It is known that, if $x_1^2 + y_1^2 = c_1$ and $x_2^2 + y_2^2 = c_2$, then $(x_1x_2 + y_1y_2)^2 + (x_1y_2 - x_2y_1)^2 = c_1 c_2$

Is there a similar analogue for general quadratic forms $Q(x, y) = ax^2 + bxy + cy^2$?

  • $\begingroup$ "Indirect" reasoning (maybe wrong!): if yes,it would already be a quite known theorem. $\endgroup$
    – Piquito
    Commented May 29, 2015 at 16:17

1 Answer 1


The actual generalization of this may not be what you expect. Your example concerns the principal form of discriminant $-4,$ and is the earliest example of Gauss composition.

There is a similar item, I think the name Brahmagupta is used in this regard, $$ (u^2 + n v^2)(x^2 + n y^2) = (ux \pm nvy)^2 + n(uy \mp vx)^2. $$

More generally, your $\langle a,b,c\rangle $ and $\langle a,-b,c\rangle $ are inverses in the class group of forms with the same discriminant $\Delta = b^2 - 4 a c$ and $\gcd(a,b,c) = 1.$ If $\Delta \geq 0,$ it is not allowed to be $0$ or $1$ or any other square, that way leads to madness.

Maybe i should just give Dirichlet's version of composition; given two "united" forms. This is in the book by Cox, Primes of the Form $x^2 + n y^2$, in the first edition page 49: given integer constants $a,a',B,C,$ with requirement $\gcd(a,a',B)=1,$ and integer variables $x,y,z,w,$ the composition identity is $$ (a x^2 + Bxy+ a'Cy^2)(a'z^2 + B zw+ a C w^2) = a a' X^2 + BXY +CY^2, $$ with new variables $$ X = xz - C yw; \; \; Y = axw + a'yz + B yw. $$ Let me check that, i think there is a typo in the book, just one letter. Yes, what I typed here is correct, one letter wrong in that book. Good, corrected in the SECOND EDITION

You also need to know about reducing forms to make much sense of this. However, if $C=1,$ we see "opposite" forms composing into the principal form (it integrally represents $1$, and is the multiplicative identity in the group). If, instead, $B=0, a=a'=1,$ we get Brahmagupta.

In case anyone gets interested, my favorite book on binary quadratic forms is BUELL; Dirichlet's brief composition law is on page 57. Most of the discussion is on how to get two forms of the same discriminant into this useful format.

  • $\begingroup$ In this mentioned book, I found the product form in case of $x^2 + ny^2$. It should be sufficient at least for now. $\endgroup$
    – Linh
    Commented Jun 1, 2015 at 23:35

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