# 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.

• Choose $m_1,m_2<0$ and $m_3,m_4>0$... Sep 11 '14 at 9:08
• Can $a_1=a_2=a_3=a_4=b_1=b_2=b_3=b_4=0$? Sep 11 '14 at 9:08
• i should modify my question. Sep 11 '14 at 9:09
• Can some of the $a_i,b_i$ be zero, or must they all be positive as well? Because if they can all simultaneously be zero it is not possible. Sep 11 '14 at 9:13
• The formula can be written, but it is bulky to be. Even for not a great equation seems cumbersome. For example, this: math.stackexchange.com/questions/738446/… Sep 11 '14 at 9:14

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.

• So AFAICT if we have a sum like $Q:=\sum_{i=1}^5m_ia_i^2$ with all $m_i$ non-zero rationals such that $m_1>0$ and $m_5<0$, then we can always find integers $a_i,i=1,\ldots,5,$ not all zero, such that $Q=0$. Sep 12 '14 at 18:19
• thank you very much.your answer is perfect.but what about: $$\sum_1^4{m_ia_i^2}-\sum_4^8{m_ia_i^2}\neq0$$ like the previous question find $m_i$'s s.t the above nonequality is always hold. Sep 13 '14 at 6:40
• Jyrki, this guy has a new question today, not sure what he really wants. However, the reason for this comment is to recommend Cassels, Rational Quadratic Forms, for you, very good job in the first few chapters on isotropy over the p-adic fields. Page 60, Lemma 2.7, dimension five or more suffices for all finite p. store.doverpublications.com/0486466701.html Mar 8 '15 at 19:35

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.

• your answer give me more insight on how to solve such problem in a simple way.thank you very much. Sep 13 '14 at 7:09

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 :)

• Tha idea seems correct. It suffices to take $m_1=1$ and $m_3=3$ for example. It becomes really interesting when we require $m_k>0$... Sep 11 '14 at 19:46
• As i said in the question, all $m_i$'s are positive.but you set $m_1=m_2=0$.but your idea is an interesting idea for me.thank you very much Sep 13 '14 at 7:21