Erdős Problem #686

Erdős Problem 686 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/686

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/686.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/686.tex.

Erdős Problem 686

1) FORMAL RESTATEMENT.

We interpret the problem in the standard number-theoretic way: $k,m,n$ are nonnegative integers, with $k\ge 2$ and $m\ge n+k$. For such $(k,m,n)$ define \[ R(k,m,n):=\frac{\prod_{i=1}^k (m+i)}{\prod_{i=1}^k (n+i)}. \] Question: Is it true that for every integer $N\ge 2$ there exist integers $k\ge 2$ and $m\ge n+k\ge 0$ such that $R(k,m,n)=N$?

Ambiguity note. If one allows negative $n$ or $m$, then denominators can vanish or signs can appear; the problem statement suggests the intended domain is nonnegative integers with $m\ge n+k$.

2) QUICK LITERATURE/CONTEXT CHECK.

No external results are used or claimed here beyond what is in the problem statement. Algebraically, $R(k,m,n)$ is a ratio of rising factorials and can be rewritten as a ratio of binomial coefficients (Lemma 686.1).

3) ATTACK PLAN.

• Proof-track idea: rewrite $R(k,m,n)$ as a ratio of binomial coefficients; attempt to realize arbitrary $N$ by forcing prime-adic valuations via choices of $(k,m,n)$.
• Disproof-track idea: search for a small $N$ not representable; then try to prove an invariant that forbids it (e.g. constraints on prime valuations forced by disjointness of the two length-$k$ intervals).

4) WORK.

Lemma 686.1 (binomial coefficient reformulation). For integers $k\ge 1$ and $m,n\ge 0$, \[ R(k,m,n)=\frac{\binom{m+k}{k}}{\binom{n+k}{k}}. \]

Proof. We have \[ \prod_{i=1}^k (m+i)=\frac{(m+k)!}{m!},\qquad \prod_{i=1}^k (n+i)=\frac{(n+k)!}{n!}. \] Therefore \[ R(k,m,n)=\frac{(m+k)!}{m!}\cdot \frac{n!}{(n+k)!} =\frac{(m+k)!}{k!\,m!}\bigg/\frac{(n+k)!}{k!\,n!} =\frac{\binom{m+k}{k}}{\binom{n+k}{k}}. \] $\square$

Lemma 686.2 (the case $k=2$ reduces to a Pell-type square condition). Fix an integer $N\ge 2$. There exist integers $m\ge n+2\ge 0$ such that \[ R(2,m,n)=N \] if and only if there exists an integer $n\ge 0$ for which \[ D(n):=4N(n+1)(n+2)+1 \] is a perfect square; in that case $m$ is uniquely determined by \[ m=\frac{-3+\sqrt{D(n)}}{2}. \]

Proof. The equality $R(2,m,n)=N$ is exactly \[ \frac{(m+1)(m+2)}{(n+1)(n+2)}=N, \] i.e. \[ (m+1)(m+2)=N(n+1)(n+2). \] Expanding gives \[ m^2+3m+2 = N(n^2+3n+2). \] View this as a quadratic equation in $m$: \[ m^2+3m+(2-N(n^2+3n+2))=0. \] Its discriminant is \[ \Delta = 3^2-4\cdot 1\cdot (2-N(n^2+3n+2)) =1+4N(n^2+3n+2)=4N(n+1)(n+2)+1. \] Thus the quadratic has an integer solution $m$ if and only if $\Delta$ is a perfect square, say $\Delta=s^2$. In that case, the quadratic formula gives \[ m=\frac{-3\pm s}{2}. \] Since $m\ge 0$ and $s\ge 1$, only the $(-3+s)/2$ branch is possible. This yields the stated formula. Finally, the constraint $m\ge n+2$ is an additional inequality one must check once $m$ is computed; it is automatic in some cases and not in others. $\square$

FAST REALITY CHECK (local computation). Brute search (bounded parameters) found many representations but also some persistent "missing" integers in the searched ranges.

• For $N=2,3,5,\dots,30$, searching over $2\le k\le 6$, $0\le n\le 30$, $n+k\le m\le 60$ found representations for all $N\in\{2,3,5,6,7,8,9,10,\dots,30\}$ except $N=4$ and $N=25$ in that box. Example outputs:

  2 -> (k,m,n)=(2,19,13)  since (20*21)/(14*15)=2
  3 -> (k,m,n)=(2, 8, 4)  since ( 9*10)/( 5* 6)=3
  9 -> (k,m,n)=(3,25,11)  since (26*27*28)/(12*13*14)=9
  16-> (k,m,n)=(3,13, 4)  since (14*15*16)/( 5* 6* 7)=16
  

• Expanding the search to $2\le k\le 8$, $0\le n\le 200$, $n+k\le m\le 2000$ still found no representation for $N=4,25,49$.
• In the special case $k=2$, Lemma 686.2 allows a fast scan of $n$ for the square condition $4N(n+1)(n+2)+1=s^2$. For $N=4,25,49$ no solutions were found for $0\le n\le 200000$ (this is not a proof of nonexistence).

