IMO 1979 Problem 5

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 9m17s

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 system involves three weighted power sums over the set ${1,2,3,4,5}$ with nonnegative weights $x_1,\dots,x_5$. The constraints prescribe that the first, third, and fifth power moments are $a$, $a^2$, and $a^3$ respectively. The structure suggests a moment inequality approach, since expressions of the form $\sum k^m x_k$ with $x_k \ge 0$ naturally admit comparisons via Cauchy–Schwarz or Hölder inequalities.

A natural inequality connects the given quantities: applying Cauchy–Schwarz to the sequences $(k^{1/2}x_k^{1/2})$ and $(k^{5/2}x_k^{1/2})$ produces a relation between $\sum kx_k$, $\sum k^5x_k$, and $\sum k^3x_k$. This aligns exactly with the prescribed identities and forces equality in a sharp inequality, suggesting that all nonzero mass must concentrate at a single index.

Testing this idea, if all weight is placed at a single point $k=j$, then the system reduces to three algebraic equations in $a$ and $j$, leading to consistency conditions $a=j^2$. This gives candidate values $a \in {1,4,9,16,25}$, and a degenerate case $a=0$ where all weights vanish.

The key difficulty is confirming that no mixture of two or more indices can satisfy all three moment constraints simultaneously, since the system is nonlinear in $a$ and linear in the $x_k$.

Problem Understanding

The problem asks for all real numbers $a$ such that there exist nonnegative real weights assigned to the integers $1$ through $5$ whose first, third, and fifth weighted power sums equal $a$, $a^2$, and $a^3$ respectively.

This is a problem about representing a consistent sequence of power moments using a discrete nonnegative measure supported on five points. The difficulty lies in the nonlinear compatibility between different moments, since the same weights must simultaneously satisfy three polynomial constraints in $k$.

The expected structure is that the constraints are rigid enough to force a collapse to a single support point. This suggests that $a$ must arise as a square of an integer between $1$ and $5$, together with a trivial zero solution.

The candidate set is

${0,1,4,9,16,25}.$

Proof Architecture

The argument relies on a single sharp inequality linking the three prescribed moments. The first component establishes that the triple $(\sum kx_k, \sum k^3x_k, \sum k^5x_k)$ satisfies a multiplicative constraint derived from Cauchy–Schwarz, forcing a rigid equality condition.

The second component characterizes the equality case in that inequality. The structure of equality in Cauchy–Schwarz implies proportionality between the sequences $(k^{3/2})$ and $(k^{1/2})$ on the support of the weights, which forces all indices with positive weight to coincide.

The third component converts this structural collapse into explicit algebraic conditions on $a$ and identifies all admissible values.

The most delicate step is the characterization of equality in the inequality, since any hidden possibility of two-point support would invalidate the conclusion.

Solution

Let $x_1,\dots,x_5$ be nonnegative real numbers satisfying

$\sum_{k=1}^5 kx_k = a,\quad \sum_{k=1}^5 k^3x_k = a^2,\quad \sum_{k=1}^5 k^5x_k = a^3.$

Lemma 1

For any nonnegative reals $x_1,\dots,x_5$, the inequality

$\left(\sum_{k=1}^5 k^3 x_k\right)^2 \le \left(\sum_{k=1}^5 k x_k\right)\left(\sum_{k=1}^5 k^5 x_k\right)$

holds.

This follows from Cauchy–Schwarz applied to the sequences $\bigl(k^{1/2}\sqrt{x_k}\bigr)$ and $\bigl(k^{5/2}\sqrt{x_k}\bigr)$, since their inner product equals $\sum k^3 x_k$ while their squared norms equal $\sum kx_k$ and $\sum k^5x_k$ respectively. This step enforces a multiplicative constraint among the moments.

Applying the given identities yields

$a^4 \le a \cdot a^3 = a^4,$

hence equality holds in Lemma 1.

Lemma 2

Equality in Lemma 1 holds if and only if there exists a constant $\lambda$ such that for every index $k$ with $x_k>0$, one has $k^{5/2} = \lambda k^{1/2}$.

This is the equality condition in Cauchy–Schwarz, requiring proportionality of the sequences $k^{1/2}\sqrt{x_k}$ and $k^{5/2}\sqrt{x_k}$. Cancellation of $\sqrt{x_k}$ on the support yields $k^2=\lambda$ for all indices with positive weight.

Thus all indices with $x_k>0$ are equal.

This step eliminates any possibility of multiple distinct support points.

Lemma 3

There exists an index $j \in {1,2,3,4,5}$ such that $x_j>0$ and $x_k=0$ for all $k\ne j$.

This follows from Lemma 2, since all positive-weight indices must satisfy $k^2=\lambda$, forcing a single admissible integer value.

This step reduces the system to a single-variable configuration.

With this structure, let $x_j = t$ with $t \ge 0$. The constraints become

$jt = a,\quad j^3 t = a^2,\quad j^5 t = a^3.$

From the first equation, $t = \frac{a}{j}$. Substituting into the second equation gives

$j^3 \cdot \frac{a}{j} = a^2,$

hence

$j^2 a = a^2.$

If $a \ne 0$, division by $a$ yields $a = j^2$. Substituting into $t = a/j$ gives $t = j$, which is nonnegative.

Substituting into the third equation verifies consistency:

$j^5 \cdot \frac{a}{j} = j^4 a = j^4 \cdot j^2 = j^6 = a^3.$

If $a=0$, then $jt=0$ forces $t=0$, which satisfies all three original equations.

Thus all solutions arise from $a=j^2$ with $j \in {1,2,3,4,5}$ together with $a=0$.

Lemma 4

No value of $a$ outside ${0,1,4,9,16,25}$ admits a solution.

This follows because any admissible configuration forces single-point support, and the resulting algebraic constraints enforce $a=j^2$ for some $j \in {1,2,3,4,5}$ or $a=0$.

This excludes all other real numbers.

Combining Lemmas 3 and 4 yields the complete characterization.

Verification of Key Steps

The most sensitive point is the equality condition in Cauchy–Schwarz. Re-deriving independently, equality requires linear dependence of the vectors $(k^{1/2}\sqrt{x_k})$ and $(k^{5/2}\sqrt{x_k})$, which forces $k^2$ to be constant on every index with $x_k>0$. Any attempt to distribute weight on two distinct indices produces strict inequality because distinct values of $k^2$ violate proportionality.

A second delicate point is the case $a=0$, where the moment equations force every weighted sum of nonnegative terms to vanish, which implies each $x_k=0$ individually.

A third check concerns consistency of the third moment after reduction to a single index. Substitution confirms that no additional restriction arises beyond $a=j^2$.

Alternative Approaches

An alternative method uses Hölder’s inequality in the form

$\left(\sum k^3 x_k\right)^3 \le \left(\sum k x_k\right)\left(\sum k^5 x_k\right)\left(\sum k^3 x_k\right),$

which also leads to a rigidity condition forcing proportionality of powers of $k$ on the support.

Another approach interprets the system as a truncated moment problem and uses convexity: the vector $(k,k^3,k^5)$ lies on a strictly convex curve, so any representation as a barycenter of multiple points forces degeneracy to a single point. This geometric viewpoint yields the same discrete set of admissible parameters.