Question

What are all the integer solutions to:

a+b+c=a*b*c

Answer 1

i got one answer !

Answer 2

1,2,3
or -1,-2,-3
these are nonzero integers.

Answer 3

oh i got a different one

Answer 4

Yeah I have a couple other answers too.

10, -10, 0 works.

Are there others?

Answer 5

Maybe I should extend for all integers, what are all solutions to:

\[\Huge \sum_{k=1}^n a_k = \prod_{k=1}^n a_k\]

Answer 6

if one is zero ,you can select any +,- pair

Answer 7

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.

Answer 8

so we have to find integers b and c such that (bc-1) divides (b+c)

Answer 9

I feel like I see that question somewhere before

Answer 10

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.

Answer 11

nevermind
it was a similar looking problem
something about if ab+1|a^2+b^2
then
\[\frac{a^2+b^2}{ab+1} \text{ is a square }\]

Answer 12

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
\[n+2+\sum_{k=1}^{n-2}1 = n*2*\prod_{k=1}^{n-2}1\]

Answer 13

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

Answer 14

i having seen this :O