# Computing representation number quad forms

• binbagsss
binbagsss

## Homework Statement

## r_{A} (n) = ## number of solutions of ## { \vec{x} \in Z^{m} ; A[\vec{x}] =n} ##
where ##A[x]= x^t A x ##, is the associated quadratic from to the matrix ##A##, where here ##A## is positive definite, of rank ##m## and even. (and I think symmetric?)

I am solving for the ##r_{A}(1) ## for the two quadratic forms:

##Q(x,y,uv)= 2(x^2+y^2+u^2+v^2)+2xu+xv+yu-2yv##
##R(x,y,uv)=x^2+4(y^2+u^2+v^2)+xu+4yu+3yv+7uv##

see above

## The Attempt at a Solution

[/B]

##Q=2(x+u/2+v/4)^2+2(y-v/2+u/4)^2+11u^2/8 + 11v^2/8 ##
##R=(x+1/2u)^2+4(y+1/2u+3/8v)^2+11/4(u+v)^2+11/16v^2 ##

Solving ##Q=1## with all ##x,y,u,v## integer, it is clear that ##u,v=0## is needed, and then ##x,y=\pm 1 ## gives ##r_{Q}(1)=4##.

Now looking at ## r_{R}(1) ## by the same reasoning as above I would have said that we require ##v=0## , and then I' m not sure what to do.

However the solution is:

Must have ##u+v=1 ## & ##|v|=0,\pm 1 ##, this gives ## \pm(1,0,0,0)##, ##\pm(1,0,-1,1) ## , ##\pm(0,0,-1,1)##

(the symbol that I interpreted as '&' in the solutions is a bit smudged, so looking at the solutions I'm not sure that this is supposed to be a 'or'? )

Either way, I'm really confused, unsure where these conditions come from, how to think about this in a logical way...

Mentor
2022 Award

## Homework Statement

## r_{A} (n) = ## number of solutions of ## { \vec{x} \in Z^{m} ; A[\vec{x}] =n} ##
where ##A[x]= x^t A x ##, is the associated quadratic from to the matrix ##A##, where here ##A## is positive definite, of rank ##m## and even. (and I think symmetric?)

I am solving for the ##r_{A}(1) ## for the two quadratic forms:

##Q(x,y,uv)= 2(x^2+y^2+u^2+v^2)+2xu+xv+yu-2yv##
##R(x,y,uv)=x^2+4(y^2+u^2+v^2)+xu+4yu+3yv+7uv##

see above

## The Attempt at a Solution

[/B]

##Q=2(x+u/2+v/4)^2+2(y-v/2+u/4)^2+11u^2/8 + 11v^2/8 ##
##R=(x+1/2u)^2+4(y+1/2u+3/8v)^2+11/4(u+v)^2+11/16v^2 ##

Solving ##Q=1## with all ##x,y,u,v## integer, it is clear that ##u,v=0## is needed, and then ##x,y=\pm 1 ## gives ##r_{Q}(1)=4##.

Now looking at ## r_{R}(1) ## by the same reasoning as above I would have said that we require ##v=0## , and then I' m not sure what to do.

However the solution is:

Must have ##u+v=1 ## & ##|v|=0,\pm 1 ##, this gives ## \pm(1,0,0,0)##, ##\pm(1,0,-1,1) ## , ##\pm(0,0,-1,1)##

(the symbol that I interpreted as '&' in the solutions is a bit smudged, so looking at the solutions I'm not sure that this is supposed to be a 'or'? )

Either way, I'm really confused, unsure where these conditions come from, how to think about this in a logical way...

Are you sure there are no typos? I calculated ##R(0,0,-1,1) = \frac{61}{64}##. It could also help to multiply the equations by ##8##, resp. ##16##, which would make the comparisons easier.

Homework Helper
Gold Member
2022 Award
Solving Q=1 with all x,y,u,v integer, it is clear that u,v=0 is needed
True, but not quite trivial.
and then x,y=±1
How do you get that? Don't those give Q=4?
by the same reasoning as above I would have said that we require v=0
Then you would be wrong. I said it wasn't quite trivial.
As fresh_42 writes, it will help to multiply through the equations to eliminate the fractions.

binbagsss
True, but not quite trivial.

How do you get that? Don't those give Q=4?

Then you would be wrong. I said it wasn't quite trivial.
.

oh right, the reason is that each term needs to be ##\leq 1 ## ?