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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.

Bart Michels
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arman sharififar
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    Choose $m_1,m_2<0$ and $m_3,m_4>0$... – sranthrop Sep 11 '14 at 09:08
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    Can $a_1=a_2=a_3=a_4=b_1=b_2=b_3=b_4=0$? – Peter Huxford Sep 11 '14 at 09:08
  • i should modify my question. – arman sharififar Sep 11 '14 at 09:09
  • 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. – Peter Huxford Sep 11 '14 at 09:13
  • The formula can be written, but it is bulky to be. Even for not a great equation seems cumbersome. For example, this: http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 – individ Sep 11 '14 at 09:14
  • yes some of the $a_i , b_i$ can be zero.but they can not be all zero in the same time. – arman sharififar Sep 11 '14 at 09:14
  • @sranthrop: Won't work if $a_{1,2,3,4}=b_{1,2,3,4}=0$. – barak manos Sep 11 '14 at 09:15
  • @barakmanos: thread has been edited, so the a's and b's cannot be 0 simultaneously. – sranthrop Sep 11 '14 at 09:19
  • @sranthrop: OP just commented that he/she can't choose $m_{1,2,3,4}$ as functions of $a_{1,2,3,4}$ and $b_{1,2,3,4}$... So I'm gonna remove my answer, and you can add yours, which seems to be correct after all these changes in the definition of the problem at hand. – barak manos Sep 11 '14 at 09:27
  • As @MichaelStocker said, you can choose $m_1, m_2 = -1$, $m_3, m_4 = 1$. I have deleted my answer and hope micheal will write an answer himself! :) – Ant Sep 11 '14 at 10:46
  • @Ant all $m_i$'s must be positive – arman sharififar Sep 11 '14 at 13:58
  • @arman: I'm a bit curious. Why did you want to find such $m_i$:s? – Jyrki Lahtonen Sep 12 '14 at 05:03
  • @JyrkiLahtonen because i'm working on Space-Time coding in wireless communication.i have an idea that work better than the other codes which known recently.so finding $m_i$'s can help me to select the best codes. – arman sharififar Sep 13 '14 at 06:31
  • @arman: That would have been my first guess :-) The number theory related to Space-Time coding is fascinating and well studied (two out of my three PhD students wrote their dissertations on such constructions). I don't think you can find this type of quadratic forms with eight variables. The Golden code (and our rival codes) work, because the quadratic forms giving the determinant of matrices in rank 8 lattices have complex values. – Jyrki Lahtonen Sep 13 '14 at 07:12
  • @JyrkiLahtonen i know that you are one of the best person in ST codes on mimo system. :D i read the papers that the authors are you and Hollanti and Vehkalahti – arman sharififar Sep 13 '14 at 07:17
  • @JyrkiLahtonen Thank you very much for your answer and guidance.I know what you say.I just want to try to construct such codes and by this i get more information about it. – arman sharififar Sep 13 '14 at 07:38

3 Answers3

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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.

Jyrki Lahtonen
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  • 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 $Q=0$. – Jyrki Lahtonen 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. – arman sharififar Sep 13 '14 at 06:40
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    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. http://store.doverpublications.com/0486466701.html – Will Jagy Mar 08 '15 at 19:35
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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.

Community
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individ
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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 :)

Ant
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  • Tha idea seems correct. It suffices to take $m_1=1$ and $m_3=3$ for example. It becomes really interesting when we require $m_k>0$... – Bart Michels 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 – arman sharififar Sep 13 '14 at 07:21