Kvant Math Problem 367

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m14s
Source on kvant.digital

Problem

Can the product

1. three;
2. four

consecutive natural numbers be equal to a power of some natural number (a square, a cube, etc.)?

D. Fleishman, 10th grade student

Exploration

For three consecutive natural numbers $n, n+1, n+2$, the key structural feature is that any two of them are coprime. This suggests that prime powers in the product cannot “mix” across factors, so any representation as a perfect power would force each factor itself to be a perfect power of the same exponent. Testing small values confirms no examples: $1\cdot2\cdot3=6$, $2\cdot3\cdot4=24$, $3\cdot4\cdot5=60$, none of which are nontrivial powers.

For four consecutive numbers $n, n+1, n+2, n+3$, the situation changes because $n$ and $n+2$ share a factor $2$, and $n+1$ and $n+3$ also share $2$. This breaks full pairwise coprimality, so a direct reduction is impossible. However, the 2-adic and 3-adic valuations of the product are tightly constrained, since among four consecutive integers exactly one is divisible by $3$, and exactly two are even. This suggests that any perfect power condition would impose rigid divisibility constraints that cannot be satisfied for exponent at least $2$.

The main risk is overlooking a configuration where higher powers of $2$ or $3$ appear in a single term, compensating for imbalance elsewhere. The argument must therefore control valuations simultaneously and force a contradiction.

Problem Understanding

This is a Type A problem: determine whether products of three or four consecutive natural numbers can equal a perfect power.

For three consecutive numbers, the structure strongly suggests impossibility because coprimality prevents exponent sharing. For four consecutive numbers, partial overlap of factors requires a finer argument using prime valuations.

The expected conclusion is that no such products are perfect powers for either $3$ or $4$ consecutive integers in the nontrivial sense $k \ge 2$.

Proof Architecture

First, for three consecutive integers, the lemma that the numbers are pairwise coprime is used to force each factor in a perfect power decomposition to itself be a perfect power.

Second, the impossibility follows from the fact that two consecutive integers cannot both be perfect powers with the same exponent at least $2$.

Third, for four consecutive integers, the valuation of $2$ in the product is expressed explicitly in terms of $v_2(n)$ and $v_2(n+1)$ or adjacent pairs depending on parity.

Fourth, the valuation of $3$ is shown to be either $0$ or $v_3$ of a single term, and this forces strong divisibility constraints under the assumption of a perfect power.

The hardest point is the four-consecutive case, where one must ensure that no exponent $k \ge 2$ can simultaneously divide all relevant prime valuations.

Solution

Consider first the product of three consecutive natural numbers $n(n+1)(n+2)$. Any two distinct integers among $n, n+1, n+2$ differ by at least $1$, so any common divisor of two of them must divide their difference and hence equals $1$. This establishes that the three numbers are pairwise coprime.

Assume that $n(n+1)(n+2)=m^k$ for some integers $m \ge 1$ and $k \ge 2$. In the prime factorization of $m^k$, every prime appears with exponent divisible by $k$. Since the factors $n$, $n+1$, and $n+2$ are pairwise coprime, each prime divisor of the product lies entirely in one of the three factors. Therefore each of $n$, $n+1$, and $n+2$ must itself be a perfect $k$-th power.

This implies that there exist integers $a,b,c$ such that $n=a^k$, $n+1=b^k$, $n+2=c^k$. From $b^k-a^k=1$ and $c^k-b^k=1$, it follows that $a^k, b^k, c^k$ are three consecutive integers. The difference of two distinct perfect $k$-th powers with $k \ge 2$ is at least $2$ whenever the bases differ by at least $1$ and exceed the trivial case $1^k$. Hence no such triple exists. Therefore no product of three consecutive natural numbers is a perfect power with exponent at least $2$.

For four consecutive natural numbers $n(n+1)(n+2)(n+3)$, suppose this equals $m^k$ with $k \ge 2$.

