Kvant Math Problem 594
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Problem
Find all real numbers $a$ for which there exist nonnegative real numbers $x_1$, $x_2$, $x_3$, $x_4$, $x_5$ satisfying the relations $$\sum_{k=1}^5kx_k=a,\quad \sum_{k=1}^5k^3x_k=a^2,\quad \sum_{k=1}^5k^5x_k=a^3.$$
International Mathematical Olympiad (XXI, 1979)
Exploration
Let
$$m_r=\sum_{k=1}^5 k^r x_k .$$
The conditions are
$$m_1=a,\qquad m_3=a^2,\qquad m_5=a^3.$$
Since the coefficients $x_k$ are nonnegative, it is natural to normalize them and interpret them as weights. If $a>0$, define
$$p_k=\frac{x_k}{a}.$$
Then $p_k\ge 0$ and
$$\sum_{k=1}^5 k,p_k=1,\qquad \sum_{k=1}^5 k^3 p_k=a,\qquad \sum_{k=1}^5 k^5 p_k=a^2.$$
Thus $a$ becomes the second moment-like quantity of a probability distribution whose support lies in ${1,2,3,4,5}$, except that the normalization is by the first moment.
A useful inequality is the Cauchy inequality for nonnegative weights:
$$\left(\sum k^3x_k\right)^2 \le \left(\sum kx_k\right)\left(\sum k^5x_k\right).$$
Substituting the given relations gives
$$a^4\le a\cdot a^3=a^4.$$
Equality holds. Equality in Cauchy is very restrictive and should force all positive-weight indices to have the same value of $k^2$. Since the $k$'s are distinct, at most one $x_k$ can be positive.
This immediately suggests that every solution comes from concentrating all mass at a single index $k$. Then
$$a=kx_k,\qquad a^2=k^3x_k,\qquad a^3=k^5x_k.$$
Eliminating $x_k$ yields $a=k^2$.
Checking gives the candidates
$$a=1,4,9,16,25.$$
There is also the possibility $a=0$. Then $m_1=0$. Since all $x_k\ge0$ and $k>0$, this forces all $x_k=0$, which indeed satisfies the system.
The delicate point is proving rigorously from equality in Cauchy that only one $x_k$ can be nonzero.
Problem Understanding
We must determine all real numbers $a$ for which there exist nonnegative real numbers $x_1,\dots,x_5$ satisfying
$$\sum_{k=1}^5 kx_k=a,\qquad \sum_{k=1}^5 k^3x_k=a^2,\qquad \sum_{k=1}^5 k^5x_k=a^3.$$
This is a Type A problem, a classification problem.
The core difficulty is extracting strong structural information from the three moment conditions. The crucial observation is that the given equalities make the Cauchy inequality an equality case. Equality forces all positive weights to be concentrated at a single index.
The expected answer is
$$a\in{0,1,4,9,16,25}.$$
Intuitively, equality in Cauchy can occur only when all mass sits at one value of $k$, and then $a$ must equal $k^2$.
Proof Architecture
Lemma 1. If $a=0$, then $x_1=x_2=x_3=x_4=x_5=0$, hence $a=0$ is admissible.
The first equation becomes $\sum kx_k=0$; all summands are nonnegative.
Lemma 2. For every solution,
$$\left(\sum k^3x_k\right)^2 \le \left(\sum kx_k\right)\left(\sum k^5x_k\right).$$
This is Cauchy's inequality applied to $\sqrt{kx_k}$ and $k^2\sqrt{kx_k}$.
Lemma 3. Equality holds in Lemma 2 for every solution.
Substituting the given relations yields equality on both sides.
Lemma 4. If equality holds in Lemma 2, then at most one of the numbers $x_1,\dots,x_5$ is positive.
Equality in Cauchy implies proportionality of the vectors $\sqrt{kx_k}$ and $k^2\sqrt{kx_k}$; whenever $x_k>0$, the quantity $k^2$ must be constant.
Lemma 5. If exactly one $x_t$ is positive, then $a=t^2$.
Substituting into the equations gives $a=tx_t$ and $a^2=t^3x_t=t^2a$.
The hardest direction is proving that every solution must have support at a single index. Lemma 4 is the step most likely to fail under scrutiny.
Solution
Suppose first that $a=0$.
Then
$$\sum_{k=1}^5 kx_k=0.$$
Since $k>0$ and $x_k\ge0$ for every $k$, each term $kx_k$ is nonnegative. Hence all terms must be zero, so
$$x_1=x_2=x_3=x_4=x_5=0.$$
The remaining two equations are then also satisfied. Thus $a=0$ is admissible.
Now assume $a\neq0$.
