IMO 1979 Problem 5
The three equalities resemble moment conditions.
Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 4m30s
Problem
Determine all real numbers a for which there exists non-negative reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $\sum_{k=1}^{5} kx_{k}=a,$ $\sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $\sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$
Exploration
The three equalities resemble moment conditions. If one defines
$$m_{1}=\sum_{k=1}^{5}kx_{k}, \qquad m_{3}=\sum_{k=1}^{5}k^{3}x_{k}, \qquad m_{5}=\sum_{k=1}^{5}k^{5}x_{k},$$
then the problem asks for all real numbers $a$ such that
$$m_{1}=a,\qquad m_{3}=a^{2},\qquad m_{5}=a^{3}.$$
The variables $x_{k}$ are nonnegative, so these are weighted sums of the odd powers of the integers $1,2,3,4,5$.
A first attempt is to search for examples supported on a single index. If $x_{j}=t$ and all other $x_{k}=0$, then
$$a=jt,\qquad a^{2}=j^{3}t,\qquad a^{3}=j^{5}t.$$
Substituting $t=a/j$ into the second equation gives
$$a^{2}=j^{2}a,$$
hence either $a=0$ or $a=j^{2}$. Then the third equation becomes
$$a^{3}=j^{4}a,$$
which is equivalent to $a^{2}=j^{4}$, again giving $a=j^{2}$ for $a\ge0$. Thus
$$a\in{0,1,4,9,16,25}$$
are immediate candidates.
The main question is whether mixtures of several indices can produce additional values. The structure strongly suggests an inequality between moments. The sequence
$$m_{1},m_{3},m_{5}$$
looks suitable for Hölder or Cauchy-Schwarz.
Trying Cauchy-Schwarz directly,
$$\left(\sum k^{3}x_{k}\right)^{2}\le \left(\sum kx_{k}\right)\left(\sum k^{5}x_{k}\right).$$
Substituting the required values gives
$$a^{4}\le a\cdot a^{3}=a^{4}.$$
Equality always holds. This is highly informative because equality in Cauchy-Schwarz is rigid.
To identify the exact equality condition, rewrite the inequality as
$$\left(\sum \sqrt{kx_{k}}\cdot k\sqrt{kx_{k}}\right)^{2} \le \left(\sum kx_{k}\right)\left(\sum k^{5}x_{k}\right).$$
Equality holds precisely when the sequences
$$\sqrt{kx_{k}} \quad\text{and}\quad k^{2}\sqrt{kx_{k}}$$
are proportional. For every $k$ with $x_{k}>0$,
$$k^{2}=\lambda$$
for a fixed constant $\lambda$. Since the numbers $1^{2},2^{2},3^{2},4^{2},5^{2}$ are distinct, at most one $x_{k}$ can be positive. This collapses the problem to the single-support case already analyzed.
A delicate point is the case $a<0$. Since all $x_{k}\ge0$ and all $k>0$,
$$a=\sum kx_{k}\ge0.$$
Hence negative $a$ are impossible. Another delicate point is $a=0$, which forces every $x_{k}=0$.
The intended proof is now clear: derive the Cauchy-Schwarz equality condition, prove only one index can occur, then solve explicitly.
Problem Understanding
We seek all real numbers $a$ for which there exist nonnegative real numbers
$$x_{1},x_{2},x_{3},x_{4},x_{5}$$
satisfying the three equations
$$\sum_{k=1}^{5}kx_{k}=a,\qquad \sum_{k=1}^{5}k^{3}x_{k}=a^{2},\qquad \sum_{k=1}^{5}k^{5}x_{k}=a^{3}.$$
This is a Type A problem because we must determine exactly which real numbers $a$ are possible. The solution must therefore contain two parts. First, every claimed value of $a$ must actually occur. Second, every other real number must be excluded.
The variables are nonnegative weights attached to the integers $1,2,3,4,5$. The equations prescribe the first, third, and fifth weighted moments. The core difficulty is that there are five unknowns but only three equations, so a direct algebraic elimination appears underdetermined. The decisive observation is that the prescribed moments satisfy the equality case of a Cauchy-Schwarz inequality, forcing a very rigid structure.
The expected answer is
$${0,1,4,9,16,25}.$$
This is plausible because if all mass is concentrated at a single index $j$, then the equations reduce to
$$a=j^{2}.$$
The equality condition in Cauchy-Schwarz suggests that no genuine mixture of several indices can occur.
Proof Architecture
The proof proceeds through three claims.
Lemma 1 states that every admissible $a$ satisfies $a\ge0$. This follows because $a=\sum kx_{k}$ is a sum of nonnegative terms.
