If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?












2















Math Olympiad Question:



If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?




I was solving some problems of the Indian National Mathematics Olympiad and I found this problem to be very intriguing. At first look it looks pretty straight forward but this problem requires the solution to be in terms of $c$ only, meaning that if you have the value of c, put that in your solution and you should get the same value as $a^2-b^{2 }$.



I tried simple algebra, trigonometry, geometry and pattern recognition but none of these methods were quite successful and could give a reasonable proof + solution.



My question is how to find the solution and prove this question? Is it possible to use Geometry and solve?



EDIT -
Sorry my bad, I think $a>b$ and $a$ and $b$ are integer values as in the starting of the exam it's written that all values will be integer and non-negative.




FINAL EDIT:
I think this question has been misprinted. Sorry for the inconvenience.







algebra-precalculus







share|cite|improve this question




















  • 8




    Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
    – sedrick
    Nov 25 '18 at 9:09












  • I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
    – Mohammad Zuhair Khan
    Nov 25 '18 at 9:10






  • 2




    Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
    – Batominovski
    Nov 25 '18 at 9:12












  • It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
    – strangeindian
    Nov 25 '18 at 9:12










  • Another example to consider $7^2+24^2=15^2+20^2=25^2$
    – Mark Bennet
    Nov 25 '18 at 9:12
















2















Math Olympiad Question:



If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?




I was solving some problems of the Indian National Mathematics Olympiad and I found this problem to be very intriguing. At first look it looks pretty straight forward but this problem requires the solution to be in terms of $c$ only, meaning that if you have the value of c, put that in your solution and you should get the same value as $a^2-b^{2 }$.



I tried simple algebra, trigonometry, geometry and pattern recognition but none of these methods were quite successful and could give a reasonable proof + solution.



My question is how to find the solution and prove this question? Is it possible to use Geometry and solve?



EDIT -
Sorry my bad, I think $a>b$ and $a$ and $b$ are integer values as in the starting of the exam it's written that all values will be integer and non-negative.




FINAL EDIT:
I think this question has been misprinted. Sorry for the inconvenience.







algebra-precalculus







share|cite|improve this question




















  • 8




    Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
    – sedrick
    Nov 25 '18 at 9:09












  • I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
    – Mohammad Zuhair Khan
    Nov 25 '18 at 9:10






  • 2




    Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
    – Batominovski
    Nov 25 '18 at 9:12












  • It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
    – strangeindian
    Nov 25 '18 at 9:12










  • Another example to consider $7^2+24^2=15^2+20^2=25^2$
    – Mark Bennet
    Nov 25 '18 at 9:12














2












2








2








Math Olympiad Question:



If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?




I was solving some problems of the Indian National Mathematics Olympiad and I found this problem to be very intriguing. At first look it looks pretty straight forward but this problem requires the solution to be in terms of $c$ only, meaning that if you have the value of c, put that in your solution and you should get the same value as $a^2-b^{2 }$.



I tried simple algebra, trigonometry, geometry and pattern recognition but none of these methods were quite successful and could give a reasonable proof + solution.



My question is how to find the solution and prove this question? Is it possible to use Geometry and solve?



EDIT -
Sorry my bad, I think $a>b$ and $a$ and $b$ are integer values as in the starting of the exam it's written that all values will be integer and non-negative.




FINAL EDIT:
I think this question has been misprinted. Sorry for the inconvenience.







algebra-precalculus







share|cite|improve this question
















Math Olympiad Question:



If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?




I was solving some problems of the Indian National Mathematics Olympiad and I found this problem to be very intriguing. At first look it looks pretty straight forward but this problem requires the solution to be in terms of $c$ only, meaning that if you have the value of c, put that in your solution and you should get the same value as $a^2-b^{2 }$.



I tried simple algebra, trigonometry, geometry and pattern recognition but none of these methods were quite successful and could give a reasonable proof + solution.



My question is how to find the solution and prove this question? Is it possible to use Geometry and solve?



EDIT -
Sorry my bad, I think $a>b$ and $a$ and $b$ are integer values as in the starting of the exam it's written that all values will be integer and non-negative.




FINAL EDIT:
I think this question has been misprinted. Sorry for the inconvenience.







