If x,y and z are positive integers and $frac 1x + frac 1y = frac 1z$ then $sqrt{x^2+y^2+z^2}$ is rational.












3












$begingroup$


To solve this problem I first started off by factoring to get $z^2=(x-z)(y-z)$ only to realise that this does nothing so I then tried squaring both sides to get the reciprocals of $x,y$ and $z$ squared minus $2$ over $xy$ but after this I have no idea what to do any form of help would be appreciated.(sorry for not using mathjax as my browser does not support it)






polynomials radicals rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    exactly @астонвіллаолофмэллбэрг
    $endgroup$
    – Mary Tom
    Jan 26 at 7:36










  • $begingroup$
    no way , but did you check x=y=z=1 @ shelby
    $endgroup$
    – Mary Tom
    Jan 26 at 7:37












  • $begingroup$
    so the question is asking to prove that ( x^2+y^2+z^2) is a perfect square.
    $endgroup$
    – Mary Tom
    Jan 26 at 7:43










  • $begingroup$
    Seen the edited question.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 26 at 7:49
















3












$begingroup$


To solve this problem I first started off by factoring to get $z^2=(x-z)(y-z)$ only to realise that this does nothing so I then tried squaring both sides to get the reciprocals of $x,y$ and $z$ squared minus $2$ over $xy$ but after this I have no idea what to do any form of help would be appreciated.(sorry for not using mathjax as my browser does not support it)






polynomials radicals rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    exactly @астонвіллаолофмэллбэрг
    $endgroup$
    – Mary Tom
    Jan 26 at 7:36










  • $begingroup$
    no way , but did you check x=y=z=1 @ shelby
    $endgroup$
    – Mary Tom
    Jan 26 at 7:37












  • $begingroup$
    so the question is asking to prove that ( x^2+y^2+z^2) is a perfect square.
    $endgroup$
    – Mary Tom
    Jan 26 at 7:43










  • $begingroup$
    Seen the edited question.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 26 at 7:49














3












3








3





$begingroup$


To solve this problem I first started off by factoring to get $z^2=(x-z)(y-z)$ only to realise that this does nothing so I then tried squaring both sides to get the reciprocals of $x,y$ and $z$ squared minus $2$ over $xy$ but after this I have no idea what to do any form of help would be appreciated.(sorry for not using mathjax as my browser does not support it)






polynomials radicals rational-numbers







share|cite|improve this question











$endgroup$




To solve this problem I first started off by factoring to get $z^2=(x-z)(y-z)$ only to realise that this does nothing so I then tried squaring both sides to get the reciprocals of $x,y$ and $z$ squared minus $2$ over $xy$ but after this I have no idea what to do any form of help would be appreciated.(sorry for not using mathjax as my browser does not support it)






polynomials radicals rational-numbers




polynomials radicals rational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 9:33









Michael Rozenberg

108k1895200




108k1895200










asked Jan 26 at 7:33









Mary TomMary Tom

326




326












  • $begingroup$
    exactly @астонвіллаолофмэллбэрг
    $endgroup$
    – Mary Tom
    Jan 26 at 7:36










  • $begingroup$
    no way , but did you check x=y=z=1 @ shelby
    $endgroup$
    – Mary Tom
    Jan 26 at 7:37












  • $begingroup$
    so the question is asking to prove that ( x^2+y^2+z^2) is a perfect square.
    $endgroup$
    – Mary Tom
    Jan 26 at 7:43










  • $begingroup$
    Seen the edited question.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 26 at 7:49


















  • $begingroup$
    exactly @астонвіллаолофмэллбэрг
    $endgroup$
    – Mary Tom
    Jan 26 at 7:36










  • $begingroup$
    no way , but did you check x=y=z=1 @ shelby
    $endgroup$
    – Mary Tom
    Jan 26 at 7:37












  • $begingroup$
    so the question is asking to prove that ( x^2+y^2+z^2) is a perfect square.
    $endgroup$
    – Mary Tom
    Jan 26 at 7:43










  • $begingroup$
    Seen the edited question.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 26 at 7:49
















$begingroup$
exactly @астонвіллаолофмэллбэрг
$endgroup$
– Mary Tom
Jan 26 at 7:36




$begingroup$
exactly @астонвіллаолофмэллбэрг
$endgroup$
– Mary Tom
Jan 26 at 7:36












