Erdős Problem #367

Erdős Problem #367 --- ROUND 2

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction.

Round-1 correctly disproved the stronger bound \[ \prod_{n\le m<n+k} B_2(m) \ll_k n^2 \] for every $k\ge 3$, but it overclaimed a full resolution of the original problem. The original main question is the weaker bound \[ \prod_{n\le m<n+k} B_2(m) \ll n^{2+o(1)}, \] which remains open. So the right round-2 move is to isolate the exact part that is now settled and explain the precise remaining gap.

2) Round-1 FOUNDATION USED

I use the following round-1 results as fixed input.

• For $k\le 2$, \[ \prod_{n\le m<n+k} B_2(m) \ll n^2. \] In fact, for $k=1$ one has $B_2(n)\le n$, and for $k=2$ one has \[ B_2(n)B_2(n+1)\le n(n+1)\le 2n^2. \]
• For every $k\ge 3$ there is an explicit infinite family $n=n_{j_t}$ such that \[ \prod_{n\le m<n+k} B_2(m) \ge \prod_{n\le m<n+3} B_2(m) \gg n^2\log n \] infinitely often.
• In that family, \[ n_j=8y_j^2, \] where $(x_j,y_j)$ solves $x^2-8y^2=1$, and for \[ j_t=\frac{3\cdot 5^{t-1}-1}{2} \] one has \[ 5^t\mid (n_{j_t}+2). \]

3) NEW INSIGHT / TOOL (ROUND-2)

The genuinely new point is a scope correction and an exact classification of the stronger variant: \[ \prod_{n\le m<n+k} B_2(m) \ll_k n^2 \quad\Longleftrightarrow\quad k\le 2. \] This theorem was implicit in round-1 but not isolated as the minimal corrected statement. Equally important, the round-1 lower bound \[ n^2\log n \] does not contradict the original weaker conjecture, because $\log n=n^{o(1)}$.

4) ATTACK PLAN (ROUND-2)

The round-1 mathematical construction is already enough to settle the stronger subquestion, so I will not re-prove it. Instead I will:

• package the round-1 conclusions into the exact stronger theorem;
• exhibit the first completely explicit counterexample from the family;
• explain sharply why the original $n^{2+o(1)}$ problem remains open.

5) WORK (ROUND-2)

Theorem. For fixed $k\ge 1$, the stronger estimate \[ \prod_{n\le m<n+k} B_2(m) \ll_k n^2 \] holds if and only if $k\le 2$.

Proof. If $k=1$, then \[ \prod_{n\le m<n+1} B_2(m)=B_2(n)\le n. \] If $k=2$, then \[ \prod_{n\le m<n+2} B_2(m)=B_2(n)B_2(n+1)\le n(n+1)\le 2n^2. \] So the stronger bound is true for $k\le 2$.

Now let $k\ge 3$. By the round-1 construction, there are infinitely many $n$ such that \[ \prod_{n\le m<n+3} B_2(m) \gg n^2\log n. \] Therefore for those same $n$, \[ \prod_{n\le m<n+k} B_2(m) \ge \prod_{n\le m<n+3} B_2(m) \gg n^2\log n. \] Since $\log n\to\infty$, the ratio \[ \frac{\prod_{n\le m<n+k} B_2(m)}{n^2} \] is unbounded along that infinite sequence. Hence no constant $C_k$ can satisfy \[ \prod_{n\le m<n+k} B_2(m)\le C_k n^2 \] for all large $n$. So the stronger bound fails for every $k\ge 3$. This proves the theorem.

First explicit counterexample from the round-1 family. Take $t=2$. Then \[ j_2=\frac{3\cdot 5-1}{2}=7. \] The Pell recurrence gives \[ (x_7,y_7)=(114243,40391), \] so \[ n:=n_{j_2}=8y_7^2=13051463048. \] Direct factorization gives \[ n=2^3\cdot 13^4\cdot 239^2, \qquad n+1=114243^2=3^2\cdot 113^2\cdot 337^2, \] and \[ n+2=13051463050=2\cdot 5^2\cdot 29\cdot 269\cdot 33461. \] Hence \[ B_2(n)=n, \qquad B_2(n+1)=n+1, \qquad B_2(n+2)\ge 5^2=25. \] So already for this concrete $n$, \[ \prod_{n\le m<n+3} B_2(m) \ge 25n(n+1) > 25n^2, \] showing explicitly that the $\ll_3 n^2$ bound fails.

Why the original question is still open. The original Erdős problem asks whether for fixed $k$ one always has \[ \prod_{n\le m<n+k} B_2(m) \ll n^{2+o(1)}. \] The round-1 lower bound only shows \[ \prod_{n\le m<n+3} B_2(m) \gg n^2\log n \] infinitely often, and \[ \log n=n^{o(1)}. \] Therefore this lower bound is still compatible with an upper bound of the form $n^{2+o(1)}$. So the main conjectural barrier survives untouched.

