Can $3p^4-3p^2+1$ be square number?












1












$begingroup$


I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?






elementary-number-theory diophantine-equations







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  • 3




    $begingroup$
    $p = 1$ is a square number.
    $endgroup$
    – Toby Mak
    Dec 20 '18 at 11:14








  • 5




    $begingroup$
    PEOPLE, please stop using $p$ for natural number! Usualy it is $n$ for natural and $p$ for prime. I did whole analisys for nothing.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 11:33






  • 1




    $begingroup$
    You must be fun at parties.
    $endgroup$
    – Lucas Henrique
    Dec 20 '18 at 11:47










  • $begingroup$
    @LucasHenrique Who are you reffering to and what does that have with a math?
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 12:37
















1












$begingroup$


I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?






elementary-number-theory diophantine-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $p = 1$ is a square number.
    $endgroup$
    – Toby Mak
    Dec 20 '18 at 11:14








  • 5




    $begingroup$
    PEOPLE, please stop using $p$ for natural number! Usualy it is $n$ for natural and $p$ for prime. I did whole analisys for nothing.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 11:33






  • 1




    $begingroup$
    You must be fun at parties.
    $endgroup$
    – Lucas Henrique
    Dec 20 '18 at 11:47










  • $begingroup$
    @LucasHenrique Who are you reffering to and what does that have with a math?
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 12:37














1












1








1





$begingroup$


I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?






elementary-number-theory diophantine-equations







share|cite|improve this question











$endgroup$




I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?






elementary-number-theory diophantine-equations




elementary-number-theory diophantine-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 11:33









Maria Mazur

48.2k1260122




48.2k1260122










asked Dec 20 '18 at 11:11









eandpiandieandpiandi

272




272








  • 3




    $begingroup$
    $p = 1$ is a square number.
    $endgroup$
    – Toby Mak
    Dec 20 '18 at 11:14








  • 5




    $begingroup$
    PEOPLE, please stop using $p$ for natural number! Usualy it is $n$ for natural and $p$ for prime. I did whole analisys for nothing.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 11:33






  • 1




    $begingroup$
    You must be fun at parties.
    $endgroup$
    – Lucas Henrique
    Dec 20 '18 at 11:47










  • $begingroup$
    @LucasHenrique Who are you reffering to and what does that have with a math?
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 12:37














  • 3




    $begingroup$
    $p = 1$ is a square number.
    $endgroup$
    – Toby Mak
    Dec 20 '18 at 11:14








  • 5




    $begingroup$
    PEOPLE, please stop using $p$ for natural number! Usualy it is $n$ for natural and $p$ for prime. I did whole analisys for nothing.
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 11:33






  • 1




    $begingroup$
    You must be fun at parties.
    $endgroup$
    – Lucas Henrique
    Dec 20 '18 at 11:47










  • $begingroup$
    @LucasHenrique Who are you reffering to and what does that have with a math?
    $endgroup$
    – Maria Mazur
    Dec 20 '18 at 12:37








3




3




$begingroup$
$p = 1$ is a square number.
$endgroup$
– Toby Mak
Dec 20 '18 at 11:14






$begingroup$
$p = 1$ is a square number.
$endgroup$
– Toby Mak
Dec 20 '18 at 11:14






5




5




$begingroup$
PEOPLE, please stop using $p$ for natural number! Usualy it is $n$ for natural and $p$ for prime. I did whole analisys for nothing.
$endgroup$
– Maria Mazur
Dec 20 '18 at 11:33




$begingroup$
PEOPLE, please stop using $p$ for natural number! Usualy it is $n$ for natural and $p$ for prime. I did whole analisys for nothing.
$endgroup$
– Maria Mazur
Dec 20 '18 at 11:33




1




1




$begingroup$
You must be fun at parties.
$endgroup$
– Lucas Henrique
Dec 20 '18 at 11:47




$begingroup$
You must be fun at parties.
$endgroup$
– Lucas Henrique
Dec 20 '18 at 11:47












$begingroup$
@LucasHenrique Who are you reffering to and what does that have with a math?
$endgroup$
– Maria Mazur
Dec 20 '18 at 12:37




$begingroup$
@LucasHenrique Who are you reffering to and what does that have with a math?
$endgroup$
– Maria Mazur
Dec 20 '18 at 12:37










3 Answers
3






active

oldest

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2












$begingroup$

Partial solution if $p$ is prime.



