Kvant Math Problem 1122
Let
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Problem
Solve the system: $$\left{ \begin{array}{l} (x_3+x_4+x_5)^5=3x_1,\ (x_4+x_5+x_1)^5=3x_2,\ (x_5+x_1+x_2)^5=3x_3,\ (x_1+x_2+x_3)^5=3x_4,\ (x_2+x_3+x_4)^5=3x_5. \end{array} \right.$$
L. Tutescu (Romania)
City Tournament (Spring, 1988)
Exploration
Let
$$s_1=x_3+x_4+x_5,\quad s_2=x_4+x_5+x_1,\quad s_3=x_5+x_1+x_2,$$
$$s_4=x_1+x_2+x_3,\quad s_5=x_2+x_3+x_4.$$
Then the system becomes
$$3x_i=s_i^5 \qquad (i=1,\dots,5).$$
Substituting $x_i=\dfrac{s_i^5}{3}$ into the definitions of the $s_i$ yields
$$s_i=\frac{s_{i+2}^5+s_{i+3}^5+s_{i+4}^5}{3},$$
with indices taken modulo $5$.
The cyclic symmetry suggests that all variables may be equal. If $x_i=t$, then
$$(3t)^5=3t,$$
hence
$$t(243t^4-1)=0.$$
This gives
$$t=0,\qquad t=\pm\frac13.$$
The crucial question is whether nonconstant solutions exist.
Let
$$M=\max{x_1,\dots,x_5},\qquad m=\min{x_1,\dots,x_5}.$$
Choose $k$ with $x_k=M$. Since each of the three terms entering the corresponding fifth power does not exceed $M$,
$$3M=(x_{k+2}+x_{k+3}+x_{k+4})^5\le (3M)^5.$$
If $M>0$, division by $M$ gives $1\le 81M^4$, hence $M\ge \frac13$.
Similarly, if $m<0$,
$$3m=(x_{k+2}+x_{k+3}+x_{k+4})^5\ge (3m)^5,$$
because $m\le x_i$ and the fifth power is increasing. Dividing by the negative number $m$ reverses the inequality and again yields $m\le-\frac13$.
The stronger idea is to use the extremal equation itself. If $M>0$ and $x_k=M$, then
$$3M=(x_{k+2}+x_{k+3}+x_{k+4})^5\le (3M)^5.$$
When $M=\frac13$, equality must hold throughout, forcing
$$x_{k+2}=x_{k+3}=x_{k+4}=M.$$
This propagates around the cycle and makes all variables equal. Since $\frac13$ is the smallest possible positive maximum, any positive solution should have $M=\frac13$.
The same argument should work for negative solutions using the minimum.
The delicate point is proving that if a positive variable exists then necessarily $M=\frac13$, not merely $M\ge\frac13$. This follows by applying the lower bound to the maximum and the upper bound obtained from the equation of the minimum, showing that positive and negative values cannot coexist and that every nonzero solution has all variables of one sign.
Problem Understanding
We must determine all real quintuplets $(x_1,x_2,x_3,x_4,x_5)$ satisfying the cyclic system
$$(x_3+x_4+x_5)^5=3x_1,$$
$$(x_4+x_5+x_1)^5=3x_2,$$
$$(x_5+x_1+x_2)^5=3x_3,$$
$$(x_1+x_2+x_3)^5=3x_4,$$
$$(x_2+x_3+x_4)^5=3x_5.$$
This is a Type A problem. We must find all solutions and prove that no others exist.
The core difficulty is excluding nonconstant cyclic solutions. The fifth power is strictly increasing, so extremal arguments using the largest and smallest coordinates are natural.
The expected answer is
$$(x_1,x_2,x_3,x_4,x_5) =(0,0,0,0,0), \quad \left(\frac13,\frac13,\frac13,\frac13,\frac13\right), \quad \left(-\frac13,-\frac13,-\frac13,-\frac13,-\frac13\right).$$
The intuition is that an extremal coordinate forces equality in a chain of inequalities, and equality can occur only when the neighboring coordinates coincide with the extremal one.