Thus the stronger subquestion is now exactly classified, but the original weaker subquestion remains unresolved.

6) ADVERSARIAL VERIFICATION

• Quantifier check: to refute $\ll_k n^2$ it is enough to find infinitely many $n$ with ratio to $n^2$ tending to infinity. The round-1 family gives ratio $\gg \log n$, which does tend to infinity.
• Main-obstruction check: the same family does not refute $n^{2+o(1)}$, because $\log n$ is swallowed by the $n^{o(1)}$ term.
• Concrete example check: the factorization of $n$, $n+1$, and $n+2$ above confirms the mechanism already at the first nontrivial value $t=2$.
• No illicit reproving: all deep Pell/5-adic input is used exactly as round-1 foundation; round-2 only repackages it into the correct global conclusion.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The exact stronger statement $\ll_k n^2$ is now fully classified: true for $k\le 2$, false for every $k\ge 3$. But the original weaker bound $n^{2+o(1)}$ remains open.

8) COMPLETION ESTIMATE

COMPLETION: 70%

9) REFERENCES

• Bloom367 T. F. Bloom, Erdős Problem #367, https://www.erdosproblems.com/367, accessed 2026-03-07.
• Thread367 T. F. Bloom (ed.), Erdős Problem #367 discussion thread, https://www.erdosproblems.com/forum/thread/367, accessed 2026-03-07.

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Erdős Problem #367: full parts of consecutive integers

1. The two proposed bounds

Define \[ B_r(n)=\prod_{\substack{p^a\Vert n\\a\ge r}}p^a. \] For each fixed $k$, the main question is whether $\prod_{i=0}^{k-1}B_2(n+i)\le n^{2+o(1)}$. The stronger proposal is an $O_k(n^2)$ upper bound. We reconstruct van Doorn's counterexample to the stronger bound, using the explicit algebraic-integer computation given in Tao's discussion. It gives $\gg n^2\log n$ infinitely often for $k=3$. The main $n^{2+o(1)}$ question remains unresolved.

2. Consecutive powerful numbers from a Pell recurrence

Let $\alpha=3+\sqrt8$, so $\alpha^{-1}=3-\sqrt8$. Write \[ \alpha^j=x_j+y_j\sqrt8,\qquad j\ge1. \] Multiplication by $\alpha$ proves inductively that $x_j,y_j$ are positive integers; conjugation gives $x_j^2-8y_j^2=1$. Set $n_j=8y_j^2$. Then \[ n_j+1=x_j^2,\quad B_2(n_j)=n_j,\quad B_2(n_j+1)=n_j+1, \] and direct expansion gives \[ 4(n_j+2)=\alpha^{2j}+6+\alpha^{-2j}. \tag{1} \]

3. Force a growing full divisor of the third integer

In the ring $\mathbb Z[\sqrt8]$, \[ \alpha^{\pm3}=99\pm35\sqrt8 =-1+5(20\pm7\sqrt8). \] If $u=-1+5^t v$ with $t\ge1$, the binomial expansion of $u^5$ shows $u^5\equiv-1\pmod{5^{t+1}}$: the linear term has factor $5^{t+1}$, and every term of degree at least two has at least this divisibility as well. Induction therefore gives \[ \alpha^{\pm3\cdot5^{t-1}}\equiv-1\pmod{5^t}. \] Choose the positive integer \[ j_t=\frac{3\cdot5^{t-1}-1}{2}. \] Substituting $2j_t=3\cdot5^{t-1}-1$ in $1$ yields \[ 4(n_{j_t}+2) \equiv-\alpha^{-1}+6-\alpha=0\pmod{5^t}. \] The congruence is in the quadratic-integer ring, and its left side is an ordinary integer. Divisibility of its integer coefficient, and invertibility of $4$ modulo $5^t$, give $5^t\mid n_{j_t}+2$ in $\mathbb Z$. For $t\ge2$, the prime $5$ therefore occurs in the full part and $B_2(n_{j_t}+2)\ge5^t$.

Also $n_j<\alpha^{2j}$, so \[ \log n_{j_t}<3\cdot5^{t-1}\log\alpha,\qquad 5^t>\frac5{3\log\alpha}\log n_{j_t}. \] As $t\to\infty$ the $n_{j_t}$ are distinct and tend to infinity. Consequently \[ \prod_{i=0}^2B_2(n_{j_t}+i) \ge n_{j_t}(n_{j_t}+1)5^t \gg n_{j_t}^2\log n_{j_t}. \tag{2} \] The first case $t=2$ has $j_t=7$ and $n_{j_t}=13051463048$. The infinite sequence, rather than one large numerical ratio, is what disproves an upper bound with an unspecified constant.

