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.


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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