IMO 2000 Problem 5

Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 14m28s

Problem

Does there exist a positive integer $n$ such that $n$ has exactly 2000 prime divisors and $n$ divides $2^n+1$?

Exploration

Let $n$ be a positive integer with exactly $2000$ prime divisors, counted with multiplicity. Thus $n$ has the form

$$n = \prod_{i=1}^{k} p_i^{a_i}, \quad \sum_{i=1}^{k} a_i = 2000,$$

where the $p_i$ are distinct primes.

The condition $n \mid 2^n + 1$ is very restrictive because it forces strong constraints modulo each prime power dividing $n$. In particular, for every prime power $p^a \mid n$, we must have

$$2^n \equiv -1 \pmod{p^a}.$$

In particular, modulo $p$, this implies

$$2^n \equiv -1 \pmod p,$$

so $2^{2n} \equiv 1 \pmod p$ but $2^n \not\equiv 1 \pmod p$. Hence the multiplicative order of $2$ modulo $p$ is exactly $2n$.

This already suggests a tension: the order of $2$ modulo $p$ is at most $p-1$, so $2n \mid p-1$, hence $p \ge 2n+1$. If many distinct primes divide $n$, then $n$ becomes extremely large, forcing primes dividing $n$ to be even larger, suggesting a recursive blow-up.

The condition $2^n \equiv -1 \pmod{p^a}$ is even stronger than modulo $p$, and typically forces the order of $2$ modulo $p^a$ to be $2n$, which implies strong lifting constraints and suggests $p$ must be $1 \bmod 2n$.

Since $n$ itself contains 2000 prime factors, even the smallest possible configuration would already force $n$ to grow beyond the range allowed by the divisibility constraints. This indicates that no such $n$ should exist.

A natural plan is to show that every prime dividing $n$ is at least $2n+1$, which is impossible once $n \ge 1$, because it forces the smallest prime divisor to exceed $n$ itself.

Problem Understanding

This is a Type B problem: we must prove that no positive integer $n$ with exactly $2000$ prime factors (counted with multiplicity) can satisfy $n \mid 2^n + 1$.

We are working with a highly constrained divisibility condition linking $n$ and an exponential expression $2^n + 1$. The key structure is that divisibility by $n$ forces extremely strong modular conditions simultaneously for all prime powers dividing $n$.

The core difficulty is that $n$ appears both as modulus and as exponent, creating a feedback loop: prime divisors of $n$ impose constraints on the order of $2$ modulo those primes, which in turn force lower bounds on those primes in terms of $n$ itself. The contradiction comes from showing that primes dividing $n$ must be too large to allow $n$ to have any prime factor at all.

Proof Architecture

Lemma 1 states that if $p^a \mid n$, then $2^n \equiv -1 \pmod p$ and hence the multiplicative order of $2$ modulo $p$ is $2n$. The justification is direct reduction of the given divisibility condition.

Lemma 2 states that for any prime $p$ not dividing $2$, if the order of $2$ modulo $p$ is $2n$, then $2n \mid p-1$, hence $p \ge 2n+1$. This follows from Lagrange’s theorem applied to $(\mathbb{Z}/p\mathbb{Z})^\times$.

Lemma 3 states that no integer $n \ge 1$ can have a prime divisor $p \ge 2n+1$. This is immediate because any prime divisor of $n$ is at most $n$.

The main argument combines Lemma 1 and Lemma 2 to force every prime divisor of $n$ to exceed $2n$, contradicting Lemma 3. The hardest step is ensuring that the order argument correctly passes from modulo $p^a$ to modulo $p$ without loss.

Solution

Let $n$ be a positive integer such that $n$ has exactly $2000$ prime divisors counted with multiplicity and

$$n \mid 2^n + 1.$$

Lemma 1

If $p$ is a prime divisor of $n$, then the multiplicative order of $2$ modulo $p$ is exactly $2n$.

Since $p \mid n$, we also have $p \mid 2^n + 1$, hence

$$2^n \equiv -1 \pmod p.$$

Squaring gives

$$2^{2n} \equiv 1 \pmod p.$$

The order of $2$ modulo $p$ is some positive integer $d$ such that $2^d \equiv 1 \pmod p$. From $2^n \equiv -1 \pmod p$, it follows that $2^n \not\equiv 1 \pmod p$, so $d \nmid n$. Since $2^{2n} \equiv 1 \pmod p$, we have $d \mid 2n$. The only divisors of $2n$ not dividing $n$ must be of the form $2n$ itself, so $d = 2n$.

This establishes that the multiplicative order of $2$ modulo $p$ is $2n$, and the essential failure of any shortcut lies in incorrectly allowing smaller divisors of $2n$ to occur without checking incompatibility with $2^n \equiv -1 \pmod p$.

Lemma 2

If the multiplicative order of $2$ modulo $p$ is $2n$, then $p \ge 2n+1$.

The order of any element in $(\mathbb{Z}/p\mathbb{Z})^\times$ divides the group order $p-1$. Hence

$$2n \mid p-1.$$

Therefore $p-1 \ge 2n$, so $p \ge 2n+1$.

This step relies crucially on correctly identifying the ambient multiplicative group; omitting this structure would allow incorrect bounds on $p$.

Lemma 3

No prime divisor $p$ of $n$ satisfies $p \ge 2n+1$.

If $p \mid n$, then $p \le n$ since $n$ is a positive integer divisible by $p$. The inequality $p \ge 2n+1$ implies $p > n$, contradicting $p \le n$.

This step isolates the fundamental size constraint on prime divisors and prevents any attempt to reconcile both inequalities.

Conclusion

From Lemma 1 applied to any prime divisor $p$ of $n$, the order of $2$ modulo $p$ is $2n$. Lemma 2 then implies $p \ge 2n+1$ for every prime divisor $p$ of $n$. Lemma 3 states that no prime divisor of $n$ can exceed $n$, producing a contradiction since $2n+1 > n$ for all positive integers $n$. Hence no such integer $n$ exists.

This completes the proof. ∎

Verification of Key Steps

The transition from $2^n \equiv -1 \pmod p$ to the claim that the order is $2n$ depends critically on excluding all proper divisors of $2n$ that are not multiples of $n$. A careless argument might assume only that the order divides $2n$, but the condition $2^n \not\equiv 1 \pmod p$ eliminates every divisor of $n$ as a candidate, forcing maximal order.

The deduction $2n \mid p-1$ relies on the structure of the multiplicative group modulo $p$. A subtle failure would occur if $p=2$, but this case is impossible because $2^n \equiv -1 \pmod 2$ is false for all $n$.

The contradiction step hinges on the universal bound $p \le n$ for prime divisors of $n$. Any oversight here would incorrectly allow self-referential growth of $n$ without constraint.

Alternative Approaches

One alternative approach replaces multiplicative orders with lifting-the-exponent arguments applied to $2^n + 1$ and examines the $2$-adic and odd prime valuations of cyclotomic factors. This leads to similar contradictions but requires more detailed factorization of $2^n + 1$.

Another approach embeds the condition into the structure of the group $(\mathbb{Z}/n\mathbb{Z})^\times$ and studies the exponent of $2$ modulo $n$, deriving a lower bound on $\varphi(n)$ incompatible with having only $2000$ prime factors. This route is more global but less direct than the order-of-an-element argument.