I need to find real numbers $a$,$b$,$c$ so $a+b+c-abc-2=0$. I know that it is true when $a=b=c=1$ but how do I know it does not have or have other solutions? Btw I have not found a tag for this question other that that.
How to solve the following equation depending on $a$,$b$,$c$?
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systems-of-equations
symmetric-polynomials
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0How about $a=b=0$ and $c=2$? You need to specify what $a,b,c$ are: reals, integers, positive integers? – 2017-02-11
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0Yeah I know there are other solutions but I was thinking if there is an infinity of solutions or not. – 2017-02-11
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0One set of solutions is $(a,a,\frac{2}{a+1})$ – 2017-02-11
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0Go to http://www.wolframalpha.com (Wolfram Alpha) and enter the command plot x+y+z-x*y*z=2 You'll see the plot with a lot of points on it. – 2017-02-11
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0$(1,1,c)$ is a solution for any $c$. In general one should not expect a single equation in *three* variables to have finitely many real solutions (it's possible, but very atypical). – 2017-02-12
1 Answers
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There are many solutions. Take an example, Let $a=0$
The given equation becomes $b+c-2=0$ or $b+c=2$
Since negative and positive numbers are possible, there are infinite values possible for (b,c) for which the above equation holds.