Erdős Problem #397
Status Solved · erdosproblems.comLLM status Solved from source status
Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolvedStatement
Statement.
Are there only finitely many identities \[ \prod_{i=1}^{r}\binom{2a_i}{a_i} = \prod_{j=1}^{s}\binom{2b_j}{b_j}, \] where all the indices on both sides are distinct?Result and attribution.
No. The current problem page records the following parametric construction of Somani, obtained with ChatGPT. Here is a direct verification, rather than a claim of a new construction.Write $B_t=\binom{2t}{t}$. For every integer $a\geq2$, put $c=8a^2+8a+1$. Then \[ B_a B_{2a+2}B_c=B_{a+1}B_{2a}B_{c+1}. \tag{1} \] Indeed, \[ \frac{B_{t+1}}{B_t}=\frac{2(2t+1)}{t+1}, \] and consequently \[ \frac{B_aB_{2a+2}}{B_{a+1}B_{2a}} =\frac{(4a+1)(4a+3)}{(2a+1)^2} =\frac{2(2c+1)}{c+1} =\frac{B_{c+1}}{B_c}. \] This proves $1$. The six indices are distinct: for $a\geq2$, \[ a<a+1<2a<2a+2<c<c+1. \] The largest index $c+1$ increases strictly with $a$, so these give infinitely many different identities. For example, $a=2$ gives $B_2B_6B_{49}=B_3B_4B_{50}$.
Scope and source.
This proves the negative answer to the stated finiteness question without an external theorem. Source and attribution: https://www.erdosproblems.com/397, checked 24 September 2026.Completion estimate.
The counterexample family and every required distinctness condition are proved.Restatement from Math-5624-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-5624
Erdős Problem 397
Statement.
Are there only finitely many identities \[ \prod_{i=1}^{r}\binom{2a_i}{a_i} = \prod_{j=1}^{s}\binom{2b_j}{b_j}, \] where all the indices on both sides are distinct?Result and attribution.
No. The current problem page records the following parametric construction of Somani, obtained with ChatGPT. Here is a direct verification, rather than a claim of a new construction.Write $B_t=\binom{2t}{t}$. For every integer $a\geq2$, put $c=8a^2+8a+1$. Then \[ B_a B_{2a+2}B_c=B_{a+1}B_{2a}B_{c+1}. \tag{1} \] Indeed, \[ \frac{B_{t+1}}{B_t}=\frac{2(2t+1)}{t+1}, \] and consequently \[ \frac{B_aB_{2a+2}}{B_{a+1}B_{2a}} =\frac{(4a+1)(4a+3)}{(2a+1)^2} =\frac{2(2c+1)}{c+1} =\frac{B_{c+1}}{B_c}. \] This proves $1$. The six indices are distinct: for $a\geq2$, \[ a<a+1<2a<2a+2<c<c+1. \] The largest index $c+1$ increases strictly with $a$, so these give infinitely many different identities. For example, $a=2$ gives $B_2B_6B_{49}=B_3B_4B_{50}$.