Help prove $p^4 + 4q^4$ is never a perfect square?

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Given that $p$ and $q$ have no common factors, how can I prove that $p^4 + 4q^4 not =z^2$, if $p,q$ and $z$ have to be positive integers?



From the comments:




I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case $n=4$, where they proved that $p^4 + q^4$ is never a perfect square.





number-theory elementary-number-theory diophantine-equations pythagorean-triples




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  • 4




    Please include some of your own thoughts on the problem. Have you seen any similar problems to this one in the past? What sort of tools do you know of that you have available to use here and what was the result of your attempt at using them? If you show that you have at least started the problem and put in effort beyond copy-pasting then you are more likely to receive help.
    – JMoravitz
    Aug 5 at 23:17






  • 2




    Well, I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case n=4, where they proved that p^4 + q^4 is never a perfect square.
    – Dean Yang
    Aug 5 at 23:24






  • 1




    The Wikipedia article on Fermat's right triangle theorem says that a square number cannot be congruent, so $1$ is not a congruent number.
    – Ross Millikan
    Aug 5 at 23:58






  • 1




    Have you tried a parity argument? What if exactly one of p,q is odd? What if both are odd?
    – Somos
    Aug 6 at 0:08






  • 2




    This gives two primitive Pythagorean triples with a common leg, which cannot be.
    – user580373
    Aug 6 at 0:58














up vote
1
down vote

favorite
1












Given that $p$ and $q$ have no common factors, how can I prove that $p^4 + 4q^4 not =z^2$, if $p,q$ and $z$ have to be positive integers?



From the comments:




I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case $n=4$, where they proved that $p^4 + q^4$ is never a perfect square.





number-theory elementary-number-theory diophantine-equations pythagorean-triples




share|cite|improve this question

















  • 4




    Please include some of your own thoughts on the problem. Have you seen any similar problems to this one in the past? What sort of tools do you know of that you have available to use here and what was the result of your attempt at using them? If you show that you have at least started the problem and put in effort beyond copy-pasting then you are more likely to receive help.
    – JMoravitz
    Aug 5 at 23:17






  • 2




    Well, I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case n=4, where they proved that p^4 + q^4 is never a perfect square.
    – Dean Yang
    Aug 5 at 23:24






  • 1




    The Wikipedia article on Fermat's right triangle theorem says that a square number cannot be congruent, so $1$ is not a congruent number.
    – Ross Millikan
    Aug 5 at 23:58






  • 1




    Have you tried a parity argument? What if exactly one of p,q is odd? What if both are odd?
    – Somos
    Aug 6 at 0:08






  • 2




    This gives two primitive Pythagorean triples with a common leg, which cannot be.
    – user580373
    Aug 6 at 0:58












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Given that $p$ and $q$ have no common factors, how can I prove that $p^4 + 4q^4 not =z^2$, if $p,q$ and $z$ have to be positive integers?



From the comments:




I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case $n=4$, where they proved that $p^4 + q^4$ is never a perfect square.





number-theory elementary-number-theory diophantine-equations pythagorean-triples




share|cite|improve this question













Given that $p$ and $q$ have no common factors, how can I prove that $p^4 + 4q^4 not =z^2$, if $p,q$ and $z$ have to be positive integers?



From the comments:




I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case $n=4$, where they proved that $p^4 + q^4$ is never a perfect square.






number-theory elementary-number-theory diophantine-equations pythagorean-triples





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 6:38









Michael Rozenberg

88.2k1579180




88.2k1579180









asked Aug 5 at 23:12









Dean Yang

1388




1388







  • 4




    Please include some of your own thoughts on the problem. Have you seen any similar problems to this one in the past? What sort of tools do you know of that you have available to use here and what was the result of your attempt at using them? If you show that you have at least started the problem and put in effort beyond copy-pasting then you are more likely to receive help.
    – JMoravitz
    Aug 5 at 23:17






