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

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          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













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          3 Answers
          3






          active

          oldest

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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          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




















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          .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 ...
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