Lemma 2 states that any admissible solution satisfies
$$\left(\sum_{k=1}^{5}k^{3}x_{k}\right)^{2} = \left(\sum_{k=1}^{5}kx_{k}\right) \left(\sum_{k=1}^{5}k^{5}x_{k}\right),$$
hence equality holds in Cauchy-Schwarz. The equality condition then implies that at most one variable $x_{k}$ is positive.
Lemma 3 states that if exactly one variable $x_{j}$ is positive, then $a=j^{2}$. This is obtained by substituting into the equations.
The hardest direction is excluding all other values of $a$. The most delicate point is applying the equality condition in Cauchy-Schwarz correctly. A careless argument might conclude merely that the sequences are proportional globally, without checking what happens at indices where $x_{k}=0$. The proof must explicitly restrict to indices with positive weight.
Solution
We determine all real numbers $a$ for which there exist nonnegative real numbers
$$x_{1},x_{2},x_{3},x_{4},x_{5}$$
such that
$$\sum_{k=1}^{5}kx_{k}=a, \qquad \sum_{k=1}^{5}k^{3}x_{k}=a^{2}, \qquad \sum_{k=1}^{5}k^{5}x_{k}=a^{3}.$$
We first prove that every admissible $a$ belongs to the set
$${0,1,4,9,16,25}.$$
Lemma 1
If $a$ is admissible, then $a\ge0$.
Proof
Since every $x_{k}\ge0$ and every $k>0$,
$$a=\sum_{k=1}^{5}kx_{k}\ge0.$$
Thus $a$ cannot be negative. ∎
This establishes that only nonnegative values of $a$ need to be considered; overlooking this would leave the sign of $a$ unjustified.
Lemma 2
If
$$\sum_{k=1}^{5}kx_{k}=a,\qquad \sum_{k=1}^{5}k^{3}x_{k}=a^{2},\qquad \sum_{k=1}^{5}k^{5}x_{k}=a^{3},$$
then at most one of the numbers $x_{1},x_{2},x_{3},x_{4},x_{5}$ is positive.
Proof
Consider the sequences
$$u_{k}=\sqrt{kx_{k}}, \qquad v_{k}=k^{2}\sqrt{kx_{k}}$$
for $k=1,2,3,4,5$.
Then
$$\sum_{k=1}^{5}u_{k}^{2} = \sum_{k=1}^{5}kx_{k} = a,$$
$$\sum_{k=1}^{5}v_{k}^{2} = \sum_{k=1}^{5}k^{5}x_{k} = a^{3},$$
and
$$\sum_{k=1}^{5}u_{k}v_{k} = \sum_{k=1}^{5}k^{3}x_{k} = a^{2}.$$
By the Cauchy-Schwarz inequality,
$$\left(\sum_{k=1}^{5}u_{k}v_{k}\right)^{2} \le \left(\sum_{k=1}^{5}u_{k}^{2}\right) \left(\sum_{k=1}^{5}v_{k}^{2}\right).$$
Substituting the above expressions yields
$$(a^{2})^{2}\le a\cdot a^{3}.$$
Hence
$$a^{4}\le a^{4},$$
so equality holds.
Equality in the Cauchy-Schwarz inequality holds precisely when the sequences $(u_{k})$ and $(v_{k})$ are proportional. Therefore there exists a real number $\lambda$ such that
$$v_{k}=\lambda u_{k}$$
for every $k$.
For every index $k$ with $x_{k}>0$, we have $u_{k}>0$. Dividing by $u_{k}$ gives
$$k^{2}=\lambda.$$
Thus every index with positive $x_{k}$ must have the same value of $k^{2}$.
Since
$$1^{2},2^{2},3^{2},4^{2},5^{2}$$
are pairwise distinct, there can be at most one index $k$ such that $x_{k}>0$. ∎
This establishes the rigidity forced by equality in Cauchy-Schwarz; merely applying the inequality without the equality condition would not determine the structure of the solution.
Lemma 3
If exactly one variable $x_{j}$ is positive, then
$$a=j^{2}.$$
Proof
Assume
$$x_{j}>0$$
for some $j\in{1,2,3,4,5}$, and
$$x_{k}=0$$
for all $k\ne j$.