$begingroup$
no way , but did you check x=y=z=1 @ shelby
$endgroup$
– Mary Tom
Jan 26 at 7:37






$begingroup$
no way , but did you check x=y=z=1 @ shelby
$endgroup$
– Mary Tom
Jan 26 at 7:37














$begingroup$
so the question is asking to prove that ( x^2+y^2+z^2) is a perfect square.
$endgroup$
– Mary Tom
Jan 26 at 7:43




$begingroup$
so the question is asking to prove that ( x^2+y^2+z^2) is a perfect square.
$endgroup$
– Mary Tom
Jan 26 at 7:43












$begingroup$
Seen the edited question.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:49




$begingroup$
Seen the edited question.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:49










3 Answers
3






active

oldest

votes


















2












$begingroup$

The solutions to $dfrac 1x + dfrac 1y = dfrac 1z$ are characterized by
(See this)



begin{align}
x &= kp(p+q) \
y &= kq(p+q) \
z &= kpq
end{align}



Then
begin{align}
dfrac{x^2+y^2+z^2}{k^2}
&= p^2(p+q)^2 + q^2(p+q)^2 + p^2q^2 \
&= p^4 + 2p^3q + 3p^2q^2 + 2pq^3 + q^4 \
&= left( begin{array}{l}
p^4 &+ &p^3q &+ &p^2q^2 &+ \
& &p^3q &+ &p^2q^2 &+ &pq^3 &+ \
& & &+ &p^2q^2 &+ &pq^3 &+ &q^4
end{array} right) \
&= p^2(p^2 + pq + q^2) + pq(p^2 + pq + q^2) + q^2(p^2 + pq + q^2) \
&= (p^2 + pq + q^2)^2
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I just love @steven gregory's solution it is so elegantly proved!
    $endgroup$
    – Mary Tom
    Jan 26 at 8:44



















4












$begingroup$

Because $xneq-y$, $z=frac{xy}{x+y}$, $x^2+xy+y^2>0$ and
$$sqrt{x^2+y^2+z^2}=sqrt{x^2+y^2+frac{x^2y^2}{(x+y)^2}}=$$
$$=frac{sqrt{(x^2+y^2+xy-xy)(x^2+y^2+xy+xy)+x^2y^2}}{x+y}=frac{x^2+xy+y^2}{x+y}.$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $(x+y-z)^2=x^2+y^2+z^2+2xy-2xz-2yz=(x^2+y^2+z^2)+2xyzunderbrace{left(dfrac 1z-dfrac 1y-dfrac 1xright)}_0$



    So $sqrt{x^2+y^2+z^2}=|x+y-z|$ so not only it is rational, but also integer.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Are you happy now? +1
      $endgroup$
      – Maria Mazur
      Jan 26 at 12:04










    • $begingroup$
      I gave you a +1 too.
      $endgroup$
      – steven gregory
      Jan 28 at 14:02











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The solutions to $dfrac 1x + dfrac 1y = dfrac 1z$ are characterized by
    (See this)



    begin{align}
    x &= kp(p+q) \
    y &= kq(p+q) \
    z &= kpq
    end{align}



    Then
    begin{align}
    dfrac{x^2+y^2+z^2}{k^2}
    &= p^2(p+q)^2 + q^2(p+q)^2 + p^2q^2 \
    &= p^4 + 2p^3q + 3p^2q^2 + 2pq^3 + q^4 \
    &= left( begin{array}{l}
    p^4 &+ &p^3q &+ &p^2q^2 &+ \
    & &p^3q &+ &p^2q^2 &+ &pq^3 &+ \
    & & &+ &p^2q^2 &+ &pq^3 &+ &q^4
    end{array} right) \
    &= p^2(p^2 + pq + q^2) + pq(p^2 + pq + q^2) + q^2(p^2 + pq + q^2) \
    &= (p^2 + pq + q^2)^2
    end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I just love @steven gregory's solution it is so elegantly proved!
      $endgroup$
      – Mary Tom
      Jan 26 at 8:44
