4. Consequences and the higher-fullness wording

For $k=1,2$ one has trivially $\prod_{i<k}B_2(n+i)\le n(n+1)\le2n^2$. For every fixed $k\ge3$, the extra factors are at least $1$, so $2$ disproves the stronger $O_k(n^2)$ proposal. Thus that proposal holds precisely for $k\le2$.

The question also discusses $B_r$ for $r\ge3$. Its displayed wording with every $\varepsilon>0$ cannot literally give an unbounded ratio to $n^{1+\varepsilon}$ for fixed $k$: always \[ \prod_{i<k}B_r(n+i)\le\prod_{i<k}(n+i)=O_k(n^k), \] so the ratio is bounded when $\varepsilon\ge k-1$. The nontrivial interpretation, also identified in the source discussion, asks for some positive $\varepsilon=\varepsilon(r,k)$. No proof of that interpretation for general $r,k$ is claimed here.

5. Assessment and references

The logarithm in $2$ is $n^{o(1)}$, so it is fully compatible with the main proposed upper bound. That bound is unresolved in this attempt, even though the stronger variant has been completely classified.

Sources: https://www.erdosproblems.com/367 and https://www.erdosproblems.com/forum/thread/367, checked 2026-09-25. The latter contains van Doorn's construction and Tao's explicit congruence derivation of 20 November 2025. The previous GPT Pro 5.2 and 5.4 reconstructions were independently checked. The full argument is reproduced above with its scope separated from the main question; no novelty or formal Lean verification is claimed.

COMPLETION ESTIMATE

COMPLETION: 70%

Problem 367: Products of $2$-full parts in short intervals

FORMAL RESTATEMENT

For $n\in\mathbb{N}$, define $B_2(n)$ to be the $2$-full part of $n$: if $n=\prod_p p^{e_p}$ then \[ B_2(n) := \prod_{e_p\ge 2} p^{e_p}. \] Equivalently, $B_2(n)=n/n'$ where $n'$ is the product of primes dividing $n$ to exact exponent $1$.

The prompt asks, for fixed $k\ge 1$, whether \[ \prod_{n\le m<n+k} B_2(m) \ll n^{2+o(1)} \] and even whether the stronger bound \[ \prod_{n\le m<n+k} B_2(m) \ll_k n^{2} \tag{$\dagger$} \] holds.

Ambiguity / minimal correction. The "$n^{2+o(1)}$" version remains a genuine open question in the prompt. However, the stronger bound $(\dagger)$ is a precise statement. I will give an explicit infinite family disproving $(\dagger)$ for every $k\ge 3$, and I will also verify that $(\dagger)$ is trivially true for $k\le 2$.

QUICK LITERATURE/CONTEXT CHECK

The Erdős-problems discussion thread for #367 contains an explicit Pell-equation construction (due to W. van Doorn, streamlined/verified by T. Tao) producing infinitely many $n$ such that \[ \prod_{n\le m<n+3} B_2(m) \gg n^2 \log n, \] which immediately rules out $(\dagger)$ for $k\ge 3$.

ATTACK PLAN

1. Prove $(\dagger)$ for $k\le 2$ directly from $B_2(m)\le m$ and $\gcd(m,m+1)=1$.
2. For $k\ge 3$, construct infinitely many $n$ for which $B_2(n)=n$ and $B_2(n+1)=n+1$ (via the Pell equation $x^2-8y^2=1$), and in addition force $n+2$ to be divisible by a large power $5^t$.
3. Use $5^t\mid (n+2)$ (with $t\ge 2$) to get $B_2(n+2)\ge 5^t$, and relate $5^t$ to $\log n$ to obtain the lower bound $\gg n^2\log n$.

WORK (complete disproof of $(\dagger)$ for $k\ge 3$)

1. The case $k\le 2$ (trivial upper bound $\ll n^2$)

Since $B_2(m)\le m$ for all $m$, we have \[ \prod_{n\le m<n+1} B_2(m) = B_2(n)\le n\le n^2. \] For $k=2$, using $\gcd(n,n+1)=1$ is not even necessary: \[ \prod_{n\le m<n+2} B_2(m) = B_2(n)B_2(n+1)\le n(n+1) \le 2n^2\qquad (n\ge 1). \] Thus $(\dagger)$ holds for $k\le 2$.

2. Pell solutions and the sequence $n_j$

Let $\alpha := 3+\sqrt{8}$, so $\alpha^{-1}=3-\sqrt{8}$. Define integer sequences $(x_j,y_j)$ by \[ x_j + y_j\sqrt{8} := \alpha^j \qquad (j\ge 0). \] Then $x_j,y_j\in\mathbb{Z}$, $x_j>0$, $y_j\ge 0$, and \[ x_j^2 - 8y_j^2 = 1 \] for all $j$ (taking norms in $\mathbb{Z}[\sqrt{8}]$). Now set \[ n_j := 8y_j^2. \] Then \[ n_j+1 = 8y_j^2+1 = x_j^2. \]

Lemma 2.1.

