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This book, which needs to be returned quite soon, has a problem I don't know where to start. How do I find a 4 parameter solution to the equation

$x^2+axy+by^2=u^2+auv+bv^2$

The title of the section this problem comes from is entitled (as this question is titled) "Numbers of the Form $x^2+axy+by^2$", yet it deals almost exclusively with numbers of the form $x^2+y^2$. It looks like almost an afterthought or a preview of what's to come where it gives the formula

$(m^2+amn+bn^2)(p^2+apq+bq^2)=r^2+ars+bs^2,r=mp-bnq,s=np+mq+anq$

Then 6 of the 7 problems use this form. The first few involve solving the form $z^k=x^2+axy+by^2$, which I quickly figured out are solved by letting $z=u^2+auv+bv^2$, then using the above formula to get higher powers. So for $z^2$ for example, I set $m=p=u$ and $n=q=v$ to get $x$ and $y$ in terms of $u$ and $v$. But for this problem, I'm drawing a blank.

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You should specify which of the letters $a,b,u,v,x,y$ are constants and which are variables. –  Greg Martin Apr 9 '12 at 5:23
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And specify what book and what pages. Your expressions for $r,s$ are the simplest description of the fact that the Gauss composition of the principal form with itself is again itself. Meanwhile, the formula with $r,s$ is a four parameter solution. –  Will Jagy Apr 9 '12 at 5:41
    
The book is one anyone is unlikely to see. Diophantine Analysis by Robert D. Carmichael. It is not mentioned anywhere which are constants and which are variables, but I have assumed a and b to be constant. –  Mike Apr 9 '12 at 7:06
    
Forgot the page number. Page 25. –  Mike Apr 9 '12 at 7:39
    
@WillJagy While it is a 4 parameter solution, it is not a 4 parameter solution for the equation in the problem I'm asked to solve. –  Mike Apr 10 '12 at 5:14

3 Answers 3

Oh, well. This is from page 57 of Binary Quadratic Forms by Duncan A. Buell. As long as $$ \gcd(a_1, a_2, B) = 1 $$ we have $$ (a_1 x_1^2 + B x_1 y_1 + a_2 C y_1^2) (a_2 x_2^2 + B x_2 y_2 + a_1 C y_2^2) = a_1 a_2 X^2 + B X Y + C Y^2, $$ where we take $$ X = x_1 x_2 - C y_1 y_1, \; \; \; \; Y = a_1 x_1 y_2 + a_2 x_2 y_1 + B y_1 y_2. $$ This is Dirichlet's "united forms" recipe for composition, with $$ \langle a_1,B, a_2 C \rangle \; \langle a_2,B, a_1 C \rangle \; = \; \langle a_1 a_2,B, C \rangle $$ in the form class group. At least, it is a group when $$ B^2 - 4 a_1 a_2 C $$ is not a square.

So a four parameter identity with parameters $x_1, y_1, x_2, y_2$ would be $$ (a_1 x_1^2 + B x_1 y_1 + a_2 C y_1^2) (a_2 x_2^2 + B x_2 y_2 + a_1 C y_2^2) = a_1 a_2 (x_1 x_2 - C y_1 y_1)^2 + B(x_1 x_2 - C y_1 y_1)(a_1 x_1 y_2 + a_2 x_2 y_1 + B y_1 y_2) + C (a_1 x_1 y_2 + a_2 x_2 y_1 + B y_1 y_2)^2. $$

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I think you did the wrong equation. That last formula was something I likely needed to use for the problem, although the section did not work with it at all. The problem was $x^2+axy+by^2=u^2+auv+bv^2$ –  Mike Apr 9 '12 at 7:38
    
@Mike, take $a_1 = a_2 = 1, B = a, C = b.$ –  Will Jagy Apr 9 '12 at 19:51
    
If I do that, won't I just get the 4 parameter solution of $(m^2+amn+bn^2)(p^2+apq+bq^2)=r^2+ars+bs^2$ from the book? –  Mike Apr 10 '12 at 5:10
    
@Mike, I do not really know the intent in Carmichael's book, which dates to 1914. I recommend you get the Buell book. The point of this is that there is an operation that combines two forms and gives a third. It goes back to Gauss and is often called Gauss composition. –  Will Jagy Apr 10 '12 at 5:19
    
Thanks for the recommendation. It looks like it's available at a less than local library. I'll have to see if I can get my hands on it somehow. By the way, if you haven't seen it, inspiration struck and I think I have an answer if you wouldn't mind taking a look at it. –  Mike Apr 10 '12 at 6:11

Over a field the space of rational solutions is three dimensional. Integer solutions can be formed as multiples of rational solutions and maybe this multiplication factor is the fourth parameter but it is not clear what the problem is asking. The number of parameters can always be increased from a known parametrization by having some of the parameters be arbitrary functions of several new parameters but this is not natural.

The problem asks for arbitrary pairs of points on a conic of the form $Q(x,y)=R$ where $Q(x,y)=x^2 + axy + by^2$. It is not specified but probably intended that the form $Q$ is fixed and $R$ is variable.

For any 3 numbers $(u,v,t)$ with $u$ and $v$ not both zero, the line through $(u,v)$ of slope $t$ intersects the conic $Q(x,y)=R$ with $R=Q(u,v)$ at a second point whose coordinates $(x,y)$ are rational functions of $(u,v,t)$. This is a 3-dimensional family of rational solutions and this is the best one can do if parameters means dimensionality of the family.

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It is rather odd that most of the problems in this section does not specify integer or rational, though I have been assuming integer. As for any sort of form, I assume it wants $u=f_1(m,n,p,q),v=f_2(m,n,p,q),x=f_3(m,n,p,q)$, and $y=f_4(m,n,p,q)$, a and b constants. So the problem asking for the 2 parameter solution $z^2=x^2+axy+by^2$ I would answer with $z=u^2+auv+bv^2,x=u^2-bv^2,y=2uv+av^2$ –  Mike Apr 9 '12 at 7:29

Okay, let's see if I can fix my previous answer. Again if i let $z=u_1^2+au_1v_1+bv_1^2$, we get

$z^2=(u_1^2+au_1v_1+bv_1^2)(u_1^2+au_1v_1+bv_1^2)=r^2+ars+bs^2$

where $r=u_1^2-bv_1^2,s=2u_1v_1+av_1^2$.

I will now multiply this by another number of the form $m^2+amn+bn^2$ to get yet another number of the same form. So $p=u_1^2-bv_1^2, q=2u_1v_1+av_1^2$. My new values for r and s are

$r=mp-bnq=m(u_1^2-bv_1^2)-bn(2u_1v_1+av_1^2)$

$s=np+mq+anq=n(u_1^2-bv_1^2)+m(2u_1v_1+av_1^2)+an(2u_1v_1+av_1^2)$

I'll let this $(r,s)$ be my $(x,y)$. So now I have

$x^2+axy+by^2=z^2(m^2+amn+bn^2)=(mz)^2+a(mz)(nz)+b(nz)^2$.

Now I finally have an equation in the form that I want. This gives my solution as

$x=m(u_1^2-bv_1^2)-bn(2u_1v_1+av_1^2)$

$y=n(u_1^2-bv_1^2)+(m+an)(2u_1v_1+av_1^2)$

$u=mz=mu_1^2+amu_1v_1+bmv_1^2$

$v=nz=nu_1^2+anu_1v_1+bnv_1^2$

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