Kvant Math Problem 882

Consider small integer triples $(a,b,c)$ satisfying $a+b+c=0$.

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Problem

The sum of three integers $a$, $b$, and $c$ is equal to 0. Prove that the number $2a^4+2b^4+2c^4$ is the square of an integer.

L. D. Kurlyandchik, A. S. Merkuryev, S. V. Fomin

Leningrad City Mathematical Olympiad (50th, 1984)

Exploration

Consider small integer triples $(a,b,c)$ satisfying $a+b+c=0$. If $a=1$, $b=1$, then $c=-2$, and $2a^4+2b^4+2c^4 = 2+2+32 = 36$, which is $6^2$. If $a=2$, $b=-1$, $c=-1$, then $2a^4+2b^4+2c^4 = 2\cdot16 + 2\cdot1 + 2\cdot1 = 36$, again a perfect square. Testing symmetric triples such as $(n,-n,0)$ gives $2n^4 + 2n^4 + 0 = 4n^4 = (2n^2)^2$, which is always a square. These examples suggest a symmetric pattern in which the sum of fourth powers reduces to a perfect square, likely through an algebraic identity exploiting $a+b+c=0$. The step most likely to hide an error is assuming such a factorization exists without proving it, because fourth powers can behave irregularly if $a$, $b$, $c$ have different magnitudes or signs.

Problem Understanding

The problem asks to show that for any integers $a$, $b$, $c$ with $a+b+c=0$, the expression $2a^4+2b^4+2c^4$ is a perfect square. This is a Type B problem, as it requests a proof of a universal property rather than a classification or optimization. The core difficulty is to manipulate $a^4+b^4+c^4$ under the linear relation $a+b+c=0$ in order to expose a hidden square, which is not immediately obvious from expansion. Intuitively, symmetry suggests rewriting one variable in terms of the other two and factorizing the sum of fourth powers.

Proof Architecture

Lemma 1: If $c = -(a+b)$, then $a^4+b^4+c^4 = a^4+b^4+(a+b)^4$. This follows from substitution.

Lemma 2: The expansion $(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$. This is standard binomial expansion.

Lemma 3: Combining like terms yields $a^4+b^4+(a+b)^4 = 2(a^2 + ab + b^2)^2$. This is verified by explicit algebraic computation.

The hardest direction is Lemma 3, since it requires careful algebra to identify the perfect square and ensure no terms are omitted. The most delicate step is combining the cubic, quadratic, and cross terms to see the square structure.

Solution

Let $a$, $b$, $c$ be integers such that $a+b+c=0$. Then $c = -(a+b)$, and the expression of interest becomes

$2a^4+2b^4+2c^4 = 2a^4 + 2b^4 + 2(a+b)^4.$

Expanding $(a+b)^4$ using the binomial theorem gives

$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.$

Adding $a^4$ and $b^4$ yields

$a^4 + b^4 + (a+b)^4 = a^4 + b^4 + a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 = 2a^4 + 2b^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 2b^4.$

Combining like terms carefully produces

$a^4 + b^4 + (a+b)^4 = 2a^4 + 2b^4 + 4a^3b + 6a^2b^2 + 4ab^3 = 2(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4).$

Observe that

$a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4 = (a^2 + ab + b^2)^2.$

Expanding $(a^2 + ab + b^2)^2$ gives

$(a^2 + ab + b^2)^2 = a^4 + 2a^3b + a^2b^2 + a^2b^2 + 2ab^3 + b^4 = a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4,$

which matches the previous expression exactly. Therefore,

$2a^4 + 2b^4 + 2c^4 = 2(a^4 + b^4 + (a+b)^4) = 2\cdot 2(a^2 + ab + b^2)^2 = 4(a^2 + ab + b^2)^2 = (2(a^2 + ab + b^2))^2.$

Since $a$ and $b$ are integers, $2(a^2 + ab + b^2)$ is an integer. Consequently, $2a^4+2b^4+2c^4$ is the square of the integer $2(a^2+ab+b^2)$. This completes the proof.

∎

Verification of Key Steps

The crucial step is identifying that $a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4$ equals $(a^2+ab+b^2)^2$. Expanding $(a^2+ab+b^2)^2$ independently gives $a^4 + a^3b + a^2b^2 + a^2b^2 + ab^3 + b^4 + 2a^3b + 2ab^3 + a^2b^2$, summing to $a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4$, confirming exact match. Testing small integer triples such as $(1,1,-2)$ and $(2,-1,-1)$ produces the same integer square as in the formula, ensuring no arithmetic oversight.

Alternative Approaches

An alternative approach rewrites $a^4+b^4+c^4$ in terms of symmetric polynomials: $a+b+c$, $ab+bc+ca$, and $abc$. With $a+b+c=0$, the sum $a^4+b^4+c^4$ can be expressed as $2(ab+bc+ca)^2 - 8abc(a+b+c)$, reducing to $2(ab+bc+ca)^2$. While elegant, this requires familiarity with Newton's identities and may obscure the direct constructive factorization. The main approach is preferable for its simplicity and explicit demonstration of the perfect square, fully accessible to high school students familiar with binomial expansion.