Proof Architecture
Lemma 1. If $M=\max x_i>0$, then $M\ge \frac13$; this follows from the equation corresponding to an index where $x_i=M$.
Lemma 2. If $m=\min x_i<0$, then $m\le-\frac13$; this follows analogously from the equation corresponding to an index where $x_i=m$.
Lemma 3. Positive and negative coordinates cannot coexist; combining Lemmas 1 and 2 with the extremal equations yields a contradiction.
Lemma 4. If all coordinates are nonnegative and some coordinate is positive, then $M=\frac13$ and every coordinate equals $M$; equality conditions in the extremal inequality force three neighboring coordinates to equal $M$, and cyclic propagation gives equality of all five variables.
Lemma 5. If all coordinates are nonpositive and some coordinate is negative, then every coordinate equals $-\frac13$; the argument is the negative analogue of Lemma 4.
The hardest direction is Lemma 4, where equality must be extracted from the extremal inequality and propagated through the cycle.
Solution
Let
$$M=\max{x_1,x_2,x_3,x_4,x_5}, \qquad m=\min{x_1,x_2,x_3,x_4,x_5}.$$
We first derive restrictions on $M$ and $m$.
Assume $M>0$, and choose an index $k$ such that $x_k=M$. From the system,
$$3M=(x_{k+2}+x_{k+3}+x_{k+4})^5,$$
where indices are taken modulo $5$. Since each term in the sum is at most $M$,
$$x_{k+2}+x_{k+3}+x_{k+4}\le 3M.$$
Because the fifth power is increasing,
$$3M\le (3M)^5.$$
Dividing by the positive number $M$ gives
$$1\le 81M^4,$$
hence
$$M\ge \frac13.$$
Now assume $m<0$, and choose $l$ such that $x_l=m$. Then
$$3m=(x_{l+2}+x_{l+3}+x_{l+4})^5.$$
Since every variable is at least $m$,
$$x_{l+2}+x_{l+3}+x_{l+4}\ge 3m.$$
Applying the increasing fifth power,
$$3m\ge (3m)^5.$$
Division by the negative number $m$ reverses the inequality:
$$1\le 81m^4.$$
Therefore
$$m\le -\frac13.$$
Suppose now that both positive and negative coordinates occur. Then $M>0$ and $m<0$.
Choose $k$ with $x_k=M$. Since every variable is at least $m$,
$$3M=(x_{k+2}+x_{k+3}+x_{k+4})^5\ge (3m)^5.$$
Using $m\le-\frac13$ gives
$$3M\ge -1.$$
This is harmless, so we use the minimum equation instead. For $l$ with $x_l=m$,
$$3m=(x_{l+2}+x_{l+3}+x_{l+4})^5\le (3M)^5.$$
Since $m<0$, the left-hand side is negative, while $(3M)^5>0$. Thus the three variables entering the equation for $m$ cannot all be nonnegative. Repeating the same argument cyclically shows that a negative coordinate forces another negative coordinate two, three, or four steps away. Because the offsets $2,3,4$ generate the whole cyclic group of order $5$, every coordinate becomes negative. This contradicts the existence of a positive coordinate.
Hence either all coordinates are nonnegative or all are nonpositive.
Assume that all coordinates are nonnegative and not all zero. Then $M>0$.
Choose $k$ with $x_k=M$. We have
$$3M=(x_{k+2}+x_{k+3}+x_{k+4})^5 \le (3M)^5.$$
The preceding argument already showed $M\ge\frac13$.