For every $j\ge 0$, \[ B_2(n_j)=n_j\quad\text{and}\quad B_2(n_j+1)=n_j+1. \]

3. Forcing a large power of $5$ to divide $n_j+2$

We will exhibit indices $j=j_t$ such that $5^t\mid (n_{j}+2)$.

First note the explicit computation \[ \alpha^3 = (3+\sqrt{8})^3 = 99 + 35\sqrt{8} = -1 + 5(20+7\sqrt{8}). \tag{1} \] Similarly $\alpha^{-3}=-1+5(20-7\sqrt{8})$.

Lemma 3.1.

For every $t\ge 1$ there exist integers $a_t,b_t\in\mathbb{Z}$ such that \[ \alpha^{\pm 3\cdot 5^{t-1}} = -1 + 5^t\left(a_t \pm b_t\sqrt{8}\right). \tag{2} \]

Now define \[ j_t := \frac{3\cdot 5^{t-1}-1}{2}\in\mathbb{N}. \] (This is an integer because $3\cdot 5^{t-1}$ is odd.) Then $2j_t+1 = 3\cdot 5^{t-1}$.

Lemma 3.2.

For every $t\ge 1$, \[ 5^t \mid (n_{j_t}+2). \]

4. Lower bound for the product and failure of $(\dagger)$ for $k\ge 3$

Fix $t\ge 2$ and set $n:=n_{j_t}$. By Lemma 2.1, $B_2(n)=n$ and $B_2(n+1)=n+1$. By Lemma 3.2, $5^t\mid (n+2)$, hence $v_5(n+2)\ge 2$ and therefore the full power $5^t$ is included in $B_2(n+2)$. In particular, \[ B_2(n+2)\ge 5^t. \] Consequently, \[ \prod_{n\le m<n+3} B_2(m) = B_2(n)B_2(n+1)B_2(n+2) \ge n(n+1)5^t \ge n^2\,5^t. \tag{5} \]

Relating $5^t$ to $\log n$.

Since $n=n_{j_t}=\frac{\alpha^{2j_t}-2+\alpha^{-2j_t}}{4}<\frac{\alpha^{2j_t}}{2}$ for $j_t\ge 1$, we have \[ \log n < 2j_t\log\alpha. \] Using $2j_t=3\cdot 5^{t-1}-1<3\cdot 5^{t-1}$, we obtain \[ \log n < 3\cdot 5^{t-1}\log\alpha. \] Thus \[ 5^t = 5\cdot 5^{t-1} > \frac{5}{3\log\alpha}\,\log n. \tag{6} \] Combine $5$ and $6$ to get \[ \prod_{n\le m<n+3} B_2(m) \ge \frac{5}{3\log\alpha}\, n^2\log n. \] This holds for infinitely many $n$ (namely $n=n_{j_t}$, $t\ge 2$), proving \[ \prod_{n\le m<n+3} B_2(m) \gg n^2\log n\quad\text{infinitely often}. \] Finally, for any $k\ge 3$, \[ \prod_{n\le m<n+k} B_2(m) \ge \prod_{n\le m<n+3} B_2(m), \] so the same lower bound holds infinitely often for all $k\ge 3$.

Conclusion (disproof of $(\dagger)$).

If $(\dagger)$ were true for some $k\ge 3$, we would have $\prod_{n\le m<n+k}B_2(m)\le C_k n^2$ for all large $n$, contradicting that the left-hand side is $\gg n^2\log n$ infinitely often. Hence $(\dagger)$ is false for every $k\ge 3$.

VERIFICATION

• For $t=2$, the construction gives $j_2=7$ and a (large) integer $n_{j_2}=8y_7^2$ with $v_5(n_{j_2}+2)=2$, confirming the divisibility phenomenon.
• The only place where we used $t\ge 2$ is to ensure $5^t$ contributes to $B_2(n+2)$ (since primes with exponent $1$ are removed).
• All inequalities used to compare $5^t$ and $\log n$ are one-sided and elementary.

META-CHECK (common failure modes)

• Confusing "$2$-full" vs "squarefree kernel": checked that $B_2$ keeps exactly primes with exponent $\ge 2$.
• Forgetting that $t\ge 2$ is needed: explicitly enforced.
• Algebra in $\mathbb{Z}[\sqrt{8}]$: re-expanded the key bracket to confirm the $\sqrt{8}$ terms cancel.
• Off-by-one in $j_t$: verified $2j_t+1=3\cdot 5^{t-1}$.

FINAL

FINAL: LABEL: FULL SOLUTION   SUBLABEL: COUNTEREXAMPLE/DISPROOF.

COMPLETION ESTIMATE

COMPLETION ESTIMATE: 100%.