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Empty
 one year ago
What are all the integer solutions to:
a+b+c=a*b*c
Empty
 one year ago
What are all the integer solutions to: a+b+c=a*b*c

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.01,2,3 or 1,2,3 these are nonzero integers.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i got a different one

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah I have a couple other answers too. 10, 10, 0 works. Are there others?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Maybe 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
 one year ago
Best ResponseYou've already chosen the best response.0if one is zero ,you can select any +, pair

Empty
 one year ago
Best ResponseYou've already chosen the best response.3True, 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
 one year ago
Best ResponseYou've already chosen the best response.1so we have to find integers b and c such that (bc1) divides (b+c)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I feel like I see that question somewhere before

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I 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 \(a+b=a*b=x\) \[x(x2) = a^2+b^2\] 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: \[(a+b+c)^3 = a^3+b^3+c^3 + 3(a^2b+ab^2+a^2c+ac^2 + b^2c+bc^2)+6abc\] \[y^2(y6) = a^3+b^3+c^3 + 3(a^2b+ab^2+a^2c+ac^2 + b^2c+bc^2)\] 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
 one year ago
Best ResponseYou've already chosen the best response.1nevermind it was a similar looking problem something about if ab+1a^2+b^2 then \[\frac{a^2+b^2}{ab+1} \text{ is a square }\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I guess generally there is some corresponding thing roughly like this for each one: \[x^{n1}(xn!) = \sum \ terms\] 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 \[n+2+\sum_{k=1}^{n2}1 = n*2*\prod_{k=1}^{n2}1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1assuming \[\frac{b+c}{bc1} \text{ is an integer } \\ \text{ I think } bc \text{ cannot be an odd \prime } \\ bc=p \cdot 1 \\ \text{ so choise } b=p \text{ and } c=1 \\ \text{ so } b+c=p+1 \\ \text{ so we have } \\ \frac{p+1}{p1}=\frac{p1}{p1}+\frac{2}{p1}=1+\frac{2}{p1} \\ \text{ so } p \text{ can be 2 } \\ \text{ but that is the only prime that works }\] but now there is all those other integers to look at :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i having seen this :O
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