  • 2




    Well, I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case n=4, where they proved that p^4 + q^4 is never a perfect square.
    – Dean Yang
    Aug 5 at 23:24






  • 1




    The Wikipedia article on Fermat's right triangle theorem says that a square number cannot be congruent, so $1$ is not a congruent number.
    – Ross Millikan
    Aug 5 at 23:58






  • 1




    Have you tried a parity argument? What if exactly one of p,q is odd? What if both are odd?
    – Somos
    Aug 6 at 0:08






  • 2




    This gives two primitive Pythagorean triples with a common leg, which cannot be.
    – user580373
    Aug 6 at 0:58












  • 4




    Please include some of your own thoughts on the problem. Have you seen any similar problems to this one in the past? What sort of tools do you know of that you have available to use here and what was the result of your attempt at using them? If you show that you have at least started the problem and put in effort beyond copy-pasting then you are more likely to receive help.
    – JMoravitz
    Aug 5 at 23:17






  • 2




    Well, I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case n=4, where they proved that p^4 + q^4 is never a perfect square.
    – Dean Yang
    Aug 5 at 23:24






  • 1




    The Wikipedia article on Fermat's right triangle theorem says that a square number cannot be congruent, so $1$ is not a congruent number.
    – Ross Millikan
    Aug 5 at 23:58






  • 1




    Have you tried a parity argument? What if exactly one of p,q is odd? What if both are odd?
    – Somos
    Aug 6 at 0:08






  • 2




    This gives two primitive Pythagorean triples with a common leg, which cannot be.
    – user580373
    Aug 6 at 0:58







4




4




Please include some of your own thoughts on the problem. Have you seen any similar problems to this one in the past? What sort of tools do you know of that you have available to use here and what was the result of your attempt at using them? If you show that you have at least started the problem and put in effort beyond copy-pasting then you are more likely to receive help.
– JMoravitz
Aug 5 at 23:17




Please include some of your own thoughts on the problem. Have you seen any similar problems to this one in the past? What sort of tools do you know of that you have available to use here and what was the result of your attempt at using them? If you show that you have at least started the problem and put in effort beyond copy-pasting then you are more likely to receive help.
– JMoravitz
Aug 5 at 23:17




2




2




Well, I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case n=4, where they proved that p^4 + q^4 is never a perfect square.
– Dean Yang
Aug 5 at 23:24




Well, I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case n=4, where they proved that p^4 + q^4 is never a perfect square.
– Dean Yang
Aug 5 at 23:24




1




1




The Wikipedia article on Fermat's right triangle theorem says that a square number cannot be congruent, so $1$ is not a congruent number.
– Ross Millikan
Aug 5 at 23:58




The Wikipedia article on Fermat's right triangle theorem says that a square number cannot be congruent, so $1$ is not a congruent number.
– Ross Millikan
Aug 5 at 23:58




1




1




Have you tried a parity argument? What if exactly one of p,q is odd? What if both are odd?
– Somos
Aug 6 at 0:08




Have you tried a parity argument? What if exactly one of p,q is odd? What if both are odd?
– Somos
Aug 6 at 0:08




2




2




This gives two primitive Pythagorean triples with a common leg, which cannot be.
– user580373
Aug 6 at 0:58




This gives two primitive Pythagorean triples with a common leg, which cannot be.
– user580373
Aug 6 at 0:58










1 Answer
1






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It's an explanation of the Ross Millikan's comment.



If $gcd(p,q)=1$ and $p^4+4q^4=z^2$ then there are $m$ and $n$ with different parity,



for which $m>n$, $gcd(m,n)=1$,
$$p^2=m^2-n^2$$ and
$$2q^2=2mn.$$
Thus, $m=x^2$, $n=y^2$ and we get that the equation
$$p^2=x^4-y^4$$ has solutions in natural numbers.



