Computing representation number quad forms

[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:

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

Homework Equations

see above

The Attempt at a Solution

[/B]
diagonalized these read:

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

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...

Many thanks for your help in advance.

 

[fresh_42]

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:

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

Homework Equations

see above

The Attempt at a Solution

[/B]
diagonalized these read:

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

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...

Many thanks for your help in advance.

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.

 

[haruspex]

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 $\leq 1$ ?

 

[haruspex]

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

Each term must be no more than 1, but I cannot say whether that is the "reason" you were wrong to conclude v=0 since I do not know how you concluded it. All I can say is that there is a solution with v not 0.