Then the three equations become
$$jx_{j}=a, \qquad j^{3}x_{j}=a^{2}, \qquad j^{5}x_{j}=a^{3}.$$
From the first equation,
$$x_{j}=\frac{a}{j}.$$
Substituting into the second equation gives
$$j^{3}\cdot \frac{a}{j}=a^{2},$$
hence
$$j^{2}a=a^{2}.$$
By Lemma 1, $a\ge0$. If $a=0$, then $x_{j}=0$, contradicting $x_{j}>0$. Therefore $a>0$, and dividing by $a$ yields
$$a=j^{2}.$$
∎
This identifies all admissible values arising from the one-support structure; skipping the case $a=0$ would make the division by $a$ invalid.
We now complete the classification.
Suppose $a$ is admissible. By Lemma 2, at most one variable $x_{j}$ is positive. If all variables are zero, then
$$a=\sum_{k=1}^{5}kx_{k}=0.$$
Otherwise exactly one variable is positive, and Lemma 3 gives
$$a=j^{2}$$
for some $j\in{1,2,3,4,5}$. Hence every admissible value belongs to
$${0,1,4,9,16,25}.$$
It remains to verify that every number in this set is indeed admissible.
For $a=0$, choose
$$x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0.$$
Then all three equations hold.
For $a=j^{2}$ with $j\in{1,2,3,4,5}$, choose
$$x_{j}=j, \qquad x_{k}=0\quad (k\ne j).$$
Then
$$\sum_{k=1}^{5}kx_{k}=j\cdot j=j^{2}=a,$$
$$\sum_{k=1}^{5}k^{3}x_{k}=j^{3}\cdot j=j^{4}=a^{2},$$
and
$$\sum_{k=1}^{5}k^{5}x_{k}=j^{5}\cdot j=j^{6}=a^{3}.$$
Thus each of these six values is admissible.
Therefore the complete set of all admissible real numbers is
$$\boxed{{0,1,4,9,16,25}}.$$
Verification of Key Steps
The most delicate step is the equality condition in Cauchy-Schwarz. Starting independently, define
$$u_{k}=\sqrt{kx_{k}},\qquad v_{k}=k^{2}\sqrt{kx_{k}}.$$
Then
$$\left(\sum u_{k}v_{k}\right)^{2} = \left(\sum k^{3}x_{k}\right)^{2} = a^{4},$$
while
$$\left(\sum u_{k}^{2}\right)\left(\sum v_{k}^{2}\right) = \left(\sum kx_{k}\right)\left(\sum k^{5}x_{k}\right) = a\cdot a^{3} = a^{4}.$$
Hence equality holds. The equality condition states that there exists $\lambda$ such that
$$v_{k}=\lambda u_{k}$$
for every $k$. If one divides by $u_{k}$ without checking whether $u_{k}=0$, the argument becomes invalid. Restricting to indices with $x_{k}>0$ avoids this error and gives
$$k^{2}=\lambda.$$
Another delicate step is deriving $a=j^{2}$ from
$$j^{2}a=a^{2}.$$
A careless proof might divide by $a$ immediately. However $a=0$ must first be excluded. Since
$$a=0$$
implies
$$x_{j}=a/j=0,$$
this contradicts the assumption that $x_{j}>0$. Only after this check may one divide by $a$.
A third point requiring verification is the construction for each admissible value. For $a=j^{2}$, taking $x_{j}=j$ gives
$$jx_{j}=j^{2}, \qquad j^{3}x_{j}=j^{4}, \qquad j^{5}x_{j}=j^{6}.$$
Since
$$(j^{2})^{2}=j^{4}, \qquad (j^{2})^{3}=j^{6},$$
all three equations are satisfied exactly.
Alternative Approaches
Another approach uses Hölder's inequality in the form
$$\left(\sum k^{3}x_{k}\right)^{2} \le \left(\sum kx_{k}\right)\left(\sum k^{5}x_{k}\right).$$
After substitution this again becomes an equality case. The equality condition for Hölder yields the same conclusion, namely that all positive weights must occur at a single index. This method is essentially equivalent to the Cauchy-Schwarz argument, but the equality condition is less transparent.
A different strategy is to interpret
$$p_{k}=\frac{kx_{k}}{a}$$
for $a>0$ as a probability distribution on ${1,2,3,4,5}$. Then
$$\sum p_{k}=1, \qquad \sum k^{2}p_{k}=a, \qquad \sum k^{4}p_{k}=a^{2}.$$
The variance of the random variable $k^{2}$ becomes
$$\sum k^{4}p_{k}-\left(\sum k^{2}p_{k}\right)^{2} = a^{2}-a^{2} = 0.$$
Hence $k^{2}$ is constant on the support of the distribution, so the support consists of a single index. This proof is elegant but introduces probabilistic language that is not necessary. The Cauchy-Schwarz approach remains more direct and entirely elementary.