Write $$3p^4-3p^2+1=n^2implies 3p^2(p^2-1) = (n-1)(n+1)$$



If $pne 2$ (which is not a solution) then $p^2mid n-1$ or $p^2mid n+1 $



First case: If $p^2mid n-1$ then $n+1mid 3p^2-3$ so $ n-1= p^2k$ and $n+1leq 3p^2-3$.



If $kgeq 3$ then $$3p^2-3geq n+1 >n-1 geq 3p^2$$ which is impossible. So $kleq 2$



$bullet$ If $k=2$ then $n= 2p^2+1$ so $$2p^2+2mid 3p^2-3 implies 2p^2+2mid 2(3p^2-3)-3(2p^2+2) = -12$$



So $p^2+1mid 6 implies p^2+1in {1,2,3,6}$ which is impssible.



$bullet$ If $k=1$ then $n= p^2+1$ so $$p^2+2mid 3p^2-3 implies p^2+2mid (3p^2-3)-3(p^2+2) =-9 $$



So $p^2+2mid 9 implies p^2+2in {1,3,9}$ which is impossible again.



Second case: If $p^2mid n+1$ then $n-1mid 3p^2-3$ so $ n+1= p^2k$ and $n-1leq 3p^2-3$.



Again, if $kgeq 3$ then $$3p^2-3geq n-1 = n+1-2geq 3p^2-2$$ which is impossible. So $kleq 2$



$bullet$ If $k=2$ then $n= 2p^2-1$ so $2p^2-2mid 3p^2-3$ which is impossible.



$bullet$ If $k=1$ then $n= p^2-1$ so $$p^2-2mid 3p^2-3 implies p^2-2mid (3p^2-3)-3(p^2-2) =3 $$



So $p^2-2mid 3 implies p^2-2in {-1,1,3}$ which is impossible again.



So the answer is negative if $p$ is prime.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    what if p is a natural number? Isn't there any way to prove it?
    $endgroup$
    – eandpiandi
    Dec 20 '18 at 12:04



















0












$begingroup$

Hint about method:



$3m^2 - 3m + 1 = n^2$ for any n $in$ Natural, for $p^2 = m$.



$$3m^2 - 3m + 1 - n^2 = 0$$
$$implies m = frac {3 +sqrt{12n^2-3}}6$$
$$implies p^2 = frac {3 +sqrt{12n^2-3}}6$$ or
$$implies p^2 = frac {3 -sqrt{12n^2-3}}6$$



Now you can just make logical comments yourself about this is valid for above given constraints or not, and make out a bit about the equation.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You can compare it with general formula for Pythagorean triples: $(p^2-k^2)^2+(2pk)^2=(p^2+k^2)^2$, we have: $(p^2-1)^2+p^2(2p^2-1)$, here $k=1$ and second term must be $(2p)^2$, which is not, so second term is not competent with that of general formula and your tri-nomial can not be square.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Partial solution if $p$ is prime.



      Write $$3p^4-3p^2+1=n^2implies 3p^2(p^2-1) = (n-1)(n+1)$$



      If $pne 2$ (which is not a solution) then $p^2mid n-1$ or $p^2mid n+1 $



      First case: If $p^2mid n-1$ then $n+1mid 3p^2-3$ so $ n-1= p^2k$ and $n+1leq 3p^2-3$.



      If $kgeq 3$ then $$3p^2-3geq n+1 >n-1 geq 3p^2$$ which is impossible. So $kleq 2$



      $bullet$ If $k=2$ then $n= 2p^2+1$ so $$2p^2+2mid 3p^2-3 implies 2p^2+2mid 2(3p^2-3)-3(2p^2+2) = -12$$



      So $p^2+1mid 6 implies p^2+1in {1,2,3,6}$ which is impssible.



