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Math-5619-A3 · revision r1 · on Math-5619

gpt pro 5.4 attempt on Math-5619

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Manuscript

Contents
  1. Problem 388. Equal products of two disjoint blocks of consecutive integers
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. 6) ADVERSARIAL VERIFICATION
  8. 7) FINAL
  9. 8) COMPLETION ESTIMATE (MANDATORY)
  10. 9) REFERENCES

Problem 388. Equal products of two disjoint blocks of consecutive integers

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. Round 1 identified a real ambiguity in the problem statement: one can ask either

  • for each fixed pair of lengths $(k_1,k_2)$, are there only finitely many solutions?, or
  • over all varying lengths and shifts, are there only finitely many solutions?

The most promising Round-2 direction is therefore to separate these two meanings rigorously. The key new fact is that the fixed-length problem is already a theorem, whereas the all-parameters problem remains open.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without reproving them.

  • Round-1 Lemma 1. Under the disjointness condition $m_1+k_1\le m_2$, any solution of \[ \prod_{i=1}^{k_1}(m_1+i)=\prod_{j=1}^{k_2}(m_2+j) \] must satisfy $k_1>k_2$.

  • Round-1 Lemma 2. If $A=m_1+k_1$, then every prime factor of every integer in the later block $\{m_2+1,\dots,m_2+k_2\}$ is at most $A$.

Only Lemma 1 is logically needed for the main correction below; Lemma 2 remains a useful structural obstruction for the still-open global version.

3) NEW INSIGHT / TOOL (ROUND-2)

The genuinely new ingredient is that the fixed-length interpretation is not open.

  • For each fixed pair of lengths, finiteness follows from a theorem of Beukers, Shorey and Tijdeman on equations of the shape \[ x(x+1)\cdots(x+k-1)=y(y+1)\cdots(y+k+l-1) \] with fixed $k,l$.

  • More generally, for each fixed rational ratio of the two lengths, finiteness follows from a theorem of Saradha, Shorey and Tijdeman on \[ d_1\,x(x+1)\cdots(x+lk-1)=d_2\,y(y+1)\cdots(y+mk-1). \]

  • In the low-ratio range $k_1/k_2\in\{2,3,4,5,6\}$ there are in fact no solutions satisfying the present constraint $k_1,k_2>3$.

So the Round-1 ambiguity can now be resolved sharply: the fixed-length reading is settled; the all-parameters reading is still open.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was this: no theorem had been supplied converting the present equation into a known finiteness result for fixed lengths.

I now do three things.

  1. Convert the present notation to the standard "shorter block versus longer block" form used in the literature, and apply the fixed-$(k,l)$ theorem.

  2. Strengthen this to fixed length ratio using the Saradha--Shorey--Tijdeman theorem.

  3. Record one further sharp consequence: if the longer length is $m$ times the shorter one with $2\le m\le 6$, then there are no solutions with both lengths $>3$.

This closes the fixed-length gap completely and isolates the true open remainder.

5) WORK (ROUND-2)

We study \[ \prod_{i=1}^{k_1}(m_1+i)=\prod_{j=1}^{k_2}(m_2+j), \qquad k_1,k_2>3,\qquad m_1+k_1\le m_2. \]

By Round-1 Lemma 1, any such solution satisfies $k_1>k_2$.

Claim 1. For each fixed pair $(k_1,k_2)$ with $k_1>k_2\ge 4$, there are only finitely many integer solutions $(m_1,m_2)$ satisfying the equation and the disjointness condition.

Proof. Fix $k_1>k_2\ge 4$ and put \[ k:=k_2,\qquad l:=k_1-k_2,\qquad x:=m_2+1,\qquad y:=m_1+1. \] Then the equation becomes \[ x(x+1)\cdots(x+k-1)=y(y+1)\cdots(y+k+l-1). \] Since $m_1+k_1\le m_2$, we have \[ x=m_2+1\ge m_1+k_1+1=y+k+l. \] In particular $x$ and $y$ are positive integers, and we are exactly in the fixed-$(k,l)$ situation treated by Beukers--Shorey--Tijdeman.

Their theorem says that for fixed positive integers $k,l$, the equation \[ x(x+1)\cdots(x+k-1)=y(y+1)\cdots(y+k+l-1) \] has only finitely many positive integer solutions $(x,y)$.

Therefore, for each fixed pair $(k_1,k_2)$, the original problem has only finitely many solutions $(m_1,m_2)$. QED.

Thus the fixed-length interpretation is fully settled.

Claim 2. Fix coprime positive integers $l<m$. Then there are only finitely many solutions whose two lengths are in the ratio $l:m$, i.e. whose lengths are $(lk,mk)$ for some positive integer $k$.

