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












2












$begingroup$


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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$endgroup$








  • 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.
    $endgroup$
    – greedoid
    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?
    $endgroup$
    – greedoid
    Dec 20 '18 at 12:37
















2












$begingroup$


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.
    $endgroup$
    – greedoid
    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?
    $endgroup$
    – greedoid
    Dec 20 '18 at 12:37














2












2








2





$begingroup$


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









greedoid

42.9k1153105




42.9k1153105










asked Dec 20 '18 at 11:11









eandpiandieandpiandi

322




322








  • 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.
    $endgroup$
    – greedoid
    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?
    $endgroup$
    – greedoid
    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.
    $endgroup$
    – greedoid
    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?
    $endgroup$
    – greedoid
    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.
$endgroup$
– greedoid
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.
$endgroup$
– greedoid
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?
$endgroup$
– greedoid
Dec 20 '18 at 12:37




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










3 Answers
3






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



















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











$endgroup$













  • $begingroup$
    Thanks, I couldn't spot that mistake.
    $endgroup$
    – PradyumanDixit
    Dec 20 '18 at 12:17



















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









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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









    greedoidgreedoid

    42.9k1153105




    42.9k1153105








    • 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











    $endgroup$













    • $begingroup$
      Thanks, I couldn't spot that mistake.
      $endgroup$
      – PradyumanDixit
      Dec 20 '18 at 12:17
















    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











    $endgroup$













    • $begingroup$
      Thanks, I couldn't spot that mistake.
      $endgroup$
      – PradyumanDixit
      Dec 20 '18 at 12:17














    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












    • $begingroup$
      Thanks, I couldn't spot that mistake.
      $endgroup$
      – PradyumanDixit
      Dec 20 '18 at 12:17


















    • $begingroup$
      Thanks, I couldn't spot that mistake.
      $endgroup$
      – PradyumanDixit
      Dec 20 '18 at 12:17
















    $begingroup$
    Thanks, I couldn't spot that mistake.
    $endgroup$
    – PradyumanDixit
    Dec 20 '18 at 12:17




    $begingroup$
    Thanks, I couldn't spot that mistake.
    $endgroup$
    – PradyumanDixit
    Dec 20 '18 at 12:17











    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.






        share|cite|improve this answer









        $endgroup$



        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

        1,6891514




        1,6891514






























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