gpt pro 5.2 attempt on Math-904

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.