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Fix $n\geq 2$. Let $p:=x_1^2+\ldots+x_{n-1}^2+1\in\mathbb{Q}[x_1,\ldots,x_{n-1}]$. Suppose $u_1,\ldots,u_n,v_1,\ldots,v_n\in\mathbb{Q}[x_1,\ldots,x_{n-1}]$ satisfy the following equations:



Prove that $u_i=0=v_i$ for all $1\leq i\leq n$.

It would even be helpful to see a reduction from the assumptions to $ap=r_1^2+\ldots+r_{n-1}^2$ for some polynomials $a,r_1,\ldots,r_{n-1}$ of the "right" form.


Some comments:

  1. $p$ is a prime element of $\mathbb{Q}[x_1,\ldots,x_{n-1}]$.
  2. The case $n=2$ is pretty easy, so maybe it sets up an inductive process. But even trying the case $n=3$, I couldn't see the reduction.

$n=2$ case. The equations are $u_1v_1+u_2v_2=0$ and $(u_1^2+u_2^2)p=v_1^2+v_2^2$.

First, suppose $u_2=0$. Then $u_1v_1=0$ and $u_1^2p=v_1^2+v_2^2$. If $u_1=0$ then we have $v_1^2+v_2^2=0$, which implies $v_1=0=v_2$ since these are polynomials over the rationals. If $v_1=0$ then $u_1^2p=v_2^2$. Then $u_1=0$ iff $v_2=0$, and in this case the result follows.

Otherwise, assume $u_1$ and $v_2$ are nonzero. Then $u_1^2p=v_2^2$ implies $v_2^2$ is in $I=(p)$, the ideal generated by $p$. But $p$ is prime so $v_2\in I$, so $v_2=p^da$ for some $a$ not divisible by $p$. So $u_1^2p=p^{2d}a^2$. So $u_1^2=p^{2d-1}a^2$. Again, this implies $u_1=p^kb$, with $2k>2d-1$. But then $p^{2k-2d+1}b^2=a^2$, contradicting that $p$ does not divide $a$.

So we can assume $u_2\neq 0$. Then $v_2=-\frac{u_1v_1}{u_2}$,

so $(u_1^2+u_2^2)p=v_1^2+\frac{u_1^2v_1^2}{u_2^2}$,

so $(u_1^2+u_2^2)u_2^2p=u_2^2v_1^2+u_1^2v_1^2$,

so $(u_1^2+u_2^2)u_2^2p=(u_1^2+u_2^2)v_1^2$.

Again, if $u_1^2+u_2^2=0$ then we get the result as above. Otherwise, $u_2^2p=v_1^2$, and we get the result as above.

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Writing $$p(u_1^2+...+u_n^2)$$ suggests that $p$ is a function of one variable. That doesn't seem like it fits with your definition of $p$ (unless of course $n=2$). Can you clarify? – Cameron Buie Oct 21 '12 at 17:26
It's just multiplication of polynomials, I hope the edit makes it more clear. – gamel Oct 21 '12 at 17:31
Ah! That does make more sense. – Cameron Buie Oct 21 '12 at 17:36
What did you do for the $n=2$ case? – Alan Guo Oct 22 '12 at 4:33
Sorry everyone. This turns out to be false, beginning with $n=3$. Counterexample for $n=3$: $$ u_1=0,~~~ u_2=1,~~~ u_3=x_2,~~~ v_1=1+x_2^2,~~~ v_2=-x_1x_2,~~~ v_3=x_1 $$ I have reason to believe that the problem may still be true when $n$ is a power of 2, but I don't know how to prove it. Should I delete the question? – gamel Nov 5 '12 at 0:10

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