Kvant Math Problem 691

Let $P(n,k)=n(n+1)\cdots(n+k-1)$ for $n\ge 2$.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m13s
Source on kvant.digital

Problem

We say that a number has property $(K)$ if it can be represented as a product of $K$ consecutive natural numbers greater than 1.

  1. Find $K$ such that there exists a number $N$ that simultaneously has properties $(K)$ and $(K+2)$.
  2. Prove that there are no numbers that simultaneously have properties (2) and (4).

V. Fedotov

All-Union Mathematical Olympiad of School Students (1981, 8th grade)

Exploration

Let $P(n,k)=n(n+1)\cdots(n+k-1)$ for $n\ge 2$. A number has property $(k)$ if it equals $P(n,k)$ for some $n$.

We seek $K$ such that there exists a number $N$ with

$$P(a,K)=P(b,K+2)$$

for some integers $a,b\ge 2$.

Testing small $K$:

For $K=1$, any number is $P(n,1)$, so any $P(b,3)$ immediately gives a common number. For instance,

$$2\cdot 3\cdot 4=24$$

shows $(1)$ and $(3)$ coexist.

For $K=2$, we would need

$$a(a+1)=b(b+1)(b+2)(b+3).$$

Trying small $b$ gives values $120,360,840,1680,\dots$ and none are products of two consecutive integers.

For $K=3$, we compare with $K+2=5$:

$$a(a+1)(a+2)=b(b+1)(b+2)(b+3)(b+4).$$

A direct check suggests a collision:

$$8\cdot 9\cdot 10=720,\qquad 2\cdot 3\cdot 4\cdot 5\cdot 6=720.$$

So $K=3$ works.

For $K=2$ vs $4$, numerical checks show no solution in small cases, and structural growth suggests no equality.

The key difficulty is controlling when a product of $2$ consecutive integers can match a product of $4$ consecutive integers.

Problem Understanding

This is a Type A classification problem.

We must determine all integers $K$ such that some number $N$ can be written both as a product of $K$ consecutive integers greater than $1$ and as a product of $K+2$ such integers.

Then we must prove that no number has both properties $(2)$ and $(4)$.

The structure is governed by strong rigidity of factorial-type expressions. The expected answer for part 1 is $K=1$ and $K=3$, while part 2 requires proving impossibility for $(2,4)$.

Proof Architecture

We introduce $P(n,k)=n(n+1)\cdots(n+k-1)$.

First, we show that for $K=1$ and $K=3$ there exist solutions to $P(a,K)=P(b,K+2)$ by explicit construction.

Second, we show that no solution exists for $K=2$ and $K=4$ by reducing the equation

$$a(a+1)=b(b+1)(b+2)(b+3)$$

to a quadratic form in $a$ and then analyzing its integrality constraints.

The key lemma is a transformation of the quartic product into a quadratic expression in $x=b^2+3b$, leading to a Diophantine equation with no integer solutions.

The most delicate step is ruling out all integer $b\ge 2$ without missing a hidden large solution.

Solution

We first solve part 1.

For $K=1$, every integer $N>1$ has property $(1)$ since $N=P(N,1)$. Choose $b=2$, then

$$P(2,3)=2\cdot 3\cdot 4=24.$$

Thus $N=24$ has both properties $(1)$ and $(3)$.

For $K=3$, we compare products of $3$ and $5$ consecutive integers. Direct computation gives

$$P(8,3)=8\cdot 9\cdot 10=720,$$

and

$$P(2,5)=2\cdot 3\cdot 4\cdot 5\cdot 6=720.$$

Hence $720$ has properties $(3)$ and $(5)$.

Now we show that no other $K$ works.

We prove that no number has properties $(2)$ and $(4)$.

Assume there exist integers $a,b\ge 2$ such that

$$a(a+1)=b(b+1)(b+2)(b+3).$$

We rewrite the right-hand side by pairing symmetric factors:

$$b(b+3)=b^2+3b,\qquad (b+1)(b+2)=b^2+3b+2.$$

Let

$$x=b^2+3b.$$

Then the equation becomes

$$a(a+1)=x(x+2)=x^2+2x.$$

We transform both sides:

$$4a^2+4a+1=4x^2+8x+1.$$

Thus

$$(2a+1)^2=(2x+1)^2+4x.$$

Rearranging gives

$$(2a+1)^2-(2x+1)^2=4x.$$

Factoring the left-hand side,

$$(2a-2x)(2a+2x+2)=4x.$$

Dividing by $4$ yields

$$(a-x)(a+x+1)=x.$$

Now $x=b^2+3b$ is strictly positive for $b\ge 2$.

We analyze possible integer factorizations of $x$. The pair $(a-x,;a+x+1)$ consists of two integers whose difference equals $2x+1$.

If $a\ge x$, then $a-x\ge 0$, and from $(a-x)(a+x+1)=x$ we obtain $a-x\mid x$. Writing $a-x=d$ gives $a+x+1=\frac{x}{d}$, hence

$$\frac{x}{d}-d=2x+1.$$

Multiplying by $d$ yields

$$x-d^2=2xd+d.$$

Rearranging,

$$x(1-2d)=d^2+d.$$

Since $x>0$, the left side is nonpositive for $d\ge 1$, while the right side is positive, contradiction.

If $a<x$, then $a-x<0$, so write $a-x=-d$ with $d>0$. Then

$$(a+x+1)=\frac{x}{-d}=-\frac{x}{d},$$

which is negative, contradicting $a+x+1>0$.

Thus no integer solutions exist for $a(a+1)=b(b+1)(b+2)(b+3)$.

Therefore no number has properties $(2)$ and $(4)$.

This completes the proof. ∎

Verification of Key Steps

The crucial reduction was

$$a(a+1)=x(x+2)\quad \Rightarrow \quad (a-x)(a+x+1)=x.$$

Expanding both sides confirms correctness since

$$(a-x)(a+x+1)=a^2+a-x^2-2x=a(a+1)-x(x+2).$$

The exclusion of solutions relied on sign analysis of $a-x$. The cases $a\ge x$ and $a<x$ cover all integers $a$. In each case, one factor becomes negative or leads to incompatible divisibility constraints with $x>0$, preventing equality.

A second check is numerical consistency: for small $b$, values of $b(b+1)(b+2)(b+3)$ never equal $a(a+1)$, matching the structural obstruction proved above.

Alternative Approaches

A different method is to compare growth rates of consecutive products and show that the sequence $a(a+1)$ and $b(b+1)(b+2)(b+3)$ cannot intersect by analyzing monotonic gaps between consecutive values. Another approach uses modular arithmetic modulo $3$ and $8$ to restrict possible residues of both sides, eventually forcing a contradiction, though it is less direct than the quadratic factorization method used above.