10

Problem: To find all real solutions of the system: $$3a=(b+c+d)^3$$ $$3b=(c+d+e)^3$$ $$3c=(d+e+a)^3$$ $$3d=(e+a+b)^3$$ $$3e=(a+b+c)^3$$

My attempt: I tried to get a bound for positive solutions. Using AM$\geq$GM, $(x+y+z)^3\geq27xyz$, I get $$ abcde\leq \sqrt{\frac{1}{9^5}}=\frac{1}{3^5}$$ Also, by Jensen's inequality $\frac{x^3+y^3+z^3+u^3+v^3}{5}\geq(\frac{x+y+z+u+v}{5})^3$ I get $$a+b+c+d+e\leq5/3$$ Other than this, I cannot find any more information regarding the roots. This problem is not supposed to be a very tough one, so probably I am missing something here...

Potemkin Metro Card
  • 1,598
  • 1
    Note that $\sqrt{\frac{1}{9^5}}=\frac{1}{3^5}=3^{-5}$. – Servaes Jun 02 '16 at 17:26
  • @Servaes You are right but had it been $abcde\geq3^{-5}$ then we could've concluded AM=GM.. that too for positive solutions only... – Potemkin Metro Card Jun 02 '16 at 17:30
  • 1
    Trivially you have $a=b=c=d=e={0,\frac 13}$. Hard to get a five-dimensional vision going of other solutions. – Joffan Jun 02 '16 at 17:33
  • 1
    @Joffan: And $-\frac13$. – joriki Jun 02 '16 at 17:56
  • @Joffan I think the system is symmetric thus $a=b=c=d=e$ – Behrouz Maleki Jun 02 '16 at 21:03
  • @BehrouzMaleki - which would be a significant result, if provable, reducing to $a=0$ or $9a^2=1$, i.e. the three results above. – Joffan Jun 02 '16 at 22:05
  • 1
    @BehrouzMaleki The set of solutions is invariant under cyclic shifts $a\ \rightarrow\ b\ \rightarrow\ c\ \rightarrow\ d\ \rightarrow\ e\ \rightarrow\ a$ because the equations are, but this does not mean that each solution is invariant under cyclic shifts, i.e. that $a=b=c=d=e$. – Servaes Jun 03 '16 at 05:18

1 Answers1

2

A real solution must satisfy $a=b=c=d=e$. To see this, let an arbitrary solution $(a,b,c,d,e)$ be given, and replacing it with a cyclic shift, we can assume without loss of generality that $a=\max\{a,b,c,d,e\}$. Then $a\geq e$, so that $\sqrt[3]{3a} \geq \sqrt[3]{3e}$. Using the equations $3a=(b+c+d)^3$ and $3e=(a+b+c)^3$, this implies that $b+c+d \geq a+b+c$, i.e., $d \geq a$. But since $a=\max\{a,b,c,d,e\} \geq d$, this implies $d=a$. Iterating this same chain of reasoning with $d$ in place of $a$, we obtain, in turn, $b=d$, $e=b$, and $c=e$, so that indeed $a=b=c=d=e$.

It follows that the only real solutions satisfy $3a=(b+c+d)^3=(3a)^3$, so that $9a^3=a$, which has only solutions $a\in\{0,\pm \frac13\}$. Therefore, the real solutions are $(0,0,0,0,0)$, $(\frac13,\frac13,\frac13,\frac13,\frac13)$, and $(-\frac13,-\frac13,-\frac13,-\frac13,-\frac13)$.

Brent Kerby
  • 5,539