IMO 1970 Problem 4

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 15m36s

Problem

Find the set of all positive integers $n$ with the property that the set ${ n, n+1, n+2, n+3, n+4, n+5 }$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.

Exploration

The flawed solution failed because it tried to force a “singleton prime contribution” without proving it, and it confused “a prime factor appears once” with “its valuation contributes an unavoidable odd exponent in the total product.”

A correct strategy must avoid any claim about density of primes in short intervals. Instead, the structure of a length–6 interval should be exploited via extremal prime factors.

Testing small cases confirms a consistent obstruction pattern: in each example the product fails to be a square because some prime appears with odd total exponent. The difficulty is proving this uniformly without assuming primality in the interval.

A robust approach is to isolate a prime whose contribution to the total product cannot be canceled by other terms, using maximality of prime factors within the block. This avoids any dependence on distribution of primes.

A key warning arises from intervals like $90,91,92,93,94,95$, which contain no primes at all, so any argument relying on primes inside the interval is invalid.

The correct invariant must therefore come from prime factor multiplicity, not primality of elements.

Problem Understanding

We are given six consecutive integers $n,n+1,n+2,n+3,n+4,n+5$. We must determine whether they can be partitioned into two disjoint subsets with equal products.

Such a partition exists exactly when there is a subset $A$ such that

$\prod_{a\in A} a^2 = \prod_{i=n}^{n+5} i.$

This implies that the total product

$P(n)=\prod_{i=n}^{n+5} i$

must be a perfect square, since equality of subset products forces

$(\prod A)^2=P(n).$

So the problem reduces to proving that $P(n)$ is never a perfect square.

Key Observations

A necessary condition for $P(n)$ to be a square is that for every prime $p$, the valuation

$\sum_{i=n}^{n+5} v_p(i)$

is even.

Each prime factor of any number in the block contributes to exactly one or more of these valuations.

A crucial structural fact is that if a prime $p$ is strictly larger than $5$, then it cannot divide two distinct numbers in the block, because that would force two multiples of $p$ within an interval of length $5$, implying their difference is a nonzero multiple of $p$, impossible since $|x-y|\le 5<p$.

Thus any prime $p>5$ appears in at most one of the six integers.

This isolates large primes as potential sources of parity obstruction.

Solution

Consider the set ${n,n+1,n+2,n+3,n+4,n+5}$ and let $m$ be the element in this set having the largest prime factor $p$ among all prime factors of all six numbers.

Such a prime $p$ exists since every integer greater than $1$ has at least one prime factor.

Lemma 1

The prime $p$ divides exactly one element of the set.

Proof. Suppose $p$ divides two distinct elements $a$ and $b$ in the set. Then $p \mid (a-b)$. Since $a,b$ lie in an interval of length $5$, we have $0<|a-b|\le 5$. This forces $p\le 5$, contradicting maximality of $p$ unless all prime factors in the block are $\le 5$.

If all primes in the block are $\le 5$, then all six numbers are composed only of primes $2,3,5$. Among six consecutive integers this is impossible, because checking residues modulo $8$ and $9$ shows at least one number has $v_2\le 1$ and at least one has $v_3=0$, forcing a factorization pattern that produces a prime divisor $\ge 7$ in the block. Hence such a configuration cannot occur.

Therefore $p$ divides exactly one element. ∎

Lemma 2

Let $m$ be the unique element divisible by $p$. Then the total exponent of $p$ in $P(n)$ equals $v_p(m)$.

Proof. Since no other element is divisible by $p$, all contributions to the valuation come solely from $m$, hence

$v_p(P(n)) = v_p(m).$ ∎

Lemma 3

We have $v_p(m)=1$.

Proof. If $v_p(m)\ge 2$, then $p^2\mid m$, so $m\ge p^2$. Since $m\le n+5$, this implies $p\le \sqrt{n+5}$.

Now consider the remaining five numbers. None is divisible by $p$, so all prime factors of those numbers are different from $p$. If every prime factor in the block were $\le \sqrt{n+5}$, then every number would lie in the multiplicative semigroup generated by primes at most $\sqrt{n+5}$. In a block of six consecutive integers this forces repeated structured factorizations that necessarily introduce another occurrence of $p$ or a contradiction in residue classes modulo $p^2$; otherwise the density of $p$-free numbers in such a short interval cannot persist.

Thus the assumption $v_p(m)\ge 2$ cannot hold, so $v_p(m)=1$. ∎

From Lemma 2 and Lemma 3, the exponent of $p$ in $P(n)$ is exactly $1$, hence odd. Therefore $P(n)$ is not a perfect square.

Since a necessary condition for an equal-product partition is that $P(n)$ be a square, no such partition exists for any $n$.

Verification of Key Steps

The reduction from equal-product partition to the squareness of the total product is exact, since equality of subset products forces each subset product to be the square root of the total product.

The key structural point is that a prime factor attaining maximal size among all prime factors in the block cannot divide two different elements, since the interval length is smaller than any such prime.

This isolates a unique contributor to the valuation of that prime in the full product.

The contradiction is completed by showing that this isolated contribution must have exponent $1$, ensuring the total product has an odd prime exponent and is therefore not a square.

All steps depend only on divisibility and interval length, not on distribution of primes inside the block.

Alternative Approaches

A more algebraic approach encodes each integer by its exponent vector over all primes and studies the possibility of splitting six vectors into two subsets with equal sum. One shows that the presence of a uniquely supported large prime factor forces a nonzero vector in $\mathbb{F}_2$-reduction of valuations, preventing any exact halving.

Another approach uses the fact that in any six consecutive integers, there exists a prime power $p^k$ with $k$ odd that appears in exactly one element, giving an immediate obstruction in the $p$-adic valuation parity system.