5) VERIFICATION.

• Lemma 686.1 is a direct factorial manipulation and is exact.
• Lemma 686.2 is an exact algebraic equivalence for $k=2$ (discriminant criterion).
• The computational searches are bounded; they only show absence of solutions within the tested ranges and cannot certify impossibility.

FINAL.

UNRESOLVED

(i) Strongest proved partial result. The representation problem is equivalent to asking whether every $N\ge 2$ is a ratio of binomial coefficients \[ N=\binom{m+k}{k}\big/\binom{n+k}{k} \] with $k\ge 2$ and $m\ge n+k\ge 0$ (Lemma 686.1). For $k=2$, existence is equivalent to the explicit square/discriminant condition of Lemma 686.2.

(ii) First gap (crisp statement). Either (a) prove that for every $N\ge 2$ there exists some $k\ge 2$ and $m\ge n+k$ with $R(k,m,n)=N$, or (b) exhibit a specific $N$ (e.g. $N=4$ or $25$ suggested by bounded search) and prove rigorously that no such $(k,m,n)$ exist.

(iii) Top 3 next moves.

1. Analyze the diophantine geometry for fixed small $k$ (starting with $k=2,3$): characterize all integer solutions to $\binom{m+k}{k}=N\binom{n+k}{k}$ and decide for which $N$ solutions exist.
2. Search for an invariant from prime-adic valuations: for each odd prime $q$, compare $v_q((m+1)_k)$ and $v_q((n+1)_k)$ and see whether $v_q(R(k,m,n))$ must take some restricted set of values.
3. If a counterexample is suspected (e.g. $N=4$), attempt to prove that $R(k,m,n)$ cannot equal that $N$ by forcing an odd prime divisor in the ratio (e.g. show existence of a prime dividing the numerator block with exponent exceeding that in the denominator block).

(iv) What a minimal counterexample would likely look like. A minimal counterexample would be a small integer $N$ whose prime factorization is "too simple" to be realized as the quotient of two length-$k$ consecutive products from disjoint intervals. Bounded search suggests $N=4$ and $25$ as candidates, but no proof is known here.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.

Erdős Problem 686

1) FORMAL RESTATEMENT.

We interpret the problem in the standard number-theoretic way: $k,m,n$ are nonnegative integers, with $k\ge 2$ and $m\ge n+k$. For such $(k,m,n)$ define \[ R(k,m,n):=\frac{\prod_{i=1}^k (m+i)}{\prod_{i=1}^k (n+i)}. \] Question: Is it true that for every integer $N\ge 2$ there exist integers $k\ge 2$ and $m\ge n+k\ge 0$ such that $R(k,m,n)=N$?

Ambiguity note. If one allows negative $n$ or $m$, then denominators can vanish or signs can appear; the problem statement suggests the intended domain is nonnegative integers with $m\ge n+k$.

2) QUICK LITERATURE/CONTEXT CHECK.

No external results are used or claimed here beyond what is in the problem statement. Algebraically, $R(k,m,n)$ is a ratio of rising factorials and can be rewritten as a ratio of binomial coefficients (Lemma 686.1).

3) ATTACK PLAN.

• Proof-track idea: rewrite $R(k,m,n)$ as a ratio of binomial coefficients; attempt to realize arbitrary $N$ by forcing prime-adic valuations via choices of $(k,m,n)$.
• Disproof-track idea: search for a small $N$ not representable; then try to prove an invariant that forbids it (e.g. constraints on prime valuations forced by disjointness of the two length-$k$ intervals).

4) WORK.

Lemma 686.1 (binomial coefficient reformulation). For integers $k\ge 1$ and $m,n\ge 0$, \[ R(k,m,n)=\frac{\binom{m+k}{k}}{\binom{n+k}{k}}. \]

Proof. We have \[ \prod_{i=1}^k (m+i)=\frac{(m+k)!}{m!},\qquad \prod_{i=1}^k (n+i)=\frac{(n+k)!}{n!}. \] Therefore \[ R(k,m,n)=\frac{(m+k)!}{m!}\cdot \frac{n!}{(n+k)!} =\frac{(m+k)!}{k!\,m!}\bigg/\frac{(n+k)!}{k!\,n!} =\frac{\binom{m+k}{k}}{\binom{n+k}{k}}. \] $\square$

