How to choose numbers $x_k$ such that ${\sum \limits_{k=1}^{n}{x_k}} = {\prod \limits_{k=1}^{n}{x_k}}$? Here $x_k \in \mathbb R$ but they could very well be generalised to $\in \mathbb C$ or Matrices. I think the case for $x_k \in \mathbb N$ was just the perfect numbers. Does allowing $x_k\in \mathbb R$ allow more unaccountably many more possibilities to choose $x_k$?
Generating $x_k$'s s.t. ${\sum \limits_{k=1}^{n}{x_k}} = {\prod \limits_{k=1}^{n}{x_k}}.$
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algebra-precalculus
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0Use the roots of any polynomial such that the coefficients of the second and last terms are equal (or opposite, depending on the parity of $n$). E.g. $x^3-x^2-3x-1=0$, $x_1=-1,x_2=1-\sqrt2,x_3=1+\sqrt2$. – 2017-02-24
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1The question of the title is different that the question inside. Both definitions are flawed, I assume you are searching for $$\sum_{k=1}^n x_k=\prod_{k=1}^n x_k$$ – 2017-02-24
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1@Masacroso: I guess they are just typos. (Which I fixed *by force*.) – 2017-02-24
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1For $x_1 \cdot \ldots \cdot x_{n-1} \ne 1$, put $x_{n}=\dfrac{x_1+\ldots+x_{n-1}}{x_1 \cdot \ldots \cdot x_{n-1}-1}$ – 2017-03-01
1 Answers
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If $n = 0$, the equation reduces to $0 = 1$.
If $n = 1$, the equation reduces to $x_1 = x_1$.
If $n = 2$, the equation is equivalent to $( x_1 - 1 ) ( x_2 - 1 ) = 1$.
If $n > 2$, the equation is equivalent to $x_1+x_2+\sum_{k=3}^n x_n = x_1 x_2 \prod_{k=3}^n x_n$, and if you fix $x_k$ for each $k \in [3..n]$, the equation would reduce to $x_1+x_2+a = x_1 x_2 b$ for some reals $a,b$, and you can pick any real $x_2$ such that $x_2 b \ne 1$ to reduce it further to the equation $x_1+c = x_1 d$ for some real $d \ne 1$, which always has a solution for $x_1$.
I'll leave you to conclude about the cardinality of the set of solutions in each case.