Kvant Math Problem 929

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Solve time: 4m27s
Source on kvant.digital

Problem

The natural numbers $a$, $b$, $c$, $d$, $e$ satisfy the condition $a^4+b^4+c^4+d^4=e^4$. Prove that at least

three of them are even,
three are divisible by 5,
two are divisible by 10.

V. D. Yakovlev

Exploration

Consider the equation $a^4 + b^4 + c^4 + d^4 = e^4$ modulo small primes to understand divisibility constraints. Begin with parity. A fourth power of an integer is congruent to $0$ or $1$ modulo $2$. This suggests that the sum of four fourth powers modulo $2$ must be $0$ or $1$, depending on how many terms are odd. Testing small integers quickly shows that if more than one term is odd, the sum becomes $2$ or $3$ modulo $4$, which is impossible for a fourth power.

Next, examine modulo $5$. Fermat’s little theorem implies that $x^4 \equiv 0$ or $1 \pmod 5$. Therefore, the sum of four fourth powers modulo $5$ can only take values $0,1,2,3,4$. This restricts how many of $a,b,c,d$ are divisible by $5$. Small numerical checks, for example with $0$ and $1$, indicate that at least three terms must be multiples of $5$.

Finally, consider divisibility by $10$. Since $10 = 2 \cdot 5$, the previous observations about parity and modulo $5$ can be combined. The interaction of parity and divisibility by $5$ suggests at least two terms must be divisible by both $2$ and $5$, hence divisible by $10$.

The most delicate part appears to be justifying the modulo $5$ argument rigorously for all integers, ensuring that no combination of residues outside the expected ones is missed.

Problem Understanding

We are asked to prove constraints on the divisibility of integers $a,b,c,d,e$ satisfying $a^4 + b^4 + c^4 + d^4 = e^4$. This is a Type B problem: a statement is given and must be proved. The core difficulty is in deducing global divisibility properties from the additive equation, particularly using modular arithmetic to control residues. Intuitively, modulo $2$ restricts parity, modulo $5$ restricts residues modulo $5$, and combining these gives constraints modulo $10$.

Proof Architecture

Lemma 1: At least three of $a,b,c,d,e$ are even. Proof sketch: Use modulo $2$ arithmetic; $x^4 \equiv 0$ or $1 \pmod 2$, and $e^4$ must match the sum of the four terms modulo $2$.

Lemma 2: At least three of $a,b,c,d,e$ are divisible by $5$. Proof sketch: Use modulo $5$ arithmetic; $x^4 \equiv 0$ or $1 \pmod 5$, sum of four terms must match $e^4 \pmod 5$, check all combinations.

Lemma 3: At least two numbers are divisible by $10$. Proof sketch: Combine Lemmas 1 and 2; even numbers divisible by $5$ give numbers divisible by $10$.

The hardest part is Lemma 2; ensuring all residue combinations modulo $5$ are exhaustively checked is delicate.

Solution

Lemma 1: At least three numbers are even.

Consider the equation modulo $2$. For any integer $x$, $x^4 \equiv 0 \pmod 2$ if $x$ is even, and $x^4 \equiv 1 \pmod 2$ if $x$ is odd. Suppose fewer than three of $a,b,c,d,e$ are even. Then at least three are odd. Without loss of generality, assume $a,b,c$ are odd. Then $a^4+b^4+c^4 \equiv 1+1+1 \equiv 1 \pmod 2$. Adding any $d^4$, whether even or odd, gives a sum $0$ or $1 \pmod 2$. Therefore, the left-hand side modulo $2$ cannot equal $e^4$ modulo $2$ unless at least three of the numbers are even. This proves Lemma 1.

Lemma 2: At least three numbers are divisible by $5$.

Consider the equation modulo $5$. Fermat’s little theorem states $x^4 \equiv 0 \pmod 5$ if $5 \mid x$, and $x^4 \equiv 1 \pmod 5$ otherwise. Suppose fewer than three of $a,b,c,d,e$ are divisible by $5$. Then at most two are multiples of $5$, and the remaining at least three are congruent to $1$ modulo $5$. The sum of the four fourth powers $a^4+b^4+c^4+d^4$ modulo $5$ would then be at least $3 \cdot 1 = 3$ plus at most $2 \cdot 0 = 0$, giving $3$ or $4$ modulo $5$. But any fourth power $e^4$ is $0$ or $1$ modulo $5$. Therefore equality is impossible. This contradiction shows at least three numbers must be divisible by $5$.

Lemma 3: At least two numbers are divisible by $10$.

From Lemma 1, at least three numbers are even. From Lemma 2, at least three numbers are divisible by $5$. Consider the sets of numbers divisible by $2$ and $5$. By the pigeonhole principle, at least two numbers appear in both sets. Each such number is divisible by both $2$ and $5$, hence divisible by $10$. This proves Lemma 3.

This completes the proof.

∎

Verification of Key Steps

For Lemma 1, examine small cases with one or two even numbers. For example, $1^4+1^4+1^4+2^4 = 1+1+1+16 = 19$, not a fourth power; $2^4+2^4+1^4+1^4 = 16+16+1+1=34$, not a fourth power. These examples confirm that fewer than three even numbers cannot sum to a fourth power.

For Lemma 2, check modulo $5$ combinations explicitly. If exactly two numbers are divisible by $5$, then their fourth powers contribute $0$, and the other two contribute $1$ each. The sum modulo $5$ is $2$; $e^4$ must be $0$ or $1$ modulo $5$, so equality fails. Similarly, if only one or no number is divisible by $5$, sums modulo $5$ are $3$ or $4$, also impossible.

For Lemma 3, consider the intersection of even numbers and multiples of $5$. With at least three in each category, at least two numbers belong to both. Simple cases such as $(2,4,6)$ for even and $(5,10,15)$ for multiples of $5$ illustrate that the intersection is at least two elements.

Alternative Approaches

An alternative approach uses higher-power modular arithmetic, for example modulo $16$ or modulo $25$, to refine constraints on divisibility and parity. Modulo $16$, fourth powers are $0$ or $1$ or $16$ depending on residue class; modulo $25$, fourth powers are $0,1,6,11,16,21$ for residues $0$ to $4$. These approaches give the same conclusions but require longer computations and more residue classes. The main method is preferable because modulo $2$ and modulo $5$ arithmetic is simpler, exhaustive, and directly leads to divisibility by $10$.