Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$

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?

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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 $p_5=812, p_6=4890$

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 $+ve$.

I do not know how to solve it, but $a=-3,b=4,c=2,d=-1$

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.