Question

Let `a`, `b`, and `c` be real numbers with `a <= b <= c`. What are all the ordered triples `(a,b,c)` such that the product of any two of the numbers is `3` more than the sum of that pair?

Answer 1

`(3,3,3)` or `(-1,-1,-1)`

Explanation:

We are given:

`{ (a <= b <= c), (ab=a+b+3), (bc=b+c+3), (ac=a+c+3) :}`

So given some number `a`, we find:

`(a-1)b = a+3`

So:

`b = (a+3)/(a-1) = (a-1+4)/(a-1) = 1+4/(a-1)`

Similarly:

`c = 1+4/(b-1)`

Similarly:

`a = 1+4/(c-1)`

`color(white)(a) = 1+4/((1+4/(b-1))-1)`

`color(white)(a) = 1+(b-1)`

`color(white)(a) = b`

Similarly `b=c`

So `a=b=c`

Putting `b=a` into `ab=a+b+3` we get:

`a^2 = 2a+3`

Subtract `2a+3` from both sides to get:

`0 = a^2-2a-3 = (a-3)(a+1)`

So `a=b=c=3` or `a=b=c=-1`

Answer 2

`(a,b,c)=(-1,-1,-1)color(white)("xxx")"or"color(white)("xxx")(a,b,c)=(3,3,3)`

Explanation:

Given
[1]`color(white)("XXX")ab=a+b+3`
[2]`color(white)("XXX")bc=b+c+3`
[3]`color(white)("XXX")ac=a+c+3`

Subtracting [1] from [3]
[4]`color(white)("XXX")a(c-b)=c-b`

`rArr` [5a]`color(white)("X")a=1color(white)("xxx")"or"color(white)("xxx")`[5b]`color(white)("X")(c-b)=0color(white)("x")rarrcolor(white)("xx")c=b`

If
`color(white)("XXX")`[5a]`color(white)("XXX")a=1`
`color(white)("XXX")`substituting `1` for `a` in [1]
`color(white)("XXX")`[6a]`color(white)("XXX")b=b+4`
`color(white)("XXX")`[7a]`color(white)("XXX")0=4`
Since this is impossible `a!=1`
and we are left with

`color(white)("XXX")`[5b]`color(white)("XXX")c=b`
`color(white)("XXX")`substituting `b` for `c` in [2]
`color(white)("XXX")`[6b]`color(white)("XXX")b^2=2b+3`
`color(white)("XXX")`[7b]`color(white)("XXX")b^2-2b-3=0`
`color(white)("XXX")`[8b]`color(white)("XXX")(b-3)(b+1)=0`
`color(white)("XXX")rarr`
`color(white)("XXX")`[9b.1]`color(white)("X")b=3color(white)("X")` or `color(white)("X")`[9b.2]`color(white)("X")b=-1`

`color(white)("XXX")`If [9b.1]`color(white)("X")b=3`
`color(white)("XXX")`Then by [5b]
`color(white)("XXX")`[10b.1]`color(white)("X")c=3`
`color(white)("XXX")`and substituting `3` for `b` in [1]
`color(white)("XXX")`[11b.1]`color(white)("X")3a=a+6`
`color(white)("XXX")`[12b.1]`color(white)("X")a=3`

Similarly
`color(white)("XXX")`If [9b.2]`color(white)("X")b=-1`
`color(white)("XXX")`repeating similar steps:
`color(white)("XXX")`[10b.2]`color(white)("X")c=-1`
`color(white)("XXX")`and substituting `(-1)` for `b` in [1]
`color(white)("XXX")`[11b.2]`color(white)("X")-a=a+2`
`color(white)("XXX")`[12b.2]`color(white)("X")a=-1`