But $$(x^4-y^4)^2+(2x^2y^2)^2=(x^4+y^4)^2,$$ which says that
$$(x^4-y^4,2x^2y^2,x^4+y^4)$$ is a Pythagorean triple and the area of this right-angled triangle it's
$$frac(x^4-y^4)cdot2x^2y^22=(pxy)^2,$$
which is impossible.






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    1 Answer
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    active

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    up vote
    6
    down vote













    It's an explanation of the Ross Millikan's comment.



    If $gcd(p,q)=1$ and $p^4+4q^4=z^2$ then there are $m$ and $n$ with different parity,



    for which $m>n$, $gcd(m,n)=1$,
    $$p^2=m^2-n^2$$ and
    $$2q^2=2mn.$$
    Thus, $m=x^2$, $n=y^2$ and we get that the equation
    $$p^2=x^4-y^4$$ has solutions in natural numbers.



    But $$(x^4-y^4)^2+(2x^2y^2)^2=(x^4+y^4)^2,$$ which says that
    $$(x^4-y^4,2x^2y^2,x^4+y^4)$$ is a Pythagorean triple and the area of this right-angled triangle it's
    $$frac(x^4-y^4)cdot2x^2y^22=(pxy)^2,$$
    which is impossible.






    share|cite|improve this answer

























      up vote
      6
      down vote













      It's an explanation of the Ross Millikan's comment.



      If $gcd(p,q)=1$ and $p^4+4q^4=z^2$ then there are $m$ and $n$ with different parity,



      for which $m>n$, $gcd(m,n)=1$,
      $$p^2=m^2-n^2$$ and
      $$2q^2=2mn.$$
      Thus, $m=x^2$, $n=y^2$ and we get that the equation
      $$p^2=x^4-y^4$$ has solutions in natural numbers.



      But $$(x^4-y^4)^2+(2x^2y^2)^2=(x^4+y^4)^2,$$ which says that
      $$(x^4-y^4,2x^2y^2,x^4+y^4)$$ is a Pythagorean triple and the area of this right-angled triangle it's
      $$frac(x^4-y^4)cdot2x^2y^22=(pxy)^2,$$
      which is impossible.






      share|cite|improve this answer























        up vote
        6
        down vote










        up vote
        6
        down vote









        It's an explanation of the Ross Millikan's comment.



        If $gcd(p,q)=1$ and $p^4+4q^4=z^2$ then there are $m$ and $n$ with different parity,



        for which $m>n$, $gcd(m,n)=1$,
        $$p^2=m^2-n^2$$ and
        $$2q^2=2mn.$$
        Thus, $m=x^2$, $n=y^2$ and we get that the equation
        $$p^2=x^4-y^4$$ has solutions in natural numbers.



        But $$(x^4-y^4)^2+(2x^2y^2)^2=(x^4+y^4)^2,$$ which says that
        $$(x^4-y^4,2x^2y^2,x^4+y^4)$$ is a Pythagorean triple and the area of this right-angled triangle it's
        $$frac(x^4-y^4)cdot2x^2y^22=(pxy)^2,$$
        which is impossible.






        share|cite|improve this answer













        It's an explanation of the Ross Millikan's comment.



        If $gcd(p,q)=1$ and $p^4+4q^4=z^2$ then there are $m$ and $n$ with different parity,



        for which $m>n$, $gcd(m,n)=1$,
        $$p^2=m^2-n^2$$ and
        $$2q^2=2mn.$$
        Thus, $m=x^2$, $n=y^2$ and we get that the equation
        $$p^2=x^4-y^4$$ has solutions in natural numbers.



        But $$(x^4-y^4)^2+(2x^2y^2)^2=(x^4+y^4)^2,$$ which says that
        $$(x^4-y^4,2x^2y^2,x^4+y^4)$$ is a Pythagorean triple and the area of this right-angled triangle it's
        $$frac(x^4-y^4)cdot2x^2y^22=(pxy)^2,$$
        which is impossible.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 6 at 6:28









        Michael Rozenberg

        88.2k1579180




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