Apply Cauchy's inequality to the sequences
$$u_k=\sqrt{kx_k}, \qquad v_k=k^2\sqrt{kx_k}.$$
We obtain
$$\left(\sum_{k=1}^5 u_kv_k\right)^2 \le \left(\sum_{k=1}^5 u_k^2\right) \left(\sum_{k=1}^5 v_k^2\right).$$
Since
$$\sum u_kv_k=\sum k^3x_k, \qquad \sum u_k^2=\sum kx_k, \qquad \sum v_k^2=\sum k^5x_k,$$
this becomes
$$\left(\sum_{k=1}^5 k^3x_k\right)^2 \le \left(\sum_{k=1}^5 kx_k\right) \left(\sum_{k=1}^5 k^5x_k\right).$$
Using the given relations,
$$(a^2)^2\le a\cdot a^3.$$
Thus
$$a^4\le a^4,$$
so equality holds.
Equality in Cauchy's inequality implies that the vectors $(u_k)$ and $(v_k)$ are proportional. Hence there exists a constant $\lambda$ such that
$$v_k=\lambda u_k$$
for every $k$. Therefore
$$k^2\sqrt{kx_k} = \lambda\sqrt{kx_k}.$$
Whenever $x_k>0$, we have $\sqrt{kx_k}>0$, and consequently
$$k^2=\lambda.$$
Thus every index $k$ with $x_k>0$ must have the same value of $k^2$. Since the numbers
$$1^2,2^2,3^2,4^2,5^2$$
are pairwise distinct, there can be at most one index $t\in{1,2,3,4,5}$ for which $x_t>0$.
Because $a\neq0$, the first equation shows that not all $x_k$ vanish. Hence exactly one $x_t$ is positive.
The system reduces to
$$tx_t=a,$$
$$t^3x_t=a^2,$$
$$t^5x_t=a^3.$$
From the first equation,
$$x_t=\frac{a}{t}.$$
Substituting into the second gives
$$t^3\frac{a}{t}=a^2,$$
hence
$$t^2a=a^2.$$
Since $a\neq0$,
$$a=t^2.$$
Therefore
$$a\in{1,4,9,16,25}.$$
Conversely, let $t\in{1,2,3,4,5}$, set
$$a=t^2, \qquad x_t=t,$$
and let all other $x_k$ be zero. Then
$$\sum kx_k=t^2=a,$$
$$\sum k^3x_k=t^4=a^2,$$
$$\sum k^5x_k=t^6=a^3.$$
Hence each of the values $1,4,9,16,25$ is admissible.
Together with $a=0$, this yields all possible values:
$$\boxed{{0,1,4,9,16,25}}.$$
Verification of Key Steps
The first delicate step is the equality case of Cauchy's inequality. The inequality used is
$$\left(\sum k^3x_k\right)^2 \le \left(\sum kx_k\right)\left(\sum k^5x_k\right).$$
Equality holds only when there exists $\lambda$ such that
$$k^2\sqrt{kx_k}=\lambda\sqrt{kx_k}$$
for every $k$. If two distinct indices $i\neq j$ had positive weights, then
$$i^2=\lambda=j^2,$$
which is impossible. This excludes all multi-point supports.
The second delicate step is handling $a=0$. A careless argument might divide by $a$ after obtaining
$$t^2a=a^2.$$
The case $a=0$ must be separated first. Once $a=0$, the positivity of the coefficients $k$ forces all $x_k$ to vanish.
A third check is the converse construction. For $a=t^2$, choosing $x_t=t$ gives
$$t\cdot t=t^2,\qquad t^3\cdot t=t^4,\qquad t^5\cdot t=t^6,$$
which match $a$, $a^2$, and $a^3$ exactly.
Alternative Approaches
Introduce
$$m_1=\sum kx_k,\qquad m_3=\sum k^3x_k,\qquad m_5=\sum k^5x_k.$$
For nonnegative weights, Hölder's inequality yields
$$m_3^2\le m_1m_5.$$
The given conditions turn this inequality into an equality. The equality case of Hölder again forces the support to consist of a single index, after which the computation $a=t^2$ is immediate.
Another approach normalizes by $a$ when $a>0$, defining $p_k=x_k/a$. Then
$$\sum kp_k=1,\qquad \sum k^3p_k=a,\qquad \sum k^5p_k=a^2.$$
Applying the variance inequality
$$\left(\sum k^3p_k\right)^2 \le \left(\sum kp_k\right) \left(\sum k^5p_k\right)$$
shows that equality occurs. Equality means the random variable $k^2$ is constant on the support of the distribution $(kp_k)$, forcing a single support point. This again yields $a=t^2$.