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I have a product formula that, given a tuple $(a_1, \dots, a_{n-1}, 0)$, computes a dimension (of a vector space) via the following formula: $$ \mathrm{dimension} = \prod_{i < j} {(a_i + \cdots + a_{j-1}) + j-i \over j-i} \qquad \text{for } 1 \leq i < j \leq n $$

I am trying to prove that the only tuples that give me a dimension equal to $p^2$ for a fixed prime $p$ are the tuples $\underbrace{((p^2 - 1)\cdot a_1, 0)}_{\text{has length $2$}}$, and $\underbrace{(a_1,0, \dots, 0)}_{\text{has length $p^2$}}$ and $\underbrace{(0,0, \dots,a_{p^2 - 1}, 0)}_{\text{has length $p^2$}}$. I'm inducting on the prime $p$, and for the base case $p = 3$ I have established this, but now I am confused about how to proceed further. Any tips?

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shouldn't you be inducting on $n$? –  wece Jun 5 '13 at 15:38
    
@wece I thought that I should induct on $p$ and then perhaps have a "nested" induction on $n$ inside the proof. How can I induct on $n$ but still establish that it is only those above tuples that give me the desired dimension? –  Mike Jun 5 '13 at 15:40
    
I have no idea :D It's just it feels weird to induct on a "fixed $p$". –  wece Jun 5 '13 at 15:52
1  
The induction feels weird actually. Did you try a case distinction? Something like: Fix a $p$ suppose that you have a solution not of the forms $(a_1,0)$, $(a_1,0,...,0)$ or $(0,...,a_{p^2-1},0)$. Show that it's not a solution. And show that for your given form, what you propose are the only one possible. –  wece Jun 5 '13 at 15:59
    
I will go down that route, thanks for your suggestions @wece –  Mike Jun 5 '13 at 18:51

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