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How can I solve this question with out find a,b,c,d

$$a+b+c+d=2$$

$$a^2+b^2+c^2+d^2=30$$

$$a^3+b^3+c^3+d^3=44$$

$$a^4+b^4+c^4+d^4=354$$

so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$

If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d

Is there any help?

Julian Kuelshammer
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mnsh
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    I'm not sure if this helps but a,b,c,d = 1,2,3,-4. Almost solves your 4x4 system. – Elements In Space May 26 '13 at 13:20
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    Look up Newton(-Girard) identities. Here's what you need to do: 1) solve for the elementary symmetric polynomials $abcd$, $ab+ac+\cdots+cd$, $abc+\cdots$ and $abcd$ in terms of the given power sums. 2) Then use those to solve for the fifth and sixth power sums. That should work (but I haven't checked). – Jyrki Lahtonen May 26 '13 at 13:54
  • By deconstructing the question, it's easy to see that ${a,b,c,d}={-1,2,-3,4}$. Beyond this, you can't know what $a,b,c,d$ are. Trying to solve the question by deductive algebra would be a horrible task. – John Bentin May 26 '13 at 14:05
  • I know the value of a,b,c,d but I think maybe there is analytical way to solve the problem – mnsh May 26 '13 at 14:08
  • Of course, the first elementary symmetric polynomial should have been $a+b+c+d$, i.e. $s_1$ in Mark Bennet's (+1) answer. Sorry about repeating the product :-/ – Jyrki Lahtonen May 26 '13 at 21:03

4 Answers4

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One way to approach such questions is to view $a, b, c, d$ as roots of a single quartic equation $x^4-s_1x^3+s_2x^2-s_3x+s_4=0$, when we have Vieta's relations $$s_1=a+b+c+d$$$$s_2=ab+ac+ad+bc+bd+cd=(a+b)(c+d)+ab+cd$$$$s_3=abc+abd+acd+bcd$$$$s_4=abcd$$

We then let $p_k=a^k+b^k+c^k+d^k$ with: $$p_0=4$$$$p_1-s_1=0$$$$p_2-p_1s_1+2s_2=0$$$$p_3-p_2s_1+p_1s_2-3s_3=0$$

With the given values of $p_1, p_2, p_3$ these determine $s_1, s_2, s_3$.

For higher powers $(k\ge4)$ we have $$a^k-s_1a^{k-1}+s_2a^{k-2}-s_3a^{k-3}+s_4a^{k-4}=0$$ using the quartic equation. Adding together the corresponding equations for $a , b, c, d$ we obtain the recurrence for $p_k$$$p_k-s_1p_{k-1}+s_2p_{k-2}-s_3p_{k-3}+s_4p_{k-4}=0$$

Since we know $p_4$ we can use the recurrence to find $s_4$, and this is then sufficient to compute $p_5, p_6$ from the recurrence.

If this looks complicated, this is misleading. The relations between the elementary symmetric functions $s_r$ and the sums of powers $p_r$ are useful to know, and render this kind of problem essentially mechanical. You can see that you can solve the problem without solving explicitly for $a, b, c, d$.

Following this scheme we find successively $s_1=2, s_2=-13, s_3=-14, s_4=24$. Then the recurrence gives:$$p_5-2p_4-13p_3+14p_2+24p_1=0$$ and$$p_6-2p_5-13p_4+14p_3+24p_2=0$$

And this gives $p_5=812, p_6=4890$

Mark Bennet
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  • The answer is $\frac {406}{2445}.$ – user64494 May 26 '13 at 16:14
  • Nice. I thought constructing polynomials wasn't a way to go.(+1) – Inceptio May 26 '13 at 16:42
  • thanks alot Mark but can you explain more how to find $p_5$ and $p_6$ ? – mnsh May 26 '13 at 19:05
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    @hmedan.mnsh - I have added something a little more explicit – Mark Bennet May 26 '13 at 19:53
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    @hmedan.mnsh It was first in a rather old-fashioned algebra text book I was given to read by a teacher at school, which had a section on roots of equations and another on recurrence relations - it was one of Ferrar's books on Algebra - can't recall which. You can find information at http://en.wikipedia.org/wiki/Newton%27s_identities and http://en.wikipedia.org/wiki/Vieta's_formulas. – Mark Bennet May 26 '13 at 20:56
  • @MarkBennet Are there other questions that use this approach? – user103828 Aug 17 '14 at 09:00
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Here's an approach which involves AM-GM inequality(To bound the numbers).

