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Let $A=\{0,1,\cdots,d-1\}$. Consider the set $P(n)=\{(x,y)\in A\times A:x+y=n\}$. Consider the function $F(X)=\sum_{n=0}^{2(d-1)} \# P(n) X^n$, where $\#$ denotes the cardinality of $P(n)$.

For the equation $F(X)=0$, is it possible to tell that multiple roots always exist whenever $d\geq2$.

I have checked it for small values of $d$. Also, I have noticed that since the equation is over integers, if $z$ is a complex root, then $\overline{z}, \frac{1}{z}, \frac{1}{\overline{z}}$ are also roots of the equation due to a symmetry among the equation coefficients.

Also I want to ask a generalisation. Let, $A_j=\{0,1,\cdots,d_j-1\}, ~j=1,\cdots,k$ and $P(n:k)= \{(x_1,\cdots,x_k)\in A_1\times\cdots \times A_k:x_1+\cdots+x_k=n\}$. Can we exactly tell when the similar equation have multiple roots and when it is not possible. I have checked for the value $d_1=2$ and $d_2=3$ for which there is no multiple root. However, I could not make a general theory for the general case.

Advanced thanks for any help. Please feel free to edit or retag, if you think it is necessary.

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For the first case, it may not be difficult to show the roots. Using the symmetry of the coefficients, you can show that the equation is actually \begin{equation} \frac{(1-x^d)^2}{(1-x)^2}=0. \end{equation} Hence your statement is true. For the case $A\times \cdots \times A$ ($k$-times), I guess the similar method will work. I am yet to find a general pattern for the $A$'s of different size. May be someone else can answer the question properly.

ADDED: The general result is also easy. Notice that the equation is actually splitting as

\begin{equation} (1+x+\cdots +x^{d_1})\cdots (1+x+\cdots +x^{d_k})=0 \end{equation} and the coefficients of $x^n$ in your definition is matching. Now it is easy to see the cases when it will have multiple roots.

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