Consider the equation: $(x-2)^6 + (x-4)^6 = 64$ This equation has two real roots $a, b$ and two pairs of complex conjugate roots $(p, q)$ and $(r, s)$. Is $p + q = r + s$ or $pq=rs$?

The trivial real roots of the equation are $(2,4)$. My question is, when there are two complex conjugate pairs of a polynomial, do the magnitude of the roots have any sort of relation? Is trivially trying to find them by substituting $x=a+ib$ (with $a=1$ in this case so that there remain no imaginary part on the LHS) the only method? This approach leads to an equation of the sixth degree in $b$.


Let's substitute $y = x-3$. Then, the equation becomes:

$$(y+1)^6+(y-1)^6 = 64$$

Expanding the left side using the Binomial Theorem and doing a bit of simplification gives us:

$$(y^6+6y^5+15y^4+20y^3+15y^2+6y+1)+(y^6-6y^5+15y^4-20y^3+15y^2-6y+1) = 64$$

$$2(y^6+15y^4+15y^2+1) = 64$$

$$y^6+15y^4+15y^2-31 = 0$$

We know that $x = 2$ and $x = 4$ are roots of the original equation, so $y = \pm 1$ are roots of the new equation. Hence, $y^2-1$ is a factor of $y^6+15y^4+15y^2-31$. Factoring this out gives us:

$$(y^2-1)(y^4+16y^2+31) = 0$$

So the complex roots are the roots of $y^4+16y^2+31$. By completing the square, we have:

$$y^4+16y^2+31 = 0$$

$$y^4+16y^2+64 = 33$$

$$(y^2+8)^2 = 33$$

$$y^2+8 = \pm\sqrt{33}$$

$$y^2 = -8\pm\sqrt{33}$$

$$y = \pm i \sqrt{8\pm \sqrt{33}}$$

Therefore, the complex roots to the original equation are $x = 3 \pm i \sqrt{8\pm \sqrt{33}}$.

By setting $p = 3+i\sqrt{8+\sqrt{33}}$, $q = 3-i\sqrt{8+\sqrt{33}}$, $r = 3+i\sqrt{8-\sqrt{33}}$, $s = 3-i\sqrt{8-\sqrt{33}}$, we see that $p+q = 6 = r+s$ but $pq = 17+\sqrt{33} \neq 17-\sqrt{33} = rs$. Therefore, $p+q=r+s$ is true while $pq=rs$ is false.

  • $\begingroup$ If I only have paper and pen in the exam, how do I find the roots? $\endgroup$ – Sat D Dec 25 '15 at 4:35
  • $\begingroup$ ^Oh, are you just trying to find the roots of the polynomial? Your question sounded like you wanted to know if there were any special relationships between the roots. $\endgroup$ – JimmyK4542 Dec 25 '15 at 4:36
  • $\begingroup$ I believed that there existed a special relationship WHICH would give me the roots. The main question is finding which is correct, P+Q = R + S or PQ = RS $\endgroup$ – Sat D Dec 25 '15 at 4:38
  • $\begingroup$ Ahh, so the problem asked you which of the two relationships were true for that specific polynomial, and you wanted to know how to figure that out. The second part of your question made it seem like you wanted to know if either relationship held for an all polynomials with 2 pairs of complex conjugate roots. I'll edit my answer so that it answers your actual question. $\endgroup$ – JimmyK4542 Dec 25 '15 at 4:55

Remark: if you want $p+q=r+s$ and $pq=rs$ to hold simultaneously, then you cannot have 2 distinct complex conjugate pairs. This is because if

$$p = a+bi, \ q=a-bi$$ and $$r = c+di, \ s=c-di,$$ then $$p+q=r+s \implies a=c,$$ and $$pq=rs \implies a^2+b^2=c^2+d^2,$$ and so $b=\pm d$.

  • $\begingroup$ So which one of the conditions is applicable for the equation, and how? $\endgroup$ – Sat D Dec 25 '15 at 4:32
  • $\begingroup$ @SatD Is the original question that you posted the whole question, or is it part of something else? When we use "and" in math we usually want both conditions to hold, and in this case they can't, unless the complex roots have multiplicity 2. $\endgroup$ – GaussTheBauss Dec 25 '15 at 4:39
  • $\begingroup$ I apologize for the 'and'. It's an 'or'. $\endgroup$ – Sat D Dec 25 '15 at 4:41
  • $\begingroup$ @SatD I think you changed the wrong "and" to an "or". You should change the one between the equations, not the pair of points. $\endgroup$ – GaussTheBauss Dec 25 '15 at 4:42

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