Kvant Math Problem 766
Consider three consecutive integers $n-1$, $n$, $n+1$ and compute the sum of their squares.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m57s
Source on kvant.digital
Problem
Prove that the sum of the squares of three consecutive integers cannot be a cube of a natural number.
Yu. I. Ionin
Exploration
Consider three consecutive integers $n-1$, $n$, $n+1$ and compute the sum of their squares. Explicitly,
$$(n-1)^2 + n^2 + (n+1)^2 = n^2 - 2n + 1 + n^2 + n^2 + 2n + 1 = 3n^2 + 2.$$
The problem asks whether this sum can equal a perfect cube, that is, whether there exists $n \in \mathbb{Z}$ and $m \in \mathbb{N}$ such that
$$3n^2 + 2 = m^3.$$
Testing small values of $n$ yields $n = 0: 2$, $n = 1: 5$, $n = 2: 14$, $n = 3: 29$, $n = 4: 50$, $n = 5: 77$, none of which are cubes. Negative $n$ produce the same values. A pattern suggests that $3n^2 + 2$ rarely aligns with a perfect cube.
Modulo analysis might reveal contradictions. For cubes modulo $3$ we have $m^3 \equiv 0,1,2 \pmod{3}$; $3n^2 + 2 \equiv 2 \pmod{3}$. This allows $m^3 \equiv 2 \pmod{3}$, so modulo $3$ alone is insufficient. Checking modulo $9$, $3n^2 + 2 \equiv 2,5,8 \pmod{9}$ depending on $n \bmod 3$. Cubes modulo $9$ are $0,1,8$. Only $n \equiv 2 \pmod 3$ gives $3n^2+2 \equiv 8 \pmod 9$, matching a cube congruence, so the congruence test narrows possibilities but does not eliminate all cases.
The key difficulty is proving that the Diophantine equation $3n^2 + 2 = m^3$ has no integer solutions. Small values show no solution; congruences narrow possibilities but require careful treatment.
Problem Understanding
The problem asks to prove that the sum of squares of three consecutive integers cannot be a cube of a natural number. This is a Type B problem: a pure proof asserting a nonexistence claim. The core difficulty is showing that the Diophantine equation $3n^2 + 2 = m^3$ admits no integer solutions. The conjecture is that no integers $n, m$ satisfy this equation. The intuitive reason is that $3n^2 + 2$ grows quadratically while cubes grow faster than quadratics, and their residue patterns modulo small integers are incompatible for all $n$.
Proof Architecture
Lemma 1: Express the sum of squares of three consecutive integers as $3n^2 + 2$. This follows from expanding $(n-1)^2 + n^2 + (n+1)^2$ and simplifying.
Lemma 2: For integers $n, m$, the Diophantine equation $3n^2 + 2 = m^3$ has no solutions modulo $9$. The idea is to compute $3n^2 + 2 \bmod 9$ for $n \equiv 0,1,2 \pmod 3$ and compare with the possible cube residues modulo $9$.
Hardest direction: Showing rigorously that no solution exists even when congruences allow one potential residue; the critical point is checking all small cases and ensuring the congruence argument is airtight.
Solution
Consider three consecutive integers $n-1$, $n$, $n+1$. Their squares sum to
$$(n-1)^2 + n^2 + (n+1)^2 = n^2 - 2n + 1 + n^2 + n^2 + 2n + 1 = 3n^2 + 2.$$
Suppose there exist integers $n$ and $m \in \mathbb{N}$ such that
$$3n^2 + 2 = m^3.$$
We examine this equation modulo $9$. Any integer $n$ satisfies $n \equiv 0,1,2 \pmod 3$.
If $n \equiv 0 \pmod 3$, then $n = 3k$ and $3n^2 + 2 = 3(9k^2) + 2 = 27k^2 + 2 \equiv 2 \pmod 9$.
If $n \equiv 1 \pmod 3$, then $n = 3k+1$ and $n^2 = 9k^2 + 6k + 1$, so $3n^2 + 2 = 27k^2 + 18k + 5 \equiv 5 \pmod 9$.
If $n \equiv 2 \pmod 3$, then $n = 3k+2$ and $n^2 = 9k^2 + 12k + 4$, so $3n^2 + 2 = 27k^2 + 36k + 14 \equiv 5 \pmod 9$.
Therefore $3n^2 + 2 \equiv 2$ if $n \equiv 0 \pmod 3$, and $\equiv 5$ if $n \equiv 1,2 \pmod 3$.
The cubes modulo $9$ are $0^3 \equiv 0$, $1^3 \equiv 1$, $2^3 \equiv 8$, $3^3 \equiv 0$, $4^3 \equiv 1$, $5^3 \equiv 8$, $6^3 \equiv 0$, $7^3 \equiv 1$, $8^3 \equiv 8 \pmod 9$, so any cube $m^3 \equiv 0,1,8 \pmod 9$.
Comparing residues modulo $9$, $3n^2 + 2 \equiv 2 \text{ or } 5$, whereas $m^3 \equiv 0,1,8$. Therefore, there is no $n$ and $m$ satisfying $3n^2 + 2 = m^3$.
Hence, the sum of the squares of three consecutive integers cannot be a cube of a natural number.
This completes the proof.
∎
Verification of Key Steps
The delicate step is computing $3n^2 + 2 \pmod 9$ for all residue classes $n \bmod 3$. Explicit substitution confirms: $n \equiv 0$ gives $2$, $n \equiv 1$ gives $5$, $n \equiv 2$ gives $5$. The cube residues modulo $9$ were checked by squaring and multiplying: $0^3,3^3,6^3 \equiv 0$, $1^3,4^3,7^3 \equiv 1$, $2^3,5^3,8^3 \equiv 8$. These sets are disjoint from ${2,5}$, confirming no solution exists.
Alternative Approaches
An alternative approach would be to treat $3n^2 + 2 = m^3$ as an elliptic curve equation and invoke Mordell's theorem, noting that integer solutions are finite. Direct trial with small $|n|$ quickly shows no solutions. The main approach using elementary congruences modulo $9$ is preferable for a Kvant audience because it avoids advanced machinery and provides a fully self-contained argument with only basic number theory.