      $bullet$ If $k=1$ then $n= p^2+1$ so $$p^2+2mid 3p^2-3 implies p^2+2mid (3p^2-3)-3(p^2+2) =-9 $$



      So $p^2+2mid 9 implies p^2+2in {1,3,9}$ which is impossible again.



      Second case: If $p^2mid n+1$ then $n-1mid 3p^2-3$ so $ n+1= p^2k$ and $n-1leq 3p^2-3$.



      Again, if $kgeq 3$ then $$3p^2-3geq n-1 = n+1-2geq 3p^2-2$$ which is impossible. So $kleq 2$



      $bullet$ If $k=2$ then $n= 2p^2-1$ so $2p^2-2mid 3p^2-3$ which is impossible.



      $bullet$ If $k=1$ then $n= p^2-1$ so $$p^2-2mid 3p^2-3 implies p^2-2mid (3p^2-3)-3(p^2-2) =3 $$



      So $p^2-2mid 3 implies p^2-2in {-1,1,3}$ which is impossible again.



      So the answer is negative if $p$ is prime.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        what if p is a natural number? Isn't there any way to prove it?
        $endgroup$
        – eandpiandi
        Dec 20 '18 at 12:04
















      2












      $begingroup$

      Partial solution if $p$ is prime.



      Write $$3p^4-3p^2+1=n^2implies 3p^2(p^2-1) = (n-1)(n+1)$$



      If $pne 2$ (which is not a solution) then $p^2mid n-1$ or $p^2mid n+1 $



      First case: If $p^2mid n-1$ then $n+1mid 3p^2-3$ so $ n-1= p^2k$ and $n+1leq 3p^2-3$.



      If $kgeq 3$ then $$3p^2-3geq n+1 >n-1 geq 3p^2$$ which is impossible. So $kleq 2$



      $bullet$ If $k=2$ then $n= 2p^2+1$ so $$2p^2+2mid 3p^2-3 implies 2p^2+2mid 2(3p^2-3)-3(2p^2+2) = -12$$



      So $p^2+1mid 6 implies p^2+1in {1,2,3,6}$ which is impssible.



      $bullet$ If $k=1$ then $n= p^2+1$ so $$p^2+2mid 3p^2-3 implies p^2+2mid (3p^2-3)-3(p^2+2) =-9 $$



      So $p^2+2mid 9 implies p^2+2in {1,3,9}$ which is impossible again.



      Second case: If $p^2mid n+1$ then $n-1mid 3p^2-3$ so $ n+1= p^2k$ and $n-1leq 3p^2-3$.



      Again, if $kgeq 3$ then $$3p^2-3geq n-1 = n+1-2geq 3p^2-2$$ which is impossible. So $kleq 2$



      $bullet$ If $k=2$ then $n= 2p^2-1$ so $2p^2-2mid 3p^2-3$ which is impossible.



      $bullet$ If $k=1$ then $n= p^2-1$ so $$p^2-2mid 3p^2-3 implies p^2-2mid (3p^2-3)-3(p^2-2) =3 $$



      So $p^2-2mid 3 implies p^2-2in {-1,1,3}$ which is impossible again.



      So the answer is negative if $p$ is prime.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        what if p is a natural number? Isn't there any way to prove it?
        $endgroup$
        – eandpiandi
        Dec 20 '18 at 12:04














      2












      2








      2





      $begingroup$

      Partial solution if $p$ is prime.



      Write $$3p^4-3p^2+1=n^2implies 3p^2(p^2-1) = (n-1)(n+1)$$



      If $pne 2$ (which is not a solution) then $p^2mid n-1$ or $p^2mid n+1 $



      First case: If $p^2mid n-1$ then $n+1mid 3p^2-3$ so $ n-1= p^2k$ and $n+1leq 3p^2-3$.



