Pythagorean Triplets with “Bounds”











up vote
3
down vote

favorite












I am interested in the algebraic/geometric way of finding the pythagorean triplets such that



$$a^2 + b^2 = c^2$$



$$a + b + c = 1000$$



I do the obvious



$$a + b = 1000 - (a^2 + b^2)^{1/2}$$



$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$



$$2a + 2b - frac{2ab}{1000} = 1000$$



$$a + b -frac{ab}{1000} = 500$$



I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing



$$a + b = 500 + frac{ab}{1000} = 1000 - c$$



But this is going backwards






algebra-precalculus







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  • Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
    – Sidd Singal
    Aug 7 '12 at 16:38












  • Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
    – user940
    Aug 7 '12 at 17:01















up vote
3
down vote

favorite












I am interested in the algebraic/geometric way of finding the pythagorean triplets such that



$$a^2 + b^2 = c^2$$



$$a + b + c = 1000$$



I do the obvious



$$a + b = 1000 - (a^2 + b^2)^{1/2}$$



$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$



$$2a + 2b - frac{2ab}{1000} = 1000$$



$$a + b -frac{ab}{1000} = 500$$



I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing



$$a + b = 500 + frac{ab}{1000} = 1000 - c$$



But this is going backwards






algebra-precalculus







share|cite|improve this question
























  • Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
    – Sidd Singal
    Aug 7 '12 at 16:38












  • Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
    – user940
    Aug 7 '12 at 17:01













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I am interested in the algebraic/geometric way of finding the pythagorean triplets such that



$$a^2 + b^2 = c^2$$



$$a + b + c = 1000$$



I do the obvious



$$a + b = 1000 - (a^2 + b^2)^{1/2}$$



$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$



$$2a + 2b - frac{2ab}{1000} = 1000$$



$$a + b -frac{ab}{1000} = 500$$



I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing



$$a + b = 500 + frac{ab}{1000} = 1000 - c$$



But this is going backwards






algebra-precalculus







share|cite|improve this question















I am interested in the algebraic/geometric way of finding the pythagorean triplets such that



$$a^2 + b^2 = c^2$$



$$a + b + c = 1000$$



I do the obvious



$$a + b = 1000 - (a^2 + b^2)^{1/2}$$



$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$



$$2a + 2b - frac{2ab}{1000} = 1000$$



$$a + b -frac{ab}{1000} = 500$$



I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing



$$a + b = 500 + frac{ab}{1000} = 1000 - c$$



But this is going backwards






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 7 '12 at 16:41









Micah

29.5k1363104




29.5k1363104










asked Aug 7 '12 at 16:33









Cactus BAMF

5441920




5441920












  • Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
    – Sidd Singal
    Aug 7 '12 at 16:38












  • Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
    – user940
    Aug 7 '12 at 17:01


















  • Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
    – Sidd Singal
    Aug 7 '12 at 16:38












  • Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
    – user940
    Aug 7 '12 at 17:01
















Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
– Sidd Singal
Aug 7 '12 at 16:38






Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple
– Sidd Singal
Aug 7 '12 at 16:38














Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
– user940
Aug 7 '12 at 17:01




Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$?
– user940
Aug 7 '12 at 17:01










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










The following is copy and pasted directly from Yahoo Answers



All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.



You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.



$m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
and
$m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.



The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
Thus, $m=20$, $n=5$, giving the answer required:
$m^2+n^2$ $= 425$;
$m^2-n^2 = 375$;
$2mn = 200$.






share|cite|improve this answer






























    up vote
    1
    down vote













    We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.



    So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$



    =>$kp(p+q)=500$



    Clearly, k can be any divisor of 500.



    If k=1,



    If p=1, p+q=500=> q = 499,



    if p=2, p+q=250=> q = 248 and so on.



    If k=2,$p(p+q)=250$



    If k=5,$p(p+q)=100$



    If k=10, $p(p+q)=50$ and so on.






    share|cite|improve this answer























    • You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
      – Qiaochu Yuan
      Aug 7 '12 at 16:43












    • @QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
      – lab bhattacharjee
      Aug 7 '12 at 18:00











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    The following is copy and pasted directly from Yahoo Answers



    All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.



    You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.



    $m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
    and
    $m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.



    The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
    Thus, $m=20$, $n=5$, giving the answer required:
    $m^2+n^2$ $= 425$;
    $m^2-n^2 = 375$;
    $2mn = 200$.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      The following is copy and pasted directly from Yahoo Answers



      All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.