Proof. Write the shorter length as $lk$ and the longer one as $mk$, with $(l,m)=1$ and $l<m$. After the same change of variables as above, the equation becomes \[ x(x+1)\cdots(x+lk-1)=y(y+1)\cdots(y+mk-1). \] This is the special case $d_1=d_2=1$ of the Saradha--Shorey--Tijdeman theorem on \[ d_1\,x(x+1)\cdots(x+lk-1)=d_2\,y(y+1)\cdots(y+mk-1). \] That theorem states that either $\max(x,y,k)$ is bounded by a constant depending only on $m,d_1,d_2$, or one falls into an exceptional case \[ m=2,\qquad k=2,\qquad d_1=2d_2^2, \] together with an auxiliary square condition. In our specialization $d_1=d_2=1$, the exceptional condition $d_1=2d_2^2$ becomes $1=2$, impossible.

Hence $\max(x,y,k)$ is bounded, and therefore only finitely many solutions exist for each fixed ratio $l:m$. QED.

Claim 3. If the longer block has length $m$ times the shorter one with $2\le m\le 6$, then there is no solution satisfying the present hypothesis $k_1,k_2>3$.

Proof. Saradha--Shorey and Mignotte--Shorey proved that for $2\le m\le 6$ the only solution to \[ x(x+1)\cdots(x+k-1)=y(y+1)\cdots(y+mk-1) \] in positive integers is \[ x=8,\qquad y=1,\qquad k=3,\qquad m=2. \] Here the two block lengths are $3$ and $6$.

Under our problem's standing condition $k_1,k_2>3$, the shorter length is at least $4$, so this exceptional solution is excluded. Therefore no solutions exist in the ratio range $k_1/k_2\in\{2,3,4,5,6\}$. QED.

Claim 4. For every fixed $N$, there are only finitely many solutions with $\max(k_1,k_2)\le N$ and $k_2\nmid 2k_1$.

Proof. By Round-1 Lemma 1, the shorter length is $k_2$ and the longer length is $k_1$. A theorem of Hajdu and Tijdeman implies that for every fixed $N$, there are only finitely many pairs of disjoint blocks of positive integers of sizes at most $N$ with equal products whenever the smaller size does not divide twice the larger size.

Our problem is a special case of that theorem, because our two blocks are themselves disjoint consecutive blocks. Thus, for fixed $N$, there are only finitely many solutions with $\max(k_1,k_2)\le N$ and $k_2\nmid 2k_1$. QED.

6) ADVERSARIAL VERIFICATION

I check the possible failure points.

  • Variable conversion. The fixed-length theorem is stated for a shorter block of length $k$ and a longer block of length $k+l$. Round-1 Lemma 1 guarantees exactly that ordering here, so the conversion is legitimate.

  • Disjointness. The literature theorems do not require the extra condition $x\ge y+k+l$; they only need positive integer variables. Hence adding disjointness can only restrict the solution set, not invalidate the theorem.

  • Exceptional case in the ratio theorem. The only exceptional branch of the Saradha--Shorey--Tijdeman theorem requires $d_1=2d_2^2$. In our application $d_1=d_2=1$, so the exceptional branch is impossible.

  • Low-ratio theorem versus the present hypotheses. The unique solution for ratios $2\le m\le 6$ has smaller length $3$, which violates the standing assumption $k_1,k_2>3$. So the conclusion "no solutions" is correct for the present problem.

  • Did Round-1 overstate openness? Yes, for the fixed-length reading. Round 1 treated that case as unresolved, but the above theorems show that interpretation is already settled. The all-parameters reading is the truly open one.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advance is the following precise correction.

  1. The fixed-length interpretation of Problem 388 is already true and can be proved rigorously from Beukers--Shorey--Tijdeman: for each fixed pair $(k_1,k_2)$ with $k_1>k_2\ge 4$, only finitely many disjoint-block solutions exist.

  2. More generally, for each fixed rational ratio of the two lengths, only finitely many solutions exist.

  3. If the ratio of lengths is an integer $m$ with $2\le m\le 6$, then under the standing hypothesis $k_1,k_2>3$ there are no solutions at all.

  4. What remains open is the all-parameters problem: are there only finitely many quadruples $(m_1,k_1,m_2,k_2)$ with \[ k_1,k_2>3,\qquad m_1+k_1\le m_2,\qquad \prod_{i=1}^{k_1}(m_1+i)=\prod_{j=1}^{k_2}(m_2+j)? \] Round-1 Lemma 2 still gives the main structural obstruction here: the later block must be exceptionally smooth.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 80%

9) REFERENCES

  1. F. Beukers, T. N. Shorey, and R. Tijdeman, results on the equation \[ x(x+1)\cdots(x+k-1)=y(y+1)\cdots(y+k+l-1) \] for fixed $k,l$; summarized in the survey "Highlights in the Research Work of T. N. Shorey."

  2. S. Saradha and T. N. Shorey, and A. Mignotte and T. N. Shorey, results on \[ x(x+1)\cdots(x+k-1)=y(y+1)\cdots(y+mk-1) \] for $2\le m\le 6$; summarized in the same survey.

  3. S. Saradha, T. N. Shorey, and R. Tijdeman, results on \[ d_1\,x(x+1)\cdots(x+lk-1)=d_2\,y(y+1)\cdots(y+mk-1). \]

  4. L. Hajdu and R. Tijdeman, The Diophantine equation $f(x)=g(y)$ for polynomials with simple rational roots, especially Theorem 10.1.