fermat gives
          $$
          -1 equiv 5^{p^2} equiv (5^{p}){^p} equiv 5^{p} equiv 5 bmod p
          $$
          and so $p$ divides $6$. Now $p=2$ does not work but $p=3$ does work.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 17:47

























          answered Dec 19 '18 at 17:44









          lhflhf

          164k10170395




          164k10170395












          • $begingroup$
            That's right, thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 17:46










          • $begingroup$
            $p=2$ does not work because $a^2not equiv 2 pmod 4$
            $endgroup$
            – Maged Saeed
            Dec 20 '18 at 4:39




















          • $begingroup$
            That's right, thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 17:46










          • $begingroup$
            $p=2$ does not work because $a^2not equiv 2 pmod 4$
            $endgroup$
            – Maged Saeed
            Dec 20 '18 at 4:39


















          $begingroup$
          That's right, thanks!
          $endgroup$
          – user626177
          Dec 19 '18 at 17:46




          $begingroup$
          That's right, thanks!
          $endgroup$
          – user626177
          Dec 19 '18 at 17:46












          $begingroup$
          $p=2$ does not work because $a^2not equiv 2 pmod 4$
          $endgroup$
          – Maged Saeed
          Dec 20 '18 at 4:39






          $begingroup$
          $p=2$ does not work because $a^2not equiv 2 pmod 4$
          $endgroup$
          – Maged Saeed
          Dec 20 '18 at 4:39













          0












          $begingroup$

          It is not correct since $varphi(p^2) = p(color{red}{p-1})$, but you can repair it easly.



          All congruences are modulo $p^2$. So we have by Euler (since $pne 5$)



          $$ 5^{p^2-p} equiv -1 implies 5^pequiv 5^{p^2}equiv -1$$



          So $pmid 5^p+1$, but by Fermat little theorem we have $pmid 5^p-5$ so $$pmid (5^p+1)-(5^p-5) = 6 implies p=2;;{rm or};;p=3$$



          Only $p=3$ works...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Holy **** that was stupid. I'll try fixing it now
            $endgroup$
            – user626177
            Dec 19 '18 at 17:36
















          0












          $begingroup$

          It is not correct since $varphi(p^2) = p(color{red}{p-1})$, but you can repair it easly.



          All congruences are modulo $p^2$. So we have by Euler (since $pne 5$)



          $$ 5^{p^2-p} equiv -1 implies 5^pequiv 5^{p^2}equiv -1$$



          So $pmid 5^p+1$, but by Fermat little theorem we have $pmid 5^p-5$ so $$pmid (5^p+1)-(5^p-5) = 6 implies p=2;;{rm or};;p=3$$



          Only $p=3$ works...






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Holy **** that was stupid. I'll try fixing it now
            $endgroup$
            – user626177
            Dec 19 '18 at 17:36














          0












          0








          0





          $begingroup$

          It is not correct since $varphi(p^2) = p(color{red}{p-1})$, but you can repair it easly.



          All congruences are modulo $p^2$. So we have by Euler (since $pne 5$)



          $$ 5^{p^2-p} equiv -1 implies 5^pequiv 5^{p^2}equiv -1$$



          So $pmid 5^p+1$, but by Fermat little theorem we have $pmid 5^p-5$ so $$pmid (5^p+1)-(5^p-5) = 6 implies p=2;;{rm or};;p=3$$



          Only $p=3$ works...






          share|cite|improve this answer











          $endgroup$



          It is not correct since $varphi(p^2) = p(color{red}{p-1})$, but you can repair it easly.



          All congruences are modulo $p^2$. So we have by Euler (since $pne 5$)



          $$ 5^{p^2-p} equiv -1 implies 5^pequiv 5^{p^2}equiv -1$$



          So $pmid 5^p+1$, but by Fermat little theorem we have $pmid 5^p-5$ so $$pmid (5^p+1)-(5^p-5) = 6 implies p=2;;{rm or};;p=3$$



          Only $p=3$ works...







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 17:46

























          answered Dec 19 '18 at 17:34









          greedoidgreedoid

          40.1k114799




          40.1k114799












          • $begingroup$
            Holy **** that was stupid. I'll try fixing it now
            $endgroup$
            – user626177
            Dec 19 '18 at 17:36


















          • $begingroup$
            Holy **** that was stupid. I'll try fixing it now
            $endgroup$
            – user626177
            Dec 19 '18 at 17:36
















          $begingroup$
          Holy **** that was stupid. I'll try fixing it now
          $endgroup$
          – user626177
          Dec 19 '18 at 17:36




          $begingroup$
          Holy **** that was stupid. I'll try fixing it now
          $endgroup$
          – user626177
          Dec 19 '18 at 17:36











          0












          $begingroup$

          If $5^{p^2} + 1$ is divisible by $p^2$, it must also be divisible by $p$



          Freshman's dream...



