Openai/69630fa9-02d4-8012-8ef2-84c443c04922

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gpt-5-2-pro

User: Are there only finitely many solutions to\\prodi \binom{2mi}{mi}\prodj \binom{2nj}{nj}\with the $mi,nj$ distinct?

Are there only finitely many solutions to\[\prod_i \binom{2m_i}{m_i}=\prod_j \binom{2n_j}{n_j}\]with the $m_i,n_j$ distinct?

Assistant: No.

No. In fact there are infinitely many such identities with all indices distinct.

Write

Bt:=(2tt)(t∈Z≥0).B_t:=\binom{2t}{t}\qquad(t\in\mathbb Z_{\ge 0}).Bt​:=(t2t​)(t∈Z≥0​). A standard ratio computation gives

BtBt+1=(2tt)(2t+2t+1)=(t+1)2(2t+2)(2t+1)=t+12(2t+1).(1)\frac{B_t}{B_{t+1}} =\frac{\binom{2t}{t}}{\binom{2t+2}{t+1}} =\frac{(t+1)^2}{(2t+2)(2t+1)} =\frac{t+1}{2(2t+1)}. \tag{1}Bt+1​Bt​​=(t+12t+2​)(t2t​)​=(2t+2)(2t+1)(t+1)2​=2(2t+1)t+1​.(1) Now fix an integer a≥2a\ge 2a≥2 and set

c:=8a2+8a+1.c:=8a^2+8a+1.c:=8a2+8a+1. Consider the six indices

m-side: {a,  2a+2,  c},n-side: {a+1,  2a,  c+1}.m\text{-side: } \{a,\;2a+2,\;c\}, \qquad n\text{-side: } \{a+1,\;2a,\;c+1\}.m-side: {a,2a+2,c},n-side: {a+1,2a,c+1}. For a≥2a\ge2a≥2 these are pairwise distinct (in particular the two sides are disjoint).

We claim

Ba B2a+2 Bc  =  Ba+1 B2a Bc+1,(2)B_a\,B_{2a+2}\,B_c \;=\; B_{a+1}\,B_{2a}\,B_{c+1}, \tag{2}Ba​B2a+2​Bc​=Ba+1​B2a​Bc+1​,(2) i.e.

(2aa)(4a+42a+2)(2cc)=(2a+2a+1)(4a2a)(2c+2c+1).\binom{2a}{a}\binom{4a+4}{2a+2}\binom{2c}{c} = \binom{2a+2}{a+1}\binom{4a}{2a}\binom{2c+2}{c+1}.(a2a​)(2a+24a+4​)(c2c​)=(a+12a+2​)(2a4a​)(c+12c+2​). To prove (2), divide both sides by Ba+1B2aBc+1B_{a+1}B_{2a}B_{c+1}Ba+1​B2a​Bc+1​; it suffices to show

BaBa+1⋅B2a+2B2a⋅BcBc+1=1.(3)\frac{B_a}{B_{a+1}}\cdot \frac{B_{2a+2}}{B_{2a}}\cdot \frac{B_c}{B_{c+1}}=1. \tag{3}Ba+1​Ba​​⋅B2a​B2a+2​​⋅Bc+1​Bc​​=1.(3) Using (1),

BaBa+1=a+12(2a+1),BcBc+1=c+12(2c+1).\frac{B_a}{B_{a+1}}=\frac{a+1}{2(2a+1)}, \qquad \frac{B_c}{B_{c+1}}=\frac{c+1}{2(2c+1)}.Ba+1​Ba​​=2(2a+1)a+1​,Bc+1​Bc​​=2(2c+1)c+1​. Also,

