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So I'm trying to compute the Galois group of family of polynomials (indexed by their degree) using the technique of the Newton polygon. In order to apply this technique I need to find some good prime number $p$.

So this is the motivation behind my question that might seem a little bit unmotivated:

Let $N$ be a large integer. Then it is not too difficult to show the following statement:

Theorem: For every prime $p$ such that $N/2< p< 2N/3$ one has that $p$ divides the following sum $$ S_N:=\sum_{k=0}^N \binom{N+k}{k}2^{N-k}(-1)^k $$

After many numerical examples, it always happens that most of the primes in the interval $N/2 < p < 2N/3$ divide $S_N$ with multiplicity one. So here is my question:

Q: How would you show that there exists at least one prime $p$ in the interval $N/2 < p < 2N/3$ that divides exactly $S_N$, i.e., $p|S_N$ but $p^2\nmid S_N$ ,?

Note that the square of the product of all primes in the interval $(\frac{N}{2},\frac{2N}{3})$ is less that $\binom{2N}{N}$, so a naive counting argument does not seem to work here.

If you think that this problem is intractable then let me know, I'll try a different strategy.

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I expect you mean there exists $p$ in this range which divides $S_N$ but $p^2$ does not? –  Todd Trimble Mar 26 '11 at 15:03
    
Yes exactly Todd –  Hugo Chapdelaine Mar 26 '11 at 15:44
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Hi GH, the key point is to notice that for $2p-N\leq k\leqp-1$ and $3p-N\leq k\leq N$ that $p$ divides $\binom{N+k}{k}$. Then one uses simple manipulations and congruences modulo $p$ and Wilson's theorem at one place and that's it. But the sketch of proof that I just explained does not say much about the residue class modulo $p^2$. –  Hugo Chapdelaine Mar 26 '11 at 21:53
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I was trying to compute $S_p \mod p^2$ but won't help as $73^2 | S_{73}$. –  Felipe Voloch Mar 27 '11 at 7:02
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@Gerhard Yes it is that much bigger. $\small S_{100}=2^5 \cdot 3 \cdot 53 \cdot 59 \cdot 61 \cdot 67 \cdot 3567917 \cdot 163655344202455746133481155996257157083231473 $ which is about $7 \cdot 10^{59}$ the square of the product of all the primes up to $67$ is about $6.2 \cdot 10^{49}$. –  Aaron Meyerowitz Mar 28 '11 at 0:12
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1 Answer

No answer, just some data. Up to $n=825$ there are 41 pairs $[p,n]$ such that $S_n \equiv 0 \mod p^2$ and $\frac n2 \lt p \lt \frac{2n}{3}$. Here they are: $ \small [7, 12], [11, 21], [29, 55], [41, 68], [43, 72], [47, 80], [61, 100], [73, 136], [\mathbf{89}, 138], [\mathbf{89}, 150], [79, 156], [89, 167]$ $ \small [109, 183] [\mathbf{127}, 206], [\mathbf{127}, 230], [131, 231], [157, 276], [181, 301], [199, 306], [197, 364], [227, 386], [257, 445] $ $\small [\mathbf{277}, 450], [\mathbf{277}, 475] [313, 482], [251, 492], [353, 538], [307, 542], [421, 654], [439, 670], [367, 701], [431, 702]$ $\small [\mathbf{359}, 703], [\mathbf{359}, 710], [373, 731] [401, 737], [467, 737], [409, 755], [431, 757], [491, 798], [419, 822] $

Over the same range the (naively ) expected number of repeat divisors of that type is about $43$ so the result seems almost certainly true, but perhaps for no special reason.

It is notable that several times one gets the same $p$ twice in a row. I don't know if it is significant however.

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Thanks Aaron for all the data. –  Hugo Chapdelaine Mar 28 '11 at 12:09
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