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  • one year ago

What are all the integer solutions to: a+b+c=a*b*c

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  1. anonymous
    • one year ago
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    i got one answer !

  2. anonymous
    • one year ago
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    1,2,3 or -1,-2,-3 these are nonzero integers.

  3. anonymous
    • one year ago
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    oh i got a different one

  4. Empty
    • one year ago
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    Yeah I have a couple other answers too. 10, -10, 0 works. Are there others?

  5. Empty
    • one year ago
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    Maybe I should extend for all integers, what are all solutions to: \[\Huge \sum_{k=1}^n a_k = \prod_{k=1}^n a_k\]

  6. anonymous
    • one year ago
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    if one is zero ,you can select any +,- pair

  7. Empty
    • one year ago
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    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.

  8. freckles
    • one year ago
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    so we have to find integers b and c such that (bc-1) divides (b+c)

  9. freckles
    • one year ago
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    I feel like I see that question somewhere before

  10. Empty
    • one year ago
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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.

  11. freckles
    • one year ago
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    nevermind it was a similar looking problem something about if ab+1|a^2+b^2 then \[\frac{a^2+b^2}{ab+1} \text{ is a square }\]

  12. Empty
    • one year ago
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    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 \[n+2+\sum_{k=1}^{n-2}1 = n*2*\prod_{k=1}^{n-2}1\]

  13. freckles
    • one year ago
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    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

  14. anonymous
    • one year ago
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    i having seen this :O

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