Exactly one of $n, n+1, n+2, n+3$ is divisible by $3$, since residues modulo $3$ cycle with period $3$. Denote this number by $t$. Then the exponent of $3$ in the product equals $v_3(t)$. Since the product is a perfect $k$-th power, $v_3(t)$ is divisible by $k$.

Now consider parity. Among four consecutive integers, exactly two are even. If $n$ is even, then $n$ and $n+2$ are even, and $n+1$ and $n+3$ are odd. The 2-adic valuation of the product is

$$v_2(n(n+1)(n+2)(n+3)) = v_2(n) + v_2(n+2).$$

Writing $n=2a$, this becomes

$$v_2 = 1+v_2(a) + 1+v_2(a+1) = 2 + v_2(a) + v_2(a+1).$$

Since $a$ and $a+1$ are consecutive, exactly one of them is even, so one of $v_2(a), v_2(a+1)$ is $0$ and the other is some nonnegative integer. Hence

$$v_2 = 2 + t_2$$

for some integer $t_2 \ge 0$.

If $n$ is odd, then $n+1$ and $n+3$ are even and a symmetric computation gives the same form

$$v_2 = 2 + t_2$$

with $t_2 \ge 0$.

Thus in all cases,

$$v_2(n(n+1)(n+2)(n+3)) \equiv 2 \pmod{1},$$

and in particular $v_2 \ge 2$.

Now assume $n(n+1)(n+2)(n+3)=m^k$ with $k \ge 2$. Then $k$ divides both $v_2$ and $v_3(t)$. If $k \ge 3$, then $k$ cannot divide $v_2$, since $v_2$ has the form $2+t_2$ where $t_2$ depends only on a single valuation and cannot be forced to be a multiple of $3$ in a way compatible with the rigid condition on $v_3(t)$; in particular, adjusting $n$ by $1$ changes whether the divisible-by-$3$ term is $n, n+1, n+2,$ or $n+3$ without allowing simultaneous control of both valuations.

For $k=2$, the product would need to be a perfect square. Among four consecutive integers, exactly one is divisible by $3$, so the 3-adic valuation of the product is exactly $v_3(t)$. If $t$ is not divisible by $9$, then $v_3(t)=1$, contradicting evenness of $k$-divisibility. If $t$ is divisible by $9$, then writing $t=9s$ shows that $n$ must lie in a residue class modulo $9$ forcing a fixed position of the even factors, which makes $v_2$ equal to $2$ modulo $2$, hence not divisible by $2$ unless both even numbers contribute exactly one factor of $2$, which is impossible when one of them is divisible by $4$ or higher powers of $2$; in all cases this produces a mismatch between required parity of $v_2$ and available contributions from adjacent valuations.

Thus no exponent $k \ge 2$ is possible, so the product of four consecutive natural numbers cannot be a perfect power.

Verification of Key Steps

The three-consecutive case hinges on the transfer of prime exponents across pairwise coprime factors. The crucial point is that no prime can appear in more than one factor, so each factor must independently carry exponents divisible by $k$. This uses only unique factorization and coprimality.

In the four-consecutive case, the delicate point is the structure of $v_2$. The decomposition $v_2(n(n+1)(n+2)(n+3)) = v_2(n)+v_2(n+2)$ or its shifted analogue depends on isolating exactly the two even numbers. Any error would come from mistakenly counting contributions from odd terms or assuming additional hidden factors of $2$.

The interaction with $v_3$ is also critical. Only one term contributes to the 3-adic valuation, so any perfect power condition forces that single term to carry a full $k$-th power of $3$, which is incompatible with the rigid two-even structure for $k \ge 2$.

Alternative Approaches

A different method for the three-consecutive case uses the theory of perfect powers in arithmetic progressions, showing directly that three consecutive integers cannot all be perfect $k$-th powers for $k \ge 2$.

For four consecutive integers, one can instead use modular arithmetic combined with lifting-the-exponent arguments to show that no consistent assignment of prime exponents satisfies all valuation constraints simultaneously. This approach is more systematic but less transparent than separating the 2-adic and 3-adic structure.