      If $kgeq 3$ then $$3p^2-3geq n+1 >n-1 geq 3p^2$$ which is impossible. So $kleq 2$



      $bullet$ If $k=2$ then $n= 2p^2+1$ so $$2p^2+2mid 3p^2-3 implies 2p^2+2mid 2(3p^2-3)-3(2p^2+2) = -12$$



      So $p^2+1mid 6 implies p^2+1in {1,2,3,6}$ which is impssible.



      $bullet$ If $k=1$ then $n= p^2+1$ so $$p^2+2mid 3p^2-3 implies p^2+2mid (3p^2-3)-3(p^2+2) =-9 $$



      So $p^2+2mid 9 implies p^2+2in {1,3,9}$ which is impossible again.



      Second case: If $p^2mid n+1$ then $n-1mid 3p^2-3$ so $ n+1= p^2k$ and $n-1leq 3p^2-3$.



      Again, if $kgeq 3$ then $$3p^2-3geq n-1 = n+1-2geq 3p^2-2$$ which is impossible. So $kleq 2$



      $bullet$ If $k=2$ then $n= 2p^2-1$ so $2p^2-2mid 3p^2-3$ which is impossible.



      $bullet$ If $k=1$ then $n= p^2-1$ so $$p^2-2mid 3p^2-3 implies p^2-2mid (3p^2-3)-3(p^2-2) =3 $$



      So $p^2-2mid 3 implies p^2-2in {-1,1,3}$ which is impossible again.



      So the answer is negative if $p$ is prime.






      share|cite|improve this answer











      $endgroup$



      Partial solution if $p$ is prime.



      Write $$3p^4-3p^2+1=n^2implies 3p^2(p^2-1) = (n-1)(n+1)$$



      If $pne 2$ (which is not a solution) then $p^2mid n-1$ or $p^2mid n+1 $



      First case: If $p^2mid n-1$ then $n+1mid 3p^2-3$ so $ n-1= p^2k$ and $n+1leq 3p^2-3$.



      If $kgeq 3$ then $$3p^2-3geq n+1 >n-1 geq 3p^2$$ which is impossible. So $kleq 2$



      $bullet$ If $k=2$ then $n= 2p^2+1$ so $$2p^2+2mid 3p^2-3 implies 2p^2+2mid 2(3p^2-3)-3(2p^2+2) = -12$$



      So $p^2+1mid 6 implies p^2+1in {1,2,3,6}$ which is impssible.



      $bullet$ If $k=1$ then $n= p^2+1$ so $$p^2+2mid 3p^2-3 implies p^2+2mid (3p^2-3)-3(p^2+2) =-9 $$



      So $p^2+2mid 9 implies p^2+2in {1,3,9}$ which is impossible again.



      Second case: If $p^2mid n+1$ then $n-1mid 3p^2-3$ so $ n+1= p^2k$ and $n-1leq 3p^2-3$.



      Again, if $kgeq 3$ then $$3p^2-3geq n-1 = n+1-2geq 3p^2-2$$ which is impossible. So $kleq 2$



      $bullet$ If $k=2$ then $n= 2p^2-1$ so $2p^2-2mid 3p^2-3$ which is impossible.



      $bullet$ If $k=1$ then $n= p^2-1$ so $$p^2-2mid 3p^2-3 implies p^2-2mid (3p^2-3)-3(p^2-2) =3 $$



      So $p^2-2mid 3 implies p^2-2in {-1,1,3}$ which is impossible again.



      So the answer is negative if $p$ is prime.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 20 '18 at 11:37

























      answered Dec 20 '18 at 11:30









      Maria MazurMaria Mazur

      48.2k1260122




      48.2k1260122








      • 1




        $begingroup$
        what if p is a natural number? Isn't there any way to prove it?
        $endgroup$
        – eandpiandi
        Dec 20 '18 at 12:04














      • 1




        $begingroup$
        what if p is a natural number? Isn't there any way to prove it?
        $endgroup$
        – eandpiandi
        Dec 20 '18 at 12:04








      1




      1




      $begingroup$
      what if p is a natural number? Isn't there any way to prove it?
      $endgroup$
      – eandpiandi
      Dec 20 '18 at 12:04




