Empty
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What are all the integer solutions to: a+b+c=a*b*c
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i got one answer !
anonymous
  • anonymous
1,2,3 or -1,-2,-3 these are nonzero integers.
anonymous
  • anonymous
oh i got a different one

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Empty
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Yeah I have a couple other answers too. 10, -10, 0 works. Are there others?
Empty
  • Empty
Maybe I should extend for all integers, what are all solutions to: \[\Huge \sum_{k=1}^n a_k = \prod_{k=1}^n a_k\]
anonymous
  • anonymous
if one is zero ,you can select any +,- pair
Empty
  • Empty
True, I am just saying are there other answers in general, not just answers of that form. How could I show there aren't more that we just happened to not find or think of.
freckles
  • freckles
so we have to find integers b and c such that (bc-1) divides (b+c)
freckles
  • freckles
I feel like I see that question somewhere before
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I was kinda looking at like the general case to see if we could use that to put something on it, by taking the terms and raising it to the power of number of terms... Like this: \[(a+b)^2 = a^2+b^2+2ab\] since for this one, the condition is \[a+b=a*b\] so we can rearrange this one at least for \(a+b=a*b=x\) \[x(x-2) = a^2+b^2\] I thought this was interesting, it's substantially worse to look at though for the next term with \(a+b+c=a*b*c = y\) since there are 27 terms haha. Here it is anyways: \[(a+b+c)^3 = a^3+b^3+c^3 + 3(a^2b+ab^2+a^2c+ac^2 + b^2c+bc^2)+6abc\] \[y^2(y-6) = a^3+b^3+c^3 + 3(a^2b+ab^2+a^2c+ac^2 + b^2c+bc^2)\] These sorts of things at the very least can tell you some things about divisibility I guess. Hmmm, maybe this isn't that useful I don't know haha.
freckles
  • freckles
nevermind it was a similar looking problem something about if ab+1|a^2+b^2 then \[\frac{a^2+b^2}{ab+1} \text{ is a square }\]
Empty
  • Empty
I guess generally there is some corresponding thing roughly like this for each one: \[x^{n-1}(x-n!) = \sum \ terms\] We might be able to look at what you're describing, it's sorta what got me thinking about this. I really don't know how to go about solving this. I did find a bunch of solutions though, they're of this form: 1+2+3=1*2*3 1+1+2+4 = 1*1*2*4 1+1+1+2+5 = 1*1*1*2*5 \[n+2+\sum_{k=1}^{n-2}1 = n*2*\prod_{k=1}^{n-2}1\]
freckles
  • freckles
assuming \[\frac{b+c}{bc-1} \text{ is an integer } \\ \text{ I think } bc \text{ cannot be an odd \prime } \\ bc=p \cdot 1 \\ \text{ so choise } b=p \text{ and } c=1 \\ \text{ so } b+c=p+1 \\ \text{ so we have } \\ \frac{p+1}{p-1}=\frac{p-1}{p-1}+\frac{2}{p-1}=1+\frac{2}{p-1} \\ \text{ so } p \text{ can be 2 } \\ \text{ but that is the only prime that works }\] but now there is all those other integers to look at :p
anonymous
  • anonymous
i having seen this :O

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