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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 $S_N$?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}$, so a naive counting argument does not seem to work here. If you think that this problem is intractable then let me know, I'll try a different strategy. 2 edited title # looking for a multipliciymultiplicity one prime in a finite sum 1 # looking for a multipliciy one prime in a finite sum So I'm trying to compute the Galois group of family of polynomials (indexed by their degree) using the technique of the Newton polygon. In order to apply this technique I need to find some good prime number$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$S_N$? Note that the square of the product of all primes in the interval 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.