    2












    $begingroup$

    The solutions to $dfrac 1x + dfrac 1y = dfrac 1z$ are characterized by
    (See this)



    begin{align}
    x &= kp(p+q) \
    y &= kq(p+q) \
    z &= kpq
    end{align}



    Then
    begin{align}
    dfrac{x^2+y^2+z^2}{k^2}
    &= p^2(p+q)^2 + q^2(p+q)^2 + p^2q^2 \
    &= p^4 + 2p^3q + 3p^2q^2 + 2pq^3 + q^4 \
    &= left( begin{array}{l}
    p^4 &+ &p^3q &+ &p^2q^2 &+ \
    & &p^3q &+ &p^2q^2 &+ &pq^3 &+ \
    & & &+ &p^2q^2 &+ &pq^3 &+ &q^4
    end{array} right) \
    &= p^2(p^2 + pq + q^2) + pq(p^2 + pq + q^2) + q^2(p^2 + pq + q^2) \
    &= (p^2 + pq + q^2)^2
    end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I just love @steven gregory's solution it is so elegantly proved!
      $endgroup$
      – Mary Tom
      Jan 26 at 8:44














    2












    2








    2





    $begingroup$

    The solutions to $dfrac 1x + dfrac 1y = dfrac 1z$ are characterized by
    (See this)



    begin{align}
    x &= kp(p+q) \
    y &= kq(p+q) \
    z &= kpq
    end{align}



    Then
    begin{align}
    dfrac{x^2+y^2+z^2}{k^2}
    &= p^2(p+q)^2 + q^2(p+q)^2 + p^2q^2 \
    &= p^4 + 2p^3q + 3p^2q^2 + 2pq^3 + q^4 \
    &= left( begin{array}{l}
    p^4 &+ &p^3q &+ &p^2q^2 &+ \
    & &p^3q &+ &p^2q^2 &+ &pq^3 &+ \
    & & &+ &p^2q^2 &+ &pq^3 &+ &q^4
    end{array} right) \
    &= p^2(p^2 + pq + q^2) + pq(p^2 + pq + q^2) + q^2(p^2 + pq + q^2) \
    &= (p^2 + pq + q^2)^2
    end{align}






    share|cite|improve this answer











    $endgroup$



    The solutions to $dfrac 1x + dfrac 1y = dfrac 1z$ are characterized by
    (See this)



    begin{align}
    x &= kp(p+q) \
    y &= kq(p+q) \
    z &= kpq
    end{align}



    Then
    begin{align}
    dfrac{x^2+y^2+z^2}{k^2}
    &= p^2(p+q)^2 + q^2(p+q)^2 + p^2q^2 \
    &= p^4 + 2p^3q + 3p^2q^2 + 2pq^3 + q^4 \
    &= left( begin{array}{l}
    p^4 &+ &p^3q &+ &p^2q^2 &+ \
    & &p^3q &+ &p^2q^2 &+ &pq^3 &+ \
    & & &+ &p^2q^2 &+ &pq^3 &+ &q^4
    end{array} right) \
    &= p^2(p^2 + pq + q^2) + pq(p^2 + pq + q^2) + q^2(p^2 + pq + q^2) \
    &= (p^2 + pq + q^2)^2
    end{align}







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 26 at 8:32

























    answered Jan 26 at 8:20









    steven gregorysteven gregory

    18.3k32358




    18.3k32358












    • $begingroup$
      I just love @steven gregory's solution it is so elegantly proved!
      $endgroup$
      – Mary Tom
      Jan 26 at 8:44


















    • $begingroup$
      I just love @steven gregory's solution it is so elegantly proved!
      $endgroup$
      – Mary Tom
      Jan 26 at 8:44
















    $begingroup$
    I just love @steven gregory's solution it is so elegantly proved!
    $endgroup$
    – Mary Tom
    Jan 26 at 8:44




    $begingroup$
    I just love @steven gregory's solution it is so elegantly proved!
    $endgroup$
    – Mary Tom
    Jan 26 at 8:44











    4












    $begingroup$

    Because $xneq-y$, $z=frac{xy}{x+y}$, $x^2+xy+y^2>0$ and
    $$sqrt{x^2+y^2+z^2}=sqrt{x^2+y^2+frac{x^2y^2}{(x+y)^2}}=$$
    $$=frac{sqrt{(x^2+y^2+xy-xy)(x^2+y^2+xy+xy)+x^2y^2}}{x+y}=frac{x^2+xy+y^2}{x+y}.$$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Because $xneq-y$, $z=frac{xy}{x+y}$, $x^2+xy+y^2>0$ and
      $$sqrt{x^2+y^2+z^2}=sqrt{x^2+y^2+frac{x^2y^2}{(x+y)^2}}=$$
      $$=frac{sqrt{(x^2+y^2+xy-xy)(x^2+y^2+xy+xy)+x^2y^2}}{x+y}=frac{x^2+xy+y^2}{x+y}.$$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Because $xneq-y$, $z=frac{xy}{x+y}$, $x^2+xy+y^2>0$ and
        $$sqrt{x^2+y^2+z^2}=sqrt{x^2+y^2+frac{x^2y^2}{(x+y)^2}}=$$
        $$=frac{sqrt{(x^2+y^2+xy-xy)(x^2+y^2+xy+xy)+x^2y^2}}{x+y}=frac{x^2+xy+y^2}{x+y}.$$






