Kvant Math Problem 879

Work is carried out in the residue field $\mathbb{F}_p$, where $p$ is odd, so $2$ is invertible.

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Problem

About five integers $a$, $b$, $c$, $d$, $e$, it is known that the sums $a+b+c+d+e$ and $a^2+b^2+c^2+d^2+e^2$ are divisible by an odd number $p$. Prove that the number $$a^5+b^5+c^5+d^5+e^5-5abcde$$ is also divisible by $p$.

S. V. Fomin

Leningrad City Mathematical Olympiad (50, 1984)

Exploration

Work is carried out in the residue field $\mathbb{F}_p$, where $p$ is odd, so $2$ is invertible. The hypotheses state that the first two power sums satisfy

$$p_1 := a+b+c+d+e \equiv 0 \pmod p, \qquad p_2 := a^2+b^2+c^2+d^2+e^2 \equiv 0 \pmod p.$$

The expression to be proved divisible by $p$ is

$$p_5 - 5s_5,$$

where $p_k = a^k+b^k+c^k+d^k+e^k$ and $s_5 = abcde$.

The natural mechanism connecting power sums and elementary symmetric polynomials is Newton’s identities. The key expectation is that once $p_1$ and $p_2$ vanish, all lower symmetric obstructions disappear in the recursion for $p_5$, leaving only the relation $p_5 = 5s_5$.

The delicate point is ensuring that no illegitimate division is used when $p=3$ or $p=5$, so the identities must be used in their integral form.

Problem Understanding

This is a Type B problem: prove that a specific symmetric expression is divisible by $p$ under conditions on the first two power sums.

The core difficulty is to connect power sums and elementary symmetric polynomials without assuming invertibility of small integers modulo $p$ in a way that could fail when $p=3$ or $p=5$. The correct structure is provided by Newton’s identities in their integer form.

Proof Architecture

The proof uses Newton’s identities relating power sums $p_k$ and elementary symmetric polynomials $s_k$ for five variables.

The first lemma establishes $s_1 \equiv 0$ and $s_2 \equiv 0$ from the given assumptions.

The second lemma computes the Newton identity for $k=5$ in a form that isolates $p_5 - 5s_5$.

The hardest point is ensuring that all intermediate terms vanish modulo $p$ without dividing by non-invertible integers.

Solution

Let $a,b,c,d,e$ be considered in the ring $\mathbb{Z}/p\mathbb{Z}$. Denote by

$$p_k = a^k+b^k+c^k+d^k+e^k$$

the power sums and by $s_k$ the elementary symmetric polynomials in $a,b,c,d,e$, with $s_5 = abcde$.

Newton’s identities in integer form state that for each $k \ge 1$,

$$k s_k = \sum_{i=1}^{k} (-1)^{i-1} s_{k-i} p_i,$$

where $s_0 = 1$.

From the case $k=1$, we obtain

$$1 \cdot s_1 = p_1,$$

hence $s_1 \equiv 0 \pmod p$.

For $k=2$, the identity becomes

$$2 s_2 = s_1 p_1 - p_2.$$

Since $s_1 \equiv 0$ and $p_1 \equiv 0$, it follows that $2 s_2 \equiv -p_2 \equiv 0 \pmod p$. Because $p$ is odd, $2$ is invertible modulo $p$, hence $s_2 \equiv 0 \pmod p$.

We now apply the identity for $k=5$:

$$5 s_5 = s_4 p_1 - s_3 p_2 + s_2 p_3 - s_1 p_4 + p_5.$$

Each term is examined modulo $p$. The assumptions give $p_1 \equiv 0$ and $p_2 \equiv 0$. The previously established relations give $s_1 \equiv 0$ and $s_2 \equiv 0$. Therefore every term except $p_5$ vanishes modulo $p$, and the identity reduces to

$$5 s_5 \equiv p_5 \pmod p.$$

Rewriting,

$$p_5 - 5abcde \equiv 0 \pmod p.$$

This completes the proof.

∎

Verification of Key Steps

The most delicate point is the use of Newton’s identities without dividing by potentially non-invertible integers. The identity for $k=2$ is used in the form $2s_2 = s_1p_1 - p_2$, which remains valid in $\mathbb{Z}$ and only requires division by $2$ after reduction modulo $p$, justified since $p$ is odd.

For $k=5$, the identity is applied directly in its integer form. Each term containing $p_1$ or $p_2$ vanishes immediately in $\mathbb{Z}/p\mathbb{Z}$, and the remaining identity $5s_5 = p_5$ holds independently of invertibility of $3$ or $4$, since no division by them is performed.

A potential failure would occur if one attempted to derive intermediate expressions such as $s_3 = p_3/3$ or $s_4 = -p_4/4$; these divisions are unnecessary and avoided entirely in the final argument.

Alternative Approaches

An alternative method uses the characteristic polynomial

$$(t-a)(t-b)(t-c)(t-d)(t-e) = t^5 - s_1 t^4 + s_2 t^3 - s_3 t^2 + s_4 t - s_5$$

and relates power sums to coefficients via traces in the quotient algebra $\mathbb{F}_p[x]/(f(x))$. One can derive the same conclusion by considering the trace of multiplication operators, but this introduces additional algebraic structure without simplifying the core cancellation mechanism provided by Newton’s identities.