B2a+2B2a=B2a+2B2a+1⋅B2a+1B2a=2a+22(4a+3)⋅2a+12(4a+1))−1=2(4a+3)(4a+1)(a+1)(2a+1).\frac{B_{2a+2}}{B_{2a}} =\frac{B_{2a+2}}{B_{2a+1}}\cdot \frac{B_{2a+1}}{B_{2a}} =\frac{2a+2}{2(4a+3)}\cdot \frac{2a+1}{2(4a+1)}\Bigg)^{-1} =\frac{2(4a+3)(4a+1)}{(a+1)(2a+1)}.B2a​B2a+2​​=B2a+1​B2a+2​​⋅B2a​B2a+1​​=2(4a+3)2a+2​⋅2(4a+1)2a+1​)−1=(a+1)(2a+1)2(4a+3)(4a+1)​. So the product of the first two factors in (3) is

a+12(2a+1)⋅2(4a+3)(4a+1)(a+1)(2a+1)=(4a+3)(4a+1)(2a+1)2.\frac{a+1}{2(2a+1)}\cdot \frac{2(4a+3)(4a+1)}{(a+1)(2a+1)} =\frac{(4a+3)(4a+1)}{(2a+1)^2}.2(2a+1)a+1​⋅(a+1)(2a+1)2(4a+3)(4a+1)​=(2a+1)2(4a+3)(4a+1)​. Thus (3) will hold provided

BcBc+1=(2a+1)2(4a+3)(4a+1).\frac{B_c}{B_{c+1}}=\frac{(2a+1)^2}{(4a+3)(4a+1)}.Bc+1​Bc​​=(4a+3)(4a+1)(2a+1)2​. But with c=8a2+8a+1c=8a^2+8a+1c=8a2+8a+1,

c+1=8a2+8a+2=2(2a+1)2,2c+1=16a2+16a+3=(4a+3)(4a+1),c+1=8a^2+8a+2=2(2a+1)^2,\qquad 2c+1=16a^2+16a+3=(4a+3)(4a+1),c+1=8a2+8a+2=2(2a+1)2,2c+1=16a2+16a+3=(4a+3)(4a+1), so indeed

BcBc+1=c+12(2c+1)=2(2a+1)22(4a+3)(4a+1)=(2a+1)2(4a+3)(4a+1).\frac{B_c}{B_{c+1}}=\frac{c+1}{2(2c+1)} =\frac{2(2a+1)^2}{2(4a+3)(4a+1)} =\frac{(2a+1)^2}{(4a+3)(4a+1)}.Bc+1​Bc​​=2(2c+1)c+1​=2(4a+3)(4a+1)2(2a+1)2​=(4a+3)(4a+1)(2a+1)2​. This completes the proof of (2).

So for every a≥2a\ge2a≥2 we get a distinct solution with all indices distinct:

(2aa)(4a+42a+2)(2(8a2+8a+1)8a2+8a+1)=(2a+2a+1)(4a2a)(2(8a2+8a+2)8a2+8a+2)\boxed{\binom{2a}{a}\binom{4a+4}{2a+2}\binom{2(8a^2+8a+1)}{8a^2+8a+1} = \binom{2a+2}{a+1}\binom{4a}{2a}\binom{2(8a^2+8a+2)}{8a^2+8a+2}}(a2a​)(2a+24a+4​)(8a2+8a+12(8a2+8a+1)​)=(a+12a+2​)(2a4a​)(8a2+8a+22(8a2+8a+2)​)​ and hence there are infinitely many solutions.

Example a=2a=2a=2:

(42)(126)(9849)=(63)(84)(10050),\binom{4}{2}\binom{12}{6}\binom{98}{49} = \binom{6}{3}\binom{8}{4}\binom{100}{50},(24​)(612​)(4998​)=(36​)(48​)(50100​), i.e. 6⋅924⋅(9849)=20⋅70⋅(10050)6\cdot924\cdot\binom{98}{49}=20\cdot70\cdot\binom{100}{50}6⋅924⋅(4998​)=20⋅70⋅(50100​), with indices {2,6,49,3,4,50}\{2,6,49,3,4,50\}{2,6,49,3,4,50} all distinct.