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Terence Tao proves that

$$A(n)=\frac n2-\frac n{2\log n}+o\!\left(\frac n{\log n}\right),$$

where $A(n)$ is the least $t$ such that $n!$ is a product of $t$ factors each at most $n^2$. The lower bound is Stirling's formula. The upper bound has two steps. Tao's post proves the variant in the site's remark: $n!$ is a product of $n-n/\log n+o(n/\log n)$ factors each at most $n$. Tao does not use the greedy decomposition but adapts the approximate-factorization method of Alexeev and others 2025: start from the integers in $[n-n/M,n]$ with no prime factor above $\varepsilon n$, each taken $M$ times, then add and remove factors to correct the surplus or deficit of each prime, keeping the total waste $\sum_i\log(n/a_i)$ at $o(n)$. Pairing consecutive factors, as Stijn Cambie observed and the site credits, then gives $n/2-n/(2\log n)+o(n/\log n)$ factors each at most $n^2$. The argument was posted in the site's discussion thread on 2026-01-03, written as the blueprint section "Erdos problem 392" of the Prime Number Theorem And (PNT+) project on 2026-01-06, and formalized there in the file Erdos392.lean, whose header says the proof is adapted from that post. The formalization's contributors, as the file's history and Alexeev's source list record them, are Tao, Pietro Monticone and Alex Kontorovich with the AI system Aristotle; Tao announced its completion on 2026-02-23. The link pins the revision that the formal-conjectures record names; the file at that revision contains no sorry, axiom or native_decide, and it has not been built or audited here, so it is a formalization link and gives no formalized evidence. The file proves the two upper bounds, Solution_1 (factors at most $n$) and Solution_2 (their pairing, factors at most $n^2$); the matching lower bound from Stirling's formula is not in it. Nat Sothanaphan posted in the thread on 2026-02-26 a write-up dated 2026-02-25, generated by GPT-5.2 Thinking in a near-autonomous process, that expands the same argument for factors at most $n$ and pairs the factors for this problem.

The thread had started from the site's remark that the variant with factors at most $n$ has $A(n)=n-n/\log n+o(n/\log n)$ by a greedy decomposition, and from Stijn Cambie's observation that pairing consecutive factors of such a decomposition answers the question for $n^2$. Participants doubted that the greedy argument proves the variant; the proof recorded here proves the variant by another method and reaches the $n^2$ asymptotic through Cambie's pairing, and Alexeev's source list names Cambie and Tao as the informal authors.

The site labels Problem 392 PROVED (LEAN); its page is linked above as a discussion. Its commentary credits Cambie's pairing reduction and does not name the author of the proof, so the label is not a curator's credit of this result and no reviewed evidence is listed. The formal-conjectures file FormalConjectures/ErdosProblems/392.lean tags the statement erdos_392 as solved and names the PNT+ file above as its formal proof, while leaving its own statement and Cambie's implication with sorry; that Lean has not been built here, so no formalized evidence is listed either. No refereed write-up is recorded. The claim is claimed, and the problem's standing is claimed, proved, through this pending full claim.

Submission note. Posted to the site's forum by Terence Tao on 3 January 2026:

Here is a sketch of how the "modified approximate factorization method" can rigorously allow one to decompose $n!$ as $n - \frac{n}{\log n} + o(\frac{n}{\log n})$ factors $a_1,\dots a_t \leq n$, as claimed by Erdos and Graham (but we will not use the greedy method). By the accounting identity and the Stirling approximation, it suffices to produce $a_1,\dots,a_t$ such that

1. All $a_i$ are bounded above by $n$.
2. The total waste $\sum_{i=1}^t \log(n/a_i)$ is $o(n)$.
3. For every prime $p$, the number of times $p$ divides $a_1 \dots a_t$ matches the number of times $p$ divides $n!$.

The strategy of modified approximate factorization is to start with an approximate factorization that obeys 1 and 2 but not 3, and then add and remove factors to fix 3 without losing 1 or 2.

Let $M$ be large, let $\varepsilon>0$ be small depending on $M$, and suppose that $n$ is large depending on $M,\varepsilon$. The initial choice of the $a_i$ will be: all the integers between $n-n/M$ and $n$ that are not divisible by a prime larger than $\varepsilon n$, with each integer repeated $M$ times. This obeys 1, and the total waste here is $O(n/M)$ which will be good for us as we will eventually take $M$ to infinity. What about 3? The situation depends on the prime p:

3a (very large primes). If $p>n$, then $p$ does not divide either $a_1 \dots a_t$ or $n!$. 3b (large primes). If $\varepsilon n < p \leq n$, then $p$ divides $n!$ $O(n/p)$ times, but does not divide $a_1 \dots a_t$ at all, thus one is short by $O(n/p)$ copies of $p$ for each such $p$. 3c (medium primes). If $\sqrt{n} < p \leq \varepsilon n$, then $p$ divides $n!$ $n/p+O(1)$ times and divides $a_1 \dots a_t$ $n/p + O(M)$ times, this one is either short or surplus by $O(M)$ copies of $p$. 3d (small primes). If $1/\varepsilon < p \leq \sqrt{n}$, then $p$ divides $n!$ $n/(p-1) + O(\log n)$ times and divides $a_1 \dots a_t$ $n/(p-1) + O(M \log n)$ times, so one is either short or surplus by $O(M \log n)$ copies of $p$. 3e (tiny primes). If $p \leq 1/\varepsilon$, then $p$ divides $n!$ $n/(p-1)+O(\log n)$ times and divides $a_1 \dots a_t)$ by $n/(p-1) + O(M \log n) - O(n/\log n)$ times, so one is either surplus by $O(M \log n)$ copies or short by $O(n/\log n)$ copies.

Now one fixes all the shortages and surpluses. For any medium, small, or tiny prime that is surplus, one simply deletes that prime from the relevant factor, generating $\log p$ in waste; because the primes are so small, the net waste in doing so adds up to $O_M(\varepsilon n) + o(n)$ which is acceptable. For the large, medium, or small primes that are short, one adds each such prime $p$ as a new factor, accepting an additional waste of $\log(n/p)$; the net cost here can be computed to be $O_{\varepsilon,M}(n/\log n) +o(n)$, which is also acceptable. The only remaining issue is with the short tiny primes $p$, which are too numerous to add as individual factors; however, one can greedily bundle them into $O_\varepsilon(n / \log^2 n)$ products of size between $\varepsilon n$ and $n$, each generating $O_\varepsilon(1)$ of waste, plus at most one remainder term with $O(\log n)$ of waste. Putting all this together, we can rebalance all the powers of $p$ while still keeping the net waste smaller than any given small multiple of $n$, giving the claim.