        share|cite|improve this answer









        $endgroup$



        Because $xneq-y$, $z=frac{xy}{x+y}$, $x^2+xy+y^2>0$ and
        $$sqrt{x^2+y^2+z^2}=sqrt{x^2+y^2+frac{x^2y^2}{(x+y)^2}}=$$
        $$=frac{sqrt{(x^2+y^2+xy-xy)(x^2+y^2+xy+xy)+x^2y^2}}{x+y}=frac{x^2+xy+y^2}{x+y}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 26 at 7:54









        Michael RozenbergMichael Rozenberg

        108k1895200




        108k1895200























            2












            $begingroup$

            $(x+y-z)^2=x^2+y^2+z^2+2xy-2xz-2yz=(x^2+y^2+z^2)+2xyzunderbrace{left(dfrac 1z-dfrac 1y-dfrac 1xright)}_0$



            So $sqrt{x^2+y^2+z^2}=|x+y-z|$ so not only it is rational, but also integer.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Are you happy now? +1
              $endgroup$
              – Maria Mazur
              Jan 26 at 12:04










            • $begingroup$
              I gave you a +1 too.
              $endgroup$
              – steven gregory
              Jan 28 at 14:02
















            2












            $begingroup$

            $(x+y-z)^2=x^2+y^2+z^2+2xy-2xz-2yz=(x^2+y^2+z^2)+2xyzunderbrace{left(dfrac 1z-dfrac 1y-dfrac 1xright)}_0$



            So $sqrt{x^2+y^2+z^2}=|x+y-z|$ so not only it is rational, but also integer.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Are you happy now? +1
              $endgroup$
              – Maria Mazur
              Jan 26 at 12:04










            • $begingroup$
              I gave you a +1 too.
              $endgroup$
              – steven gregory
              Jan 28 at 14:02














            2












            2








            2





            $begingroup$

            $(x+y-z)^2=x^2+y^2+z^2+2xy-2xz-2yz=(x^2+y^2+z^2)+2xyzunderbrace{left(dfrac 1z-dfrac 1y-dfrac 1xright)}_0$



            So $sqrt{x^2+y^2+z^2}=|x+y-z|$ so not only it is rational, but also integer.






            share|cite|improve this answer









            $endgroup$



            $(x+y-z)^2=x^2+y^2+z^2+2xy-2xz-2yz=(x^2+y^2+z^2)+2xyzunderbrace{left(dfrac 1z-dfrac 1y-dfrac 1xright)}_0$



            So $sqrt{x^2+y^2+z^2}=|x+y-z|$ so not only it is rational, but also integer.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 26 at 10:32









            zwimzwim

            12.6k831




            12.6k831












            • $begingroup$
              Are you happy now? +1
              $endgroup$
              – Maria Mazur
              Jan 26 at 12:04










            • $begingroup$
              I gave you a +1 too.
              $endgroup$
              – steven gregory
              Jan 28 at 14:02


















            • $begingroup$
              Are you happy now? +1
              $endgroup$
              – Maria Mazur
              Jan 26 at 12:04










            • $begingroup$
              I gave you a +1 too.
              $endgroup$
              – steven gregory
              Jan 28 at 14:02
















            $begingroup$
            Are you happy now? +1
            $endgroup$
            – Maria Mazur
            Jan 26 at 12:04




            $begingroup$
            Are you happy now? +1
            $endgroup$
            – Maria Mazur
            Jan 26 at 12:04












            $begingroup$
            I gave you a +1 too.
            $endgroup$
            – steven gregory
            Jan 28 at 14:02




            $begingroup$
            I gave you a +1 too.
            $endgroup$
            – steven gregory
            Jan 28 at 14:02


















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            A Topological Invariant for $pi_3(U(n))$

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            3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
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