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 Nov 25 '18 at 9:23

























asked Nov 25 '18 at 9:05









strangeindian

455




455








  • 8




    Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
    – sedrick
    Nov 25 '18 at 9:09












  • I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
    – Mohammad Zuhair Khan
    Nov 25 '18 at 9:10






  • 2




    Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
    – Batominovski
    Nov 25 '18 at 9:12












  • It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
    – strangeindian
    Nov 25 '18 at 9:12










  • Another example to consider $7^2+24^2=15^2+20^2=25^2$
    – Mark Bennet
    Nov 25 '18 at 9:12














  • 8




    Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
    – sedrick
    Nov 25 '18 at 9:09












  • I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
    – Mohammad Zuhair Khan
    Nov 25 '18 at 9:10






  • 2




    Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
    – Batominovski
    Nov 25 '18 at 9:12












  • It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
    – strangeindian
    Nov 25 '18 at 9:12










  • Another example to consider $7^2+24^2=15^2+20^2=25^2$
    – Mark Bennet
    Nov 25 '18 at 9:12








8




8




Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
– sedrick
Nov 25 '18 at 9:09






Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
– sedrick
Nov 25 '18 at 9:09














I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
– Mohammad Zuhair Khan
Nov 25 '18 at 9:10




I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
– Mohammad Zuhair Khan
Nov 25 '18 at 9:10




2




2




Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
– Batominovski
Nov 25 '18 at 9:12






Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
– Batominovski
Nov 25 '18 at 9:12














It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
– strangeindian
Nov 25 '18 at 9:12




It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
– strangeindian
Nov 25 '18 at 9:12












Another example to consider $7^2+24^2=15^2+20^2=25^2$
– Mark Bennet
Nov 25 '18 at 9:12




Another example to consider $7^2+24^2=15^2+20^2=25^2$
– Mark Bennet
Nov 25 '18 at 9:12










2 Answers
2






active

oldest

votes


















4














You are essentially looking at points on a circle of radius $c$.



If $x=arctan left(frac baright)$



then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$



while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.



$cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.






share|cite|improve this answer





















  • I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
    – strangeindian
    Nov 25 '18 at 9:25



















3














This is not possible.



Take, for example, the following:



$$
4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
$$



but
$$
4^2 - 3^2 = 7\
5^2 - 0^2 =25\
sqrt{21}^2-2^2 = 17
$$



So the value of $a^2-b^2$ does not depend only on $c$.






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









    4














    You are essentially looking at points on a circle of radius $c$.



    If $x=arctan left(frac baright)$



    then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$



    while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.



    $cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.






    share|cite|improve this answer





















    • I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
      – strangeindian
      Nov 25 '18 at 9:25
















    4














    You are essentially looking at points on a circle of radius $c$.



    If $x=arctan left(frac baright)$



    then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$



    while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.



    $cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.






    share|cite|improve this answer





















    • I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
      – strangeindian
      Nov 25 '18 at 9:25














    4












    4








    4






    You are essentially looking at points on a circle of radius $c$.



    If $x=arctan left(frac baright)$



    then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$



    while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.



    $cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.






    share|cite|improve this answer












    You are essentially looking at points on a circle of radius $c$.



    If $x=arctan left(frac baright)$



    then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$



    while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.



    $cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 25 '18 at 9:23









    Mark Bennet

    80.5k981179




    80.5k981179












    • I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
      – strangeindian
      Nov 25 '18 at 9:25


















    • I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
      – strangeindian
      Nov 25 '18 at 9:25
















    I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
    – strangeindian
    Nov 25 '18 at 9:25




    I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
    – strangeindian
    Nov 25 '18 at 9:25











    3














    This is not possible.



    Take, for example, the following:



    $$
    4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
    $$



    but
    $$
    4^2 - 3^2 = 7\
    5^2 - 0^2 =25\
    sqrt{21}^2-2^2 = 17
    $$



    So the value of $a^2-b^2$ does not depend only on $c$.






    share|cite|improve this answer


























      3














      This is not possible.



      Take, for example, the following:



      $$
      4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
      $$



      but
      $$
      4^2 - 3^2 = 7\
      5^2 - 0^2 =25\
      sqrt{21}^2-2^2 = 17
      $$



      So the value of $a^2-b^2$ does not depend only on $c$.






      share|cite|improve this answer
























        3












        3








        3






        This is not possible.



        Take, for example, the following:



        $$
        4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
        $$



        but
        $$
        4^2 - 3^2 = 7\
        5^2 - 0^2 =25\
        sqrt{21}^2-2^2 = 17
        $$



        So the value of $a^2-b^2$ does not depend only on $c$.






        share|cite|improve this answer












        This is not possible.



        Take, for example, the following:



        $$
        4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
        $$



        but
        $$
        4^2 - 3^2 = 7\
        5^2 - 0^2 =25\
        sqrt{21}^2-2^2 = 17
        $$



        So the value of $a^2-b^2$ does not depend only on $c$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 9:14









        bruderjakob17

        1587




        1587






























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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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