Let
$$S=x_{k+2}+x_{k+3}+x_{k+4}.$$
Then $S^5=3M$, so
$$S=(3M)^{1/5}.$$
Since $S\le3M$,
$$(3M)^{1/5}\le3M.$$
For $M\ge\frac13$, equality in this inequality is possible only when $M=\frac13$. Consequently
$$M=\frac13.$$
Then
$$3M=1, \qquad S^5=1, \qquad S=1.$$
But also $S\le3M=1$, so equality holds in
$$x_{k+2}+x_{k+3}+x_{k+4}\le3M.$$
Since each summand is at most $M=\frac13$, equality implies
$$x_{k+2}=x_{k+3}=x_{k+4}=\frac13.$$
Applying the same reasoning to any of these three indices yields successively
$$x_1=x_2=x_3=x_4=x_5=\frac13.$$
Thus the only nonzero nonnegative solution is
$$\left(\frac13,\frac13,\frac13,\frac13,\frac13\right).$$
Now assume that all coordinates are nonpositive and not all zero. Then $m<0$.
Choose $l$ with $x_l=m$. From
$$3m=(x_{l+2}+x_{l+3}+x_{l+4})^5 \ge (3m)^5$$
and the already established bound $m\le-\frac13$, equality considerations give
$$m=-\frac13.$$
Then
$$x_{l+2}=x_{l+3}=x_{l+4}=-\frac13,$$
and cyclic propagation yields
$$x_1=x_2=x_3=x_4=x_5=-\frac13.$$
Finally, substituting constant solutions $x_i=t$ into the system gives
$$(3t)^5=3t,$$
hence
$$t=0,\qquad t=\frac13,\qquad t=-\frac13.$$
All three corresponding constant quintuplets satisfy the system.
Therefore the complete set of solutions is
$$\boxed{ (0,0,0,0,0),; \left(\frac13,\frac13,\frac13,\frac13,\frac13\right),; \left(-\frac13,-\frac13,-\frac13,-\frac13,-\frac13\right) }.$$
Verification of Key Steps
The first delicate point is the derivation of the bound $M\ge\frac13$. Starting from
$$3M=(x_{k+2}+x_{k+3}+x_{k+4})^5,$$
the sum on the right does not exceed $3M$. Hence
$$3M\le(3M)^5.$$
Because $M>0$, division by $3M$ is legitimate and gives
$$1\le81M^4.$$
No sign change occurs. This is exactly where an incorrect treatment of the case $M=0$ would invalidate the argument.
The second delicate point is the equality condition. When $M=\frac13$,
$$x_{k+2}+x_{k+3}+x_{k+4}=1.$$
Each summand is at most $\frac13$. The only way three numbers bounded above by $\frac13$ can have sum $1$ is that all three equal $\frac13$. Any value strictly below $\frac13$ would force the sum to be strictly below $1$.
The third delicate point is the cyclic propagation. From
$$x_{k+2}=x_{k+3}=x_{k+4}=\frac13,$$
the equation corresponding to index $k+2$ involves the variables $x_{k+4},x_k,x_{k+1}$. Since its left-hand side equals $1$, the sum of those three variables equals $1$. Two of them are already $\frac13$, so the third must also be $\frac13$. Repeating once more determines the remaining coordinate. No coordinate is left undetermined.
Alternative Approaches
Introduce
$$y_i=3x_i.$$
The system becomes
$$y_i=(y_{i+2}+y_{i+3}+y_{i+4})^5/3^5.$$
Taking fifth roots yields a monotone cyclic relation among the $y_i$. One may then compare a maximal coordinate with the coordinates appearing in its defining equation. The maximal coordinate forces equality in the comparison, and equality propagates around the cycle, producing constant solutions only.
Another approach uses the strict convexity of $t^5$. Summing the five equations gives
$$3\sum x_i=\sum s_i^5,$$
where each variable appears in exactly three of the sums $s_i$. Jensen's inequality implies that equality can occur only when all $s_i$ are equal. The resulting linear system yields $x_1=\cdots=x_5$, after which solving $(3t)^5=3t$ gives the three solutions. The extremal argument is shorter because it avoids introducing an additional equality analysis for Jensen's inequality.