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Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials in the polynomial ring $\mathbf{F}_{q}[X]$ over the finite field $\mathbf{F}_{q}$. Since \begin{equation*} M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d} \end{equation*} we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \cdots + n_se_s$. Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} \end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?

Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials of degree $n$ in the polynomial ring $\mathbf{F}_{q}[X]$ over the finite field $\mathbf{F}_{q}$. Since \begin{equation*} M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d} \end{equation*} we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \cdots + n_se_s$. Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} \end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?

Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials in the polynomial ring $\mathbf{F}_{q}[X]$ over the finite field $\mathbf{F}_{q}$. Since \begin{equation*} M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d} \end{equation*} we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \cdots + n_se_s$. Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} =0 \end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?

Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials in the polynomial ring $\mathbf{F}_{q}[X]$ over the finite field $\mathbf{F}_{q}$. Since \begin{equation*} M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d} \end{equation*} we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \cdots + n_se_s$. Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} \end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?

Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials in the polynomial ring $\mathbf{F}_{q^n}[X]$ over the finite field $\mathbf{F}_{q}$. Since \begin{equation*} M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d} \end{equation*} we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \cdots + n_se_s$. Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} =0 \end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?

Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials in the polynomial ring $\mathbf{F}_{q}[X]$ over the finite field $\mathbf{F}_{q}$. Since \begin{equation*} M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d} \end{equation*} we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \cdots + n_se_s$. Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} =0 \end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?