we-solve (beta)

Math-783-A2 · revision r1 · on Math-783

gpt pro 5.2 attempt on Math-783

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem 367: Products of $2$-full parts in short intervals
  2. FORMAL RESTATEMENT
  3. QUICK LITERATURE/CONTEXT CHECK
  4. ATTACK PLAN
  5. WORK (complete disproof of $(\dagger)$ for $k\ge 3$)
  6. 1. The case $k\le 2$ (trivial upper bound $\ll n^2$)
  7. 2. Pell solutions and the sequence $n_j$
  8. Lemma 2.1.
  9. 3. Forcing a large power of $5$ to divide $n_j+2$
  10. Lemma 3.1.
  11. Lemma 3.2.
  12. 4. Lower bound for the product and failure of $(\dagger)$ for $k\ge 3$
  13. Relating $5^t$ to $\log n$.
  14. Conclusion (disproof of $(\dagger)$).
  15. VERIFICATION
  16. META-CHECK (common failure modes)
  17. FINAL
  18. COMPLETION ESTIMATE

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 $j\ge 0$, \[ B_2(n_j)=n_j\quad\text{and}\quad B_2(n_j+1)=n_j+1. \]
Proof: We have $n_j=8y_j^2=2^3\cdot y_j^2$. In the prime factorization of $y_j^2$, every exponent is even; multiplying by $2^3$ makes the $2$-exponent at least $3$. Hence every prime exponent in $n_j$ is $\ge 2$, so $n_j$ is powerful and $B_2(n_j)=n_j$. Also $n_j+1=x_j^2$ is a perfect square, hence powerful, so $B_2(n_j+1)=n_j+1$. ∎

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 \[ \alpha^{\pm 3\cdot 5^{t-1}} = -1 + 5^t\left(a_t \pm b_t\sqrt{8}\right). \tag{2} \]
Proof: We argue by induction on $t$. For $t=1$, $2$ is exactly $1$, with $(a_1,b_1)=(20,7)$. Assume $2$ holds for some $t\ge 1$ with the $+$ sign, i.e. $\alpha^{3\cdot 5^{t-1}}=-1+5^t w$ with $w\in\mathbb{Z}[\sqrt{8}]$. Then \[ \alpha^{3\cdot 5^{t}} = \left(\alpha^{3\cdot 5^{t-1}}\right)^5 = (-1+5^t w)^5. \] Expanding by the binomial theorem, \[ (-1+5^t w)^5 = -1 + 5\cdot 5^t w + \sum_{j=2}^5 \binom{5}{j} (-1)^{5-j} (5^t w)^j. \] Every term in the sum over $j\ge 2$ is divisible by $5^{2t}$ in $\mathbb{Z}[\sqrt{8}]$, hence (since $t\ge 1$) divisible by $5^{t+1}$. Therefore \[ (-1+5^t w)^5 = -1 + 5^{t+1} w' \] for some $w'\in\mathbb{Z}[\sqrt{8}]$. Writing $w'=a_{t+1}+b_{t+1}\sqrt{8}$ gives $2$ for $t+1$. The statement for the $-$ sign follows by applying the same argument to $\alpha^{-1}$ (or by conjugation $\sqrt{8}\mapsto -\sqrt{8}$). ∎

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 $t\ge 1$, \[ 5^t \mid (n_{j_t}+2). \]
Proof: We first record a closed form for $n_j$. From $x_j+y_j\sqrt{8}=\alpha^j$ and conjugation, \[ x_j = \frac{\alpha^j+\alpha^{-j}}{2},\qquad y_j = \frac{\alpha^j-\alpha^{-j}}{2\sqrt{8}}. \] Hence \[ n_j = 8y_j^2 = 8\left(\frac{\alpha^j-\alpha^{-j}}{2\sqrt{8}}\right)^2 = \frac{\alpha^{2j}-2+\alpha^{-2j}}{4}. \] Therefore \[ 4(n_j+2)=\alpha^{2j}+6+\alpha^{-2j}. \tag{3} \] Rewrite $\alpha^{2j}=\alpha^{-1}\alpha^{2j+1}$ and $\alpha^{-2j}=\alpha\alpha^{-(2j+1)}$ to get \[ 4(n_j+2)=\alpha^{-1}\alpha^{2j+1}+6+\alpha\alpha^{-(2j+1)}. \tag{4} \] Now take $j=j_t$ so $2j+1=3\cdot 5^{t-1}$. Using Lemma 3.1, \[ \alpha^{\pm(2j_t+1)}=\alpha^{\pm 3\cdot 5^{t-1}}=-1+5^t(a_t\pm b_t\sqrt{8}). \] Substitute into $4$ and use $\alpha^{\pm 1}=3\pm\sqrt{8}$: \begin{align*} 4(n_{j_t}+2) &=(3-\sqrt{8})\bigl(-1+5^t(a_t+b_t\sqrt{8})\bigr)+6+(3+\sqrt{8})\bigl(-1+5^t(a_t-b_t\sqrt{8})\bigr)\\ &=\underbrace{(3-\sqrt{8})(-1)+(3+\sqrt{8})(-1)+6}_{=0} +5^t\Bigl((3-\sqrt{8})(a_t+b_t\sqrt{8})+(3+\sqrt{8})(a_t-b_t\sqrt{8})\Bigr). \end{align*} A direct expansion shows the bracket equals $6a_t-16b_t\in\mathbb{Z}$, so \[ 4(n_{j_t}+2)=5^t(6a_t-16b_t), \] which implies $5^t\mid 4(n_{j_t}+2)$. Since $\gcd(4,5)=1$, we conclude $5^t\mid (n_{j_t}+2)$. ∎

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%.