Erdős Problem #388

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.

Erdős Problem #388: independent continuation

1) TARGET AND SOURCE AUDIT

All supplied versions considered: 388.tex. The source correctly separates fixed lengths from the global question, although its cited theorems need hypotheses beyond the overbroad phrase "fixed positive integers. I avoid relying on that phrasing and give an exact reduction and exhaustive check of the four-term later block.

2) WORK

The saved script checks every $0\le m_1\le10000$ and $5\le k_1\le12$. It finds exactly \[ (m_1,k_1,m_2,k_2)=(7,7,62,4), \] corresponding to \[ 8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14 =63\cdot64\cdot65\cdot66=17297280. \] The reduction is complete for this bounded parameter box because every four-term later product satisfies the displayed identity and every admissible pair of squares reconstructs its unique $m_2$.

3) ADVERSARIAL VERIFICATION

4) FINAL

5) SOURCES CHECKED

6) COMPLETION ESTIMATE

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

Formal restatement

• (A) Fixed lengths: for each fixed pair $(k_1,k_2)$ with $k_1,k_2>3$, are there only finitely many pairs $(m_1,m_2)$ satisfying (eq:P388-main) and $m_1+k_1\le m_2$?
• (B) All parameters: are there only finitely many quadruples $(m_1,k_1,m_2,k_2)$ with $k_1,k_2>3$ and $m_1+k_1\le m_2$ satisfying (eq:P388-main)?

A more general variant mentioned is \begin{equation} \label{wsref-0-65-71-3a-50-33-38-38-2d-61-62} a\,P_{k_1}(m_1)=b\,P_{k_2}(m_2), \end{equation} with fixed nonzero integers $a,b$ and $k_1>2$.

Quick literature/context check (browsing available)

Attack plan

1. Prove structural necessary conditions for any solution (e.g. compare lengths; prime-factor "smoothness" constraints).
2. Do a bounded brute-force search to verify the smallest solutions and gain confidence about the landscape.
3. Attempt to leverage general finiteness theorems for polynomial Diophantine equations of the form $f(x)=g(y)$ (e.g. Siegel/Bilu--Tichy) to prove finiteness for fixed $(k_1,k_2)$; identify the precise obstruction to converting this into a full classification across all lengths.

Work

Lemma 1 (the longer block must be the earlier one).

Lemma 2 ("smoothness" obstruction).

A computational check (small search).

Where a full solution would likely come from.

• quantitative bounds on the maximal length of a run of $A$-smooth integers in $[x,x+L]$ with $x$ much larger than $A$,
• together with the size constraint imposed by (eq:P388-main), which forces the growth rates of $P_{k_1}(m_1)$ and $P_{k_2}(m_2)$ to match.

Verification

• Lemma 1 uses only monotone inequalities and the disjointness condition $m_1+k_1\le m_2$; the strict inequality step $(A+1)^{k_1}>(A)^{k_1}$ is immediate.
• Lemma 2 is immediate from prime-factor containment.
• The identity (eq:P388-known) was verified by direct multiplication (in a separate computation).

Final

UNRESOLVED.

1. (i) Strongest proved partial result: Any solution of (eq:P388-main) with $m_1+k_1\le m_2$ must satisfy $k_1>k_2$ and the entire right block $\{m_2+1,\dots,m_2+k_2\}$ is $(m_1+k_1)$-smooth (Lemmas 1--2). In addition, a bounded brute-force search (specified above) finds only the solution (eq:P388-known) with $k_1,k_2>3$ in that range.
2. (ii) First gap / obstruction: Turning the smoothness obstruction into a global finiteness or uniqueness theorem requires nontrivial control of long runs of $A$-smooth numbers above $A$ that is uniform in the parameters and compatible with the size constraint from (eq:P388-main).
3. (iii) Top 3 next moves: $1$ Translate (eq:P388-main) into a statement about smoothness of a short interval and try to apply strongest available bounds on smooth numbers in short intervals. $2$ For fixed $(k_1,k_2)$, treat $P_{k_1}(x)=P_{k_2}(y)$ as an integral-point problem on an algebraic curve and attempt an effective determination of all integral points (in small degrees this reduces to elliptic/hyperelliptic methods). $3$ Expand the computational search using modular sieving informed by Lemma 2 (restricting to candidate intervals whose elements are all $A$-smooth).
4. (iv) What a minimal counterexample (to "only finitely many solutions") would look like: It would require an infinite family of pairs of disjoint blocks with $k_1>k_2\ge 4$ in which the later block forms unusually long runs of $A$-smooth numbers relative to $A=m_1+k_1$, while simultaneously matching the magnitude constraint $\log P_{k_1}(m_1)\approx \log P_{k_2}(m_2)$.

Completion estimate