I have considered the positive values of $(a,b,c,d)$. I can consider $a^2,b^2,c^2$ and $d^2$, since they are all strictly positive.

$\dfrac{a^2+b^2+c^2+d^2}{4} \ge \sqrt{abcd}$

$\dfrac{30}{4} \ge \sqrt{abcd} \implies 56.25 \ge abcd$

Since $a^2+b^2+c^2+d^2=30$ and $56 >|abcd|$, One of the $|a|,|b|,|c|,|d|$ is at most $5$.

Considering $a^4+b^4+c^4+d^4$, one of the $|a|,|b|,|c|,|d|$ is atmost $4$, since $5^4=625$ .

Now we have $a^2+b^2+c^2+d^2=30$, the value of $(|a|,|b|,|c|,|d|)$ is from set $(1,2,3,4)$.

And also:

Considering $4th$ degree polynomial such that $a,b,c,d$ are the roots of the equation.

$x^4+px^3+qx^2+rx+s=0$

$a+b+c+d=2=-p$

$(a+b+c+d)^2-a^2+b^2+c^2+d^2=2(ab+bc+cd+ad+ac+bd)=q=-13$

Finding $\sum abc$ and $abcd$ with the help of other inequalities and finding the roots of equation is another way to go. (Not quite sure whether constructing polynomial is a great way to go. )

Edit: I have considered the absolute values of $(a,b,c,d)$ . I didn't assume them to be $+ve$.

Inceptio
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  • -1 You cannot assume that the variables are positive, or even real for that matter. – Calvin Lin May 27 '13 at 19:05
  • Yes, but I rather found their absolute value. It is quite clear from the absolute value notation. $\max (|a|,|b|,|c|,|d|)=4$. – Inceptio May 28 '13 at 03:45
  • You may not assume that they are real. Note that $\frac{i^2 + i^2 + i^2 + i^2 } { 4} \not \geq \sqrt{i^4} = \sqrt{1} = 1$. – Calvin Lin May 28 '13 at 13:44
  • Why not think it as a Diophantine in 4 variables and solve? If you say so, alright. I did make note of constructing polynomial. This is typical Newton identities question. – Inceptio May 28 '13 at 13:46
  • A diophantine, by definition, is restricted to integer solutions. Why do you make this unnecessary and unwarranted assumption? How do you know that there are not more sets of solutions? – Calvin Lin May 28 '13 at 13:47
  • @CalvinLin: I did assume Diophantine and solved as I was interested in integer solutions only. – Inceptio May 28 '13 at 13:49
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    That's fine. As I said, you made an unwarranted assumption. With that assumption, your solution is correct. However, it doesn't tell you much more. You might as well have said "I wanted to find solutions over the positive reals, and by AM-GM, there are none". – Calvin Lin May 28 '13 at 13:56
  • Yes. And I knew, just assuming the Diophantine wouldn't give me the whole lot of solutions. Thanks for reviewing. – Inceptio May 28 '13 at 14:02
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I do not know how to solve it, but $a=-3,b=4,c=2,d=-1$

Rosa
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    He doesn't want to find the value of $(a,b,c,d)$. Question says it all. – Inceptio May 26 '13 at 13:25
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    @Hakhak: It would also help to explain how you cam to this answer. Was there a method ot guessing or trial and error? – Amzoti May 26 '13 at 13:49
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If you inspect the system you can see that the solutions for a,b,c,d are all going to have to be small integers. This constrains the system heavily. I managed to solve the system guessing some small numbers, testing them and tweaking minus signs to get the desired result. After we have found a,b,c,d the fraction becomes trivial.

Any analytical method to solving this system would have to accommodate high order solutions and would inevitably be very complicated. I would like to see analytical solution but doubt any could be as efficient as trial and error in this case.

Elements In Space
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  • it was obvouis that the answer is integer and when the power is odd,the sum increases by low er speed, so we should negative answer, so find the answer by guessing is not difficult – Rosa May 26 '13 at 14:00
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    How can you see that the variables are integers? – Anna May 26 '13 at 14:39