Lemma 686.2 (the case $k=2$ reduces to a Pell-type square condition). Fix an integer $N\ge 2$. There exist integers $m\ge n+2\ge 0$ such that \[ R(2,m,n)=N \] if and only if there exists an integer $n\ge 0$ for which \[ D(n):=4N(n+1)(n+2)+1 \] is a perfect square; in that case $m$ is uniquely determined by \[ m=\frac{-3+\sqrt{D(n)}}{2}. \]

Proof. The equality $R(2,m,n)=N$ is exactly \[ \frac{(m+1)(m+2)}{(n+1)(n+2)}=N, \] i.e. \[ (m+1)(m+2)=N(n+1)(n+2). \] Expanding gives \[ m^2+3m+2 = N(n^2+3n+2). \] View this as a quadratic equation in $m$: \[ m^2+3m+(2-N(n^2+3n+2))=0. \] Its discriminant is \[ \Delta = 3^2-4\cdot 1\cdot (2-N(n^2+3n+2)) =1+4N(n^2+3n+2)=4N(n+1)(n+2)+1. \] Thus the quadratic has an integer solution $m$ if and only if $\Delta$ is a perfect square, say $\Delta=s^2$. In that case, the quadratic formula gives \[ m=\frac{-3\pm s}{2}. \] Since $m\ge 0$ and $s\ge 1$, only the $(-3+s)/2$ branch is possible. This yields the stated formula. Finally, the constraint $m\ge n+2$ is an additional inequality one must check once $m$ is computed; it is automatic in some cases and not in others. $\square$

FAST REALITY CHECK (local computation). Brute search (bounded parameters) found many representations but also some persistent "missing" integers in the searched ranges.

• For $N=2,3,5,\dots,30$, searching over $2\le k\le 6$, $0\le n\le 30$, $n+k\le m\le 60$ found representations for all $N\in\{2,3,5,6,7,8,9,10,\dots,30\}$ except $N=4$ and $N=25$ in that box. Example outputs:

  2 -> (k,m,n)=(2,19,13)  since (20*21)/(14*15)=2
  3 -> (k,m,n)=(2, 8, 4)  since ( 9*10)/( 5* 6)=3
  9 -> (k,m,n)=(3,25,11)  since (26*27*28)/(12*13*14)=9
  16-> (k,m,n)=(3,13, 4)  since (14*15*16)/( 5* 6* 7)=16
  

• Expanding the search to $2\le k\le 8$, $0\le n\le 200$, $n+k\le m\le 2000$ still found no representation for $N=4,25,49$.
• In the special case $k=2$, Lemma 686.2 allows a fast scan of $n$ for the square condition $4N(n+1)(n+2)+1=s^2$. For $N=4,25,49$ no solutions were found for $0\le n\le 200000$ (this is not a proof of nonexistence).

5) VERIFICATION.

• Lemma 686.1 is a direct factorial manipulation and is exact.
• Lemma 686.2 is an exact algebraic equivalence for $k=2$ (discriminant criterion).
• The computational searches are bounded; they only show absence of solutions within the tested ranges and cannot certify impossibility.

FINAL.

UNRESOLVED

(i) Strongest proved partial result. The representation problem is equivalent to asking whether every $N\ge 2$ is a ratio of binomial coefficients \[ N=\binom{m+k}{k}\big/\binom{n+k}{k} \] with $k\ge 2$ and $m\ge n+k\ge 0$ (Lemma 686.1). For $k=2$, existence is equivalent to the explicit square/discriminant condition of Lemma 686.2.

(ii) First gap (crisp statement). Either (a) prove that for every $N\ge 2$ there exists some $k\ge 2$ and $m\ge n+k$ with $R(k,m,n)=N$, or (b) exhibit a specific $N$ (e.g. $N=4$ or $25$ suggested by bounded search) and prove rigorously that no such $(k,m,n)$ exist.

(iii) Top 3 next moves.

1. Analyze the diophantine geometry for fixed small $k$ (starting with $k=2,3$): characterize all integer solutions to $\binom{m+k}{k}=N\binom{n+k}{k}$ and decide for which $N$ solutions exist.
2. Search for an invariant from prime-adic valuations: for each odd prime $q$, compare $v_q((m+1)_k)$ and $v_q((n+1)_k)$ and see whether $v_q(R(k,m,n))$ must take some restricted set of values.
3. If a counterexample is suspected (e.g. $N=4$), attempt to prove that $R(k,m,n)$ cannot equal that $N$ by forcing an odd prime divisor in the ratio (e.g. show existence of a prime dividing the numerator block with exponent exceeding that in the denominator block).

(iv) What a minimal counterexample would likely look like. A minimal counterexample would be a small integer $N$ whose prime factorization is "too simple" to be realized as the quotient of two length-$k$ consecutive products from disjoint intervals. Bounded search suggests $N=4$ and $25$ as candidates, but no proof is known here.