Any product of primes in the form of 4n+1 is the sum of 2 relatively prime squares












0












$begingroup$


How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?



EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.






elementary-number-theory







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  • $begingroup$
    See also this question. The "unique" squares must be coprime, because $p$ is prime.
    $endgroup$
    – Dietrich Burde
    Dec 25 '18 at 17:53








  • 1




    $begingroup$
    Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
    $endgroup$
    – Michael Behrend
    Dec 25 '18 at 18:08










  • $begingroup$
    Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
    $endgroup$
    – OmicronGamma
    Dec 25 '18 at 18:13










  • $begingroup$
    @DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
    $endgroup$
    – Bill Dubuque
    Dec 25 '18 at 19:59


















0












$begingroup$


How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?



EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    See also this question. The "unique" squares must be coprime, because $p$ is prime.
    $endgroup$
    – Dietrich Burde
    Dec 25 '18 at 17:53








  • 1




    $begingroup$
    Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
    $endgroup$
    – Michael Behrend
    Dec 25 '18 at 18:08










  • $begingroup$
    Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
    $endgroup$
    – OmicronGamma
    Dec 25 '18 at 18:13










  • $begingroup$
    @DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
    $endgroup$
    – Bill Dubuque
    Dec 25 '18 at 19:59
















0












0








0





$begingroup$


How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?



EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.






elementary-number-theory







share|cite|improve this question











$endgroup$




How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?



EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 18:26







OmicronGamma

















asked Dec 25 '18 at 17:48









OmicronGammaOmicronGamma

536




536












  • $begingroup$
    See also this question. The "unique" squares must be coprime, because $p$ is prime.
    $endgroup$
    – Dietrich Burde
    Dec 25 '18 at 17:53








  • 1




    $begingroup$
    Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
    $endgroup$
    – Michael Behrend
    Dec 25 '18 at 18:08










  • $begingroup$
    Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
    $endgroup$
    – OmicronGamma
    Dec 25 '18 at 18:13










  • $begingroup$
    @DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
    $endgroup$
    – Bill Dubuque
    Dec 25 '18 at 19:59




















  • $begingroup$
    See also this question. The "unique" squares must be coprime, because $p$ is prime.
    $endgroup$
    – Dietrich Burde
    Dec 25 '18 at 17:53








  • 1




    $begingroup$
    Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
    $endgroup$
    – Michael Behrend
    Dec 25 '18 at 18:08










  • $begingroup$
    Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
    $endgroup$
    – OmicronGamma
    Dec 25 '18 at 18:13










  • $begingroup$
    @DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
    $endgroup$
    – Bill Dubuque
    Dec 25 '18 at 19:59


















$begingroup$
See also this question. The "unique" squares must be coprime, because $p$ is prime.
$endgroup$
– Dietrich Burde
Dec 25 '18 at 17:53






$begingroup$
See also this question. The "unique" squares must be coprime, because $p$ is prime.
$endgroup$
– Dietrich Burde
Dec 25 '18 at 17:53






1




1




$begingroup$
Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
$endgroup$
– Michael Behrend
Dec 25 '18 at 18:08




$begingroup$
Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
$endgroup$
– Michael Behrend
Dec 25 '18 at 18:08












$begingroup$
Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
$endgroup$
– OmicronGamma
Dec 25 '18 at 18:13




$begingroup$
Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
$endgroup$
– OmicronGamma
Dec 25 '18 at 18:13












$begingroup$
@DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 19:59






$begingroup$
@DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 19:59












2 Answers
2






active

oldest

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2












$begingroup$

Since every prime
of the form 4n+1
is the sum of two squares,
and



$begin{array}\
(a^2+b^2)(c^2+d^2)
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
&=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
&=(acpm bd)^2+(admp bc)^2\
end{array}
$



the product of
any number of primes
of the form 4n+1
is the sum of two squares.



If the two squares
are not relatively prime,
some prime divides both of them,
and so the square of that prime
divides each square and so divides
the product of the primes.



But the product of the primes
(unless two of them are the same)
is only divisible by
primes to the first power,
a contradiction.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.


    For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
    $$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
    where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.


    Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
    $$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
    Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.


    If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Since every prime
      of the form 4n+1
      is the sum of two squares,
      and



      $begin{array}\
      (a^2+b^2)(c^2+d^2)
      &=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
      &=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
      &=(acpm bd)^2+(admp bc)^2\
      end{array}
      $



      the product of
      any number of primes
      of the form 4n+1
      is the sum of two squares.



      If the two squares
      are not relatively prime,
      some prime divides both of them,
      and so the square of that prime
      divides each square and so divides
      the product of the primes.



      But the product of the primes
      (unless two of them are the same)
      is only divisible by
      primes to the first power,
      a contradiction.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Since every prime
        of the form 4n+1
        is the sum of two squares,
        and



        $begin{array}\
        (a^2+b^2)(c^2+d^2)
        &=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
        &=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
        &=(acpm bd)^2+(admp bc)^2\
        end{array}
        $



        the product of
        any number of primes
        of the form 4n+1
        is the sum of two squares.



        If the two squares
        are not relatively prime,
        some prime divides both of them,
        and so the square of that prime
        divides each square and so divides
        the product of the primes.



        But the product of the primes
        (unless two of them are the same)
        is only divisible by
        primes to the first power,
        a contradiction.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Since every prime
          of the form 4n+1
          is the sum of two squares,
          and



          $begin{array}\
          (a^2+b^2)(c^2+d^2)
          &=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
          &=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
          &=(acpm bd)^2+(admp bc)^2\
          end{array}
          $



          the product of
          any number of primes
          of the form 4n+1
          is the sum of two squares.



          If the two squares
          are not relatively prime,
          some prime divides both of them,
          and so the square of that prime
          divides each square and so divides
          the product of the primes.



          But the product of the primes
          (unless two of them are the same)
          is only divisible by
          primes to the first power,
          a contradiction.






          share|cite|improve this answer











          $endgroup$



          Since every prime
          of the form 4n+1
          is the sum of two squares,
          and



          $begin{array}\
          (a^2+b^2)(c^2+d^2)
          &=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
          &=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
          &=(acpm bd)^2+(admp bc)^2\
          end{array}
          $



          the product of
          any number of primes
          of the form 4n+1
          is the sum of two squares.



          If the two squares
          are not relatively prime,
          some prime divides both of them,
          and so the square of that prime
          divides each square and so divides
          the product of the primes.



          But the product of the primes
          (unless two of them are the same)
          is only divisible by
          primes to the first power,
          a contradiction.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 26 '18 at 21:27

























          answered Dec 25 '18 at 22:25









          marty cohenmarty cohen

          73.8k549128




          73.8k549128























              1












              $begingroup$

              I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.


              For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
              $$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
              where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.


              Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
              $$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
              Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.


              If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.


                For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
                $$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
                where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.


                Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
                $$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
                Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.


                If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.


                  For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
                  $$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
                  where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.


                  Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
                  $$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
                  Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.


                  If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.






                  share|cite|improve this answer









                  $endgroup$



                  I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.


                  For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
                  $$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
                  where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.


                  Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
                  $$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
                  Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.


                  If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '18 at 10:04









                  Michael BehrendMichael Behrend

                  1,22746




                  1,22746






























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                      Mont Emei

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                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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