<s>[INST] 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.
 [/INST] 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.
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