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Is there an existing parametrization of the equation above that is similar to Brahmagupta's identity for $a^2+b^2=c^2+d^2$? I need either a reference to look it up or a hint to solve it. Thanks.

  • There is no $3$-square theorem, I you mean that. – Dietrich Burde Apr 04 '16 at 12:03
  • Could you give a link to Bramagupta's identity? (I'm interested in whether it gives all the solutions of $a^2+b^2=c^2+d^2$ in a parametric form, like the known one for $x^2+y^2=z^2.$] – coffeemath Apr 04 '16 at 12:04
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    http://math.stackexchange.com/questions/889062/solve-f2-e2-d2-c2-b2a2/889139#889139 – individ Apr 04 '16 at 12:54
  • Like Brahmagupta's identity:

    $$\left( {p}^{2}+{k}^{2}+{b}^{2}\right) ,\left( {s}^{2}+{h}^{2}+{f}^{2}\right) ={\left( p,s+b,h\right) }^{2}+{\left( k,s+b,f\right) }^{2}+{\left( b,s-h,p-f,k\right) }^{2}+{\left( f,p-h,k\right) }^{2}$$

    taking into account different signs will get four equations.

     – AlexSam Apr 24 '16 at 13:32

2 Answers2

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There is a parameterization for every equal sums of squares equation $$ X_1^2 + \dotsb + X_m^2 = Y_1^2 + \dotsb + Y_n^2 $$ with $n,m$ positive integers and all $X_i,Y_i$ integers. The papers by Barnett and Bradley are my first recommendations.

Kieren MacMillan
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  • Thanks. I'll check them out. I can't open those files now on the mobile app. Just curious, do they give all the possible solutions? –  Apr 04 '16 at 12:21
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    Yes. They give formulas (with derivations) for general $n,m$, and then work out specific cases of $n$ and $m$. – Kieren MacMillan Apr 04 '16 at 12:36
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Above equation (a^2+b^2+c^2)=(d^2+e^2+f^2) has parametric solution. Refer to Tito piezas on line book "collection of algebraic Identities".
Section sum of squares. The answer is given below;

(a,b,c)=[(p+q),(r+s),(t+u)] and

(d,e,f)=[(p-q),(r-s),(t-u)]

Condition is (pq+rs+tu)=0

After parametrization of (pq+rs+tu)=0, the parametric solution in one variable is given below:

(a,b,c)=[(4k^2+12k+1),(3k-8),(7k+28)] and

(d,e,f)=[(4k^2+12k-13),(k+2),(13k+26)]

Sam
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