          Over a finite fields:



          $(a + b)^{p^n}equiv a^{p^n} + b^{p^n} pmod{p}$



          $5^{p^2} + 1^{p^2} equiv (5+1)^{p^2}equiv 0 pmod p$



          $p$ divides $6$



          leaving us with only 2 candidates to check.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Never heard of it, will check it out. Seems like an overkill for problems I need to do. Thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 19:13






          • 1




            $begingroup$
            Usually, the Freshman's dream is written as $(a+b)^2 = a^2+ b^2$ which is of course the mistake that a high-school freshman would make in his algebra class. But, oddly enough is true over rings with the right characteristic.
            $endgroup$
            – Doug M
            Dec 19 '18 at 20:55


















          0












          $begingroup$

          If $5^{p^2} + 1$ is divisible by $p^2$, it must also be divisible by $p$



          Freshman's dream...



          Over a finite fields:



          $(a + b)^{p^n}equiv a^{p^n} + b^{p^n} pmod{p}$



          $5^{p^2} + 1^{p^2} equiv (5+1)^{p^2}equiv 0 pmod p$



          $p$ divides $6$



          leaving us with only 2 candidates to check.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Never heard of it, will check it out. Seems like an overkill for problems I need to do. Thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 19:13






          • 1




            $begingroup$
            Usually, the Freshman's dream is written as $(a+b)^2 = a^2+ b^2$ which is of course the mistake that a high-school freshman would make in his algebra class. But, oddly enough is true over rings with the right characteristic.
            $endgroup$
            – Doug M
            Dec 19 '18 at 20:55
















          0












          0








          0





          $begingroup$

          If $5^{p^2} + 1$ is divisible by $p^2$, it must also be divisible by $p$



          Freshman's dream...



          Over a finite fields:



          $(a + b)^{p^n}equiv a^{p^n} + b^{p^n} pmod{p}$



          $5^{p^2} + 1^{p^2} equiv (5+1)^{p^2}equiv 0 pmod p$



          $p$ divides $6$



          leaving us with only 2 candidates to check.






          share|cite|improve this answer









          $endgroup$



          If $5^{p^2} + 1$ is divisible by $p^2$, it must also be divisible by $p$



          Freshman's dream...



          Over a finite fields:



          $(a + b)^{p^n}equiv a^{p^n} + b^{p^n} pmod{p}$



          $5^{p^2} + 1^{p^2} equiv (5+1)^{p^2}equiv 0 pmod p$



          $p$ divides $6$



          leaving us with only 2 candidates to check.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 17:57









          Doug MDoug M

          44.5k31854




          44.5k31854












          • $begingroup$
            Never heard of it, will check it out. Seems like an overkill for problems I need to do. Thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 19:13






          • 1




            $begingroup$
            Usually, the Freshman's dream is written as $(a+b)^2 = a^2+ b^2$ which is of course the mistake that a high-school freshman would make in his algebra class. But, oddly enough is true over rings with the right characteristic.
            $endgroup$
            – Doug M
            Dec 19 '18 at 20:55




















          • $begingroup$
            Never heard of it, will check it out. Seems like an overkill for problems I need to do. Thanks!
            $endgroup$
            – user626177
            Dec 19 '18 at 19:13






          • 1




            $begingroup$
            Usually, the Freshman's dream is written as $(a+b)^2 = a^2+ b^2$ which is of course the mistake that a high-school freshman would make in his algebra class. But, oddly enough is true over rings with the right characteristic.
            $endgroup$
            – Doug M
            Dec 19 '18 at 20:55


















          $begingroup$
          Never heard of it, will check it out. Seems like an overkill for problems I need to do. Thanks!
          $endgroup$
          – user626177
          Dec 19 '18 at 19:13




          $begingroup$
          Never heard of it, will check it out. Seems like an overkill for problems I need to do. Thanks!
          $endgroup$
          – user626177
          Dec 19 '18 at 19:13




          1




          1




          $begingroup$
          Usually, the Freshman's dream is written as $(a+b)^2 = a^2+ b^2$ which is of course the mistake that a high-school freshman would make in his algebra class. But, oddly enough is true over rings with the right characteristic.
          $endgroup$
          – Doug M
          Dec 19 '18 at 20:55






          $begingroup$
          Usually, the Freshman's dream is written as $(a+b)^2 = a^2+ b^2$ which is of course the mistake that a high-school freshman would make in his algebra class. But, oddly enough is true over rings with the right characteristic.
          $endgroup$
          – Doug M
          Dec 19 '18 at 20:55




















          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%2f3046648%2fchecking-the-proof-of-find-all-primes-p-such-that-p2-mid-5p2-1%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
          Read more