Find $m_1 , m_2,m_3,m_4\in\mathbb{Q}$ s.t. $\forall a_k,b_k\in\mathbb Z,\,m_1(a_1^2+a_2^2)+m_2(a_3^2+a_4^2)\neq m_3(b_1^2+b_2^2)+m_4(b_3^2+b_4^2)$ Let us assume that $a_1 , a_2 , a_3 ,a_4,b_1,b_2,b_3,b_4\in\mathbb{Z}$.
If $m_1 , m_2,m_3,m_4\in\mathbb{Q}$, then how can I choose $m_1,m_2,m_3,m_4$, such that the following equation is $never$ satisfied? (all $a_i$'s and $b_i$'s can not be all zero at the same time) $$m_1(a_1^2+a_2^2)+m_2(a_3^2+a_4^2)=m_3(b_1^2+b_2^2)+m_4(b_3^2+b_4^2)$$
Note that the $m_i$'s are all $positive$ numbers and can not be varying with $a_i$'s and $b_i$. Thank you.
 A: This is impossible. 
Consider the quadratic form
$$
Q(a_1,\ldots,a_4,b_1,\ldots,b_4):=m_1(a_1^2+a_2^2)+m_2(a_3^2+a_4^2)-m_3(b_1^2+b_2^2)-m_4(b_3^2+b_4^2).
$$
The Hasse-Minkowski theorem states that $Q$ takes the value zero non-trivially with $(a_1,\ldots,b_4)\in\Bbb{Q})$ if and only if it takes the value zero non-trivially with the parameters ranging over A) the reals, and B) over the $p$-adics $\Bbb{Q}_p$ for all primes $p$.
With all the coefficients $m_i$ non-zero the answer is affirmative for all the $p$-adics. The number of variables is the key, Borevich-Shafarevich state that five is enough irrespective of how cleverly you choose the coefficients $m_1,m_2,m_3,m_4$. I haven't checked the details, but it is easy to believe that expanding the techniques outlined here (congruences, quadratic residues, Hensel lifts and such) lead to such a result.
With all the coefficients $m_i$ positive, the form $Q$ trivially represents zero non-trivially over the reals. Therefore Hasse-Minkowski implies that
$Q(a_1,\ldots,a_4,b_1,\ldots,b_4)=0$ for some rational numbers $a_1,\ldots,b_4$. Of course, we can then clear the denominators by multiplying the variables with the least common multiple of the denominators and make them integers.
A: As I said, for 8 unknown parameters and the formula goes bulky.
$$a(z_1^2+z_2^2)+b(z_3^2+z_4^2)=c(z_5^2+z_6^2)+j(z_7^2+z_8^2)$$
3 - the formula looks like this: Solutions to $ax^2 + by^2 = cz^2$
Will consider here the special case when:  $a+b=c+j$ $(1)$
Then the solutions are of the form:
$$z_1=js^2+jt^2+ck^2+cp^2-bq^2-bx^2-ay^2$$
$$z_2=js^2+jt^2+ck^2+cp^2-bq^2-bx^2+ay^2+2(bx+bq-js-jt-ck-cp)y$$
$$z_3=js^2+jt^2+ck^2+cp^2-bq^2+bx^2-ay^2+2(ay+bq-js-jt-ck-cp)x$$
$$z_4=js^2+jt^2+ck^2+cp^2+bq^2-bx^2-ay^2+2(ay+bx-js-jt-ck-cp)q$$
$$z_5=js^2+jt^2+ck^2-cp^2-bq^2-bx^2-ay^2+2(ay+bx+bq-js-jt-ck)p$$
$$z_6=js^2+jt^2-ck^2+cp^2-bq^2-bx^2-ay^2+2(ay+bx+bq-js-jt-cp)k$$
$$z_7=js^2-jt^2+ck^2+cp^2-bq^2-bx^2-ay^2+2(ay+bx+bq-js-ck-cp)t$$
$$z_8=jt^2-js^2+ck^2+cp^2-bq^2-bx^2-ay^2+2(ay+bx+bq-jt-ck-cp)s$$
$s,t,k,p,q,x,y$ - integers asked us.
It is clear that if you will satisfy the condition $(1)$ we can always write such a simple solution. It is easy enough to see how it turns out.
A: This is gonna lack some formality (and to be honest I'm not entirely sure it is correct), but the idea is the following:
set $m_2 = m_3 = 0$, and write $$m_1(a_1^2 + a_2^2) = m_4(b_3^2 + b_4^2) \implies \frac pq = \frac{b_3^2 + b_4^2}{a_1^2 + a_2^2} \tag{$\star$}$$ where $\displaystyle \frac pq = \frac{m_1}{m_4}$
The idea is to choose primes that cannot be expressed as sum of squares; for example, let's take $p=3$, $q=7$.
Now the only possibility for $(\star)$ to hold is $\begin{cases} b_3^2 + b_4^2 = 3k, \\ a_1^2 + a_2^2 = 7k \end{cases}$
It is easy to see that if $3 \mid b_3^2 + b_4^2$, then $3 \mid b_3$, $3 \mid b_4$. Same thing for $a_i$. This yields 
$$\begin{cases} 3\beta_3^2 + 3\beta_4^2 = k \\ 7 \alpha_1^2 + 7\alpha_2^2 = k\end{cases}$$ ($b_i = 3\beta_i$, $a_i = 7\alpha_i$)
This can be simplified if we set in the first equation $k = 3k_1$, $k = 7k_2$ in the second to yield 
$$\begin{cases} \beta_3'^2 + \beta_4'^2 = k_1 \\  \alpha_1'^2 + \alpha_2'^2 = k_2\end{cases}$$
with the condition $3k_1 = 7k_2$. This again implies $k_1 = 7k_3$, $k_2 = 3k_4$ and subsituting 
$$\begin{cases} \beta_3'^2 + \beta_4'^2 = 7k_3 \\  \alpha_1'^2 + \alpha_2'^2 = 3k_4\end{cases}$$
with the condition $21k_3 = 21k_4 \implies k_3 = k_4$
But then this is equivalent to the first system we tried to solve! 
This is basically a proof by infinite descent (http://en.wikipedia.org/wiki/Proof_by_infinite_descent), though I admit I'm not particularly familiar with it, so I may be wrong.
Let me know what you think :)