      You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.



      $m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
      and
      $m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.



      The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
      Thus, $m=20$, $n=5$, giving the answer required:
      $m^2+n^2$ $= 425$;
      $m^2-n^2 = 375$;
      $2mn = 200$.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        The following is copy and pasted directly from Yahoo Answers



        All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.



        You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.



        $m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
        and
        $m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.



        The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
        Thus, $m=20$, $n=5$, giving the answer required:
        $m^2+n^2$ $= 425$;
        $m^2-n^2 = 375$;
        $2mn = 200$.






        share|cite|improve this answer














        The following is copy and pasted directly from Yahoo Answers



        All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $mgt n$.



        You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.



        $m+ngt m$, so $m(m+n)gt m^2$, so $mlt sqrt{(500)}$,
        and
        $m+nlt2m$ (since $mgt n$), so $m(m+n)lt2m^2$, so $mgt sqrt{(250)}$.



        The prime factorisation of 500 is $2 cdot 2 cdot 5 cdot 5 cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$.
        Thus, $m=20$, $n=5$, giving the answer required:
        $m^2+n^2$ $= 425$;
        $m^2-n^2 = 375$;
        $2mn = 200$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 19 at 20:38









        Ben Brian

        1034




        1034










        answered Aug 7 '12 at 16:47









        Sidd Singal

        2,56231630




        2,56231630






















            up vote
            1
            down vote













            We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.



            So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$



            =>$kp(p+q)=500$



            Clearly, k can be any divisor of 500.



            If k=1,



            If p=1, p+q=500=> q = 499,



            if p=2, p+q=250=> q = 248 and so on.



            If k=2,$p(p+q)=250$



            If k=5,$p(p+q)=100$



            If k=10, $p(p+q)=50$ and so on.






            share|cite|improve this answer























            • You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
              – Qiaochu Yuan
              Aug 7 '12 at 16:43












            • @QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
              – lab bhattacharjee
              Aug 7 '12 at 18:00















            up vote
            1
            down vote













            We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.



            So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$



            =>$kp(p+q)=500$



            Clearly, k can be any divisor of 500.



            If k=1,



            If p=1, p+q=500=> q = 499,



            if p=2, p+q=250=> q = 248 and so on.



            If k=2,$p(p+q)=250$



            If k=5,$p(p+q)=100$



            If k=10, $p(p+q)=50$ and so on.






            share|cite|improve this answer























            • You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
              – Qiaochu Yuan
              Aug 7 '12 at 16:43












            • @QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
              – lab bhattacharjee
              Aug 7 '12 at 18:00













            up vote
            1
            down vote










            up vote
            1
            down vote









            We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.



            So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$



            =>$kp(p+q)=500$



            Clearly, k can be any divisor of 500.



            If k=1,



            If p=1, p+q=500=> q = 499,



            if p=2, p+q=250=> q = 248 and so on.



            If k=2,$p(p+q)=250$



            If k=5,$p(p+q)=100$



            If k=10, $p(p+q)=50$ and so on.






            share|cite|improve this answer














            We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.



            So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$



            =>$kp(p+q)=500$



            Clearly, k can be any divisor of 500.



            If k=1,



            If p=1, p+q=500=> q = 499,



            if p=2, p+q=250=> q = 248 and so on.



            If k=2,$p(p+q)=250$



            If k=5,$p(p+q)=100$



            If k=10, $p(p+q)=50$ and so on.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 7 '12 at 16:53

























            answered Aug 7 '12 at 16:41









            lab bhattacharjee

            222k15155273




            222k15155273












            • You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
              – Qiaochu Yuan
              Aug 7 '12 at 16:43












            • @QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
              – lab bhattacharjee
              Aug 7 '12 at 18:00


















            • You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
              – Qiaochu Yuan
              Aug 7 '12 at 16:43












            • @QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
              – lab bhattacharjee
              Aug 7 '12 at 18:00
















            You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
            – Qiaochu Yuan
            Aug 7 '12 at 16:43






            You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$).
            – Qiaochu Yuan
            Aug 7 '12 at 16:43














            @QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
            – lab bhattacharjee
            Aug 7 '12 at 18:00




            @QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted.
            – lab bhattacharjee
            Aug 7 '12 at 18:00


















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