      $begingroup$
      what if p is a natural number? Isn't there any way to prove it?
      $endgroup$
      – eandpiandi
      Dec 20 '18 at 12:04











      0












      $begingroup$

      Hint about method:



      $3m^2 - 3m + 1 = n^2$ for any n $in$ Natural, for $p^2 = m$.



      $$3m^2 - 3m + 1 - n^2 = 0$$
      $$implies m = frac {3 +sqrt{12n^2-3}}6$$
      $$implies p^2 = frac {3 +sqrt{12n^2-3}}6$$ or
      $$implies p^2 = frac {3 -sqrt{12n^2-3}}6$$



      Now you can just make logical comments yourself about this is valid for above given constraints or not, and make out a bit about the equation.






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        Hint about method:



        $3m^2 - 3m + 1 = n^2$ for any n $in$ Natural, for $p^2 = m$.



        $$3m^2 - 3m + 1 - n^2 = 0$$
        $$implies m = frac {3 +sqrt{12n^2-3}}6$$
        $$implies p^2 = frac {3 +sqrt{12n^2-3}}6$$ or
        $$implies p^2 = frac {3 -sqrt{12n^2-3}}6$$



        Now you can just make logical comments yourself about this is valid for above given constraints or not, and make out a bit about the equation.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Hint about method:



          $3m^2 - 3m + 1 = n^2$ for any n $in$ Natural, for $p^2 = m$.



          $$3m^2 - 3m + 1 - n^2 = 0$$
          $$implies m = frac {3 +sqrt{12n^2-3}}6$$
          $$implies p^2 = frac {3 +sqrt{12n^2-3}}6$$ or
          $$implies p^2 = frac {3 -sqrt{12n^2-3}}6$$



          Now you can just make logical comments yourself about this is valid for above given constraints or not, and make out a bit about the equation.






          share|cite|improve this answer











          $endgroup$



          Hint about method:



          $3m^2 - 3m + 1 = n^2$ for any n $in$ Natural, for $p^2 = m$.



          $$3m^2 - 3m + 1 - n^2 = 0$$
          $$implies m = frac {3 +sqrt{12n^2-3}}6$$
          $$implies p^2 = frac {3 +sqrt{12n^2-3}}6$$ or
          $$implies p^2 = frac {3 -sqrt{12n^2-3}}6$$



          Now you can just make logical comments yourself about this is valid for above given constraints or not, and make out a bit about the equation.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 12:17

























          answered Dec 20 '18 at 11:19









          PradyumanDixitPradyumanDixit

          847214




          847214























              0












              $begingroup$

              You can compare it with general formula for Pythagorean triples: $(p^2-k^2)^2+(2pk)^2=(p^2+k^2)^2$, we have: $(p^2-1)^2+p^2(2p^2-1)$, here $k=1$ and second term must be $(2p)^2$, which is not, so second term is not competent with that of general formula and your tri-nomial can not be square.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You can compare it with general formula for Pythagorean triples: $(p^2-k^2)^2+(2pk)^2=(p^2+k^2)^2$, we have: $(p^2-1)^2+p^2(2p^2-1)$, here $k=1$ and second term must be $(2p)^2$, which is not, so second term is not competent with that of general formula and your tri-nomial can not be square.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You can compare it with general formula for Pythagorean triples: $(p^2-k^2)^2+(2pk)^2=(p^2+k^2)^2$, we have: $(p^2-1)^2+p^2(2p^2-1)$, here $k=1$ and second term must be $(2p)^2$, which is not, so second term is not competent with that of general formula and your tri-nomial can not be square.






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                  $endgroup$



                  You can compare it with general formula for Pythagorean triples: $(p^2-k^2)^2+(2pk)^2=(p^2+k^2)^2$, we have: $(p^2-1)^2+p^2(2p^2-1)$, here $k=1$ and second term must be $(2p)^2$, which is not, so second term is not competent with that of general formula and your tri-nomial can not be square.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 15:32









                  siroussirous

                  1,6851514




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