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@Lopsy  
We have
f(n) = \sum_{k=0}^N m_k n^k.
Therefore
f(f(n)+1)/f(n) = 1/f(n) * \sum_{k=0}^N m_k (f(n)+1)^k = \sum_{k=0}^N \sum_{l=0}^k (k choose l) m_k f(n)^{l-1},
by Newton’s formula.
Change the order of summation to get
f(f(n)+1)/f(n) = \sum_{l=0}^N \sum_{k=l}^N (…)
and then split the sum over l into the l=0 term and the l>0 terms. The latter sum to an integer as each term is individually an integer, so we only need to check the l=0 term.
This is just
\sum_{k=0}^N m_k/f(n) = f(1)/f(n).
By hypothesis, f(n) is strictly larger than f(1) (in which case f(1)/f(n) is not an integer and f(n) does not divide f(f(n)+1)) unless n = 1 or f is a zero-degree polynomial (a constant).
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Lopsy

This is problem B1 from the 2007 Putnam competition. “Let f be a polynomial with positive integer coefficients. Prove that if n is a positive integer, then f(n) divides f(f(n)+1